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See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/332863667 PROBLEM SET & SOLUTIONS: DIFFERENTIAL EQUATION Chapter Β· May 2016 CITATIONS READS 0 246,010 1 author: Ibnu Rafi Universitas Negeri Yogyakarta 16 PUBLICATIONS 46 CITATIONS SEE PROFILE Some of the authors of this publication are also working on these related projects: Post-examination analysis View project Developing open approach-based lesson plan and worksheet View project All content following this page was uploaded by Ibnu Rafi on 30 March 2021. The user has requested enhancement of the downloaded file.
PROBLEM SET & SOLUTIONS DIFFERENTIAL EQUATION By: Ibnu Rafi e-mail: ibnu257fmipa@student.uny.ac.id
Table of Contents Table of Contents ...................................................................................................................................................2 Solution of Exercise 1 (Linear and Nonlinear D.E)................................................................................3 Solution of Exercise 2 (Ordinary and Partial D.E) .................................................................................4 Solution of Exercise 3 (Solution of D.E) ......................................................................................................6 Solution of Exercise 4 (Initial Value Problem) ..................................................................................... 11 Solution of Exercise 5 (Separable D.E) ..................................................................................................... 13 Solution of Exercise 6 (General Solution of Separable D.E) ........................................................... 15 Solution of Quiz.................................................................................................................................................... 21 Solution of Exercise 7 (Homogeneous D.E) ........................................................................................... 22 Solution of Exercise 8 (Solution of Homogeneous D.E) ................................................................... 23 Solution of Exercise 9 (Non Homogeneous D.E) ................................................................................. 28 Solution of Exercise 10 (Solution of Non Homogeneous D.E) ...................................................... 29 Solution of Exercise 11 (Exact and Non Exact D.E) ............................................................................ 35 Solution of Exercise 12 (Integrating Factor) ......................................................................................... 42 Solution of Exercise 13 (Grouping Method) .......................................................................................... 44 Solution of Quiz.................................................................................................................................................... 45 Solution of Exercise 14 (Linear D.E) ......................................................................................................... 48 Solution of Exercise 15 (Solution of Linear D.E) ................................................................................. 50 Solution of Exercise 16 (Properties of Linear D.E) ............................................................................ 51 Solution of Exercise 17 (Integrating Factor of Linear D.E) ............................................................ 53 Solution of Exercise 18 (Orthogonal and Oblique Trajectories).................................................. 57 Solution of Exercise 19 (Problem in Mechanics (Frictional Forces)) ....................................... 64 Solution of Exercise 20 (Rate Problems (Rate of Growth and Decay and Population) Growth) ................................................................................................................................................................... 66 Solution of Exercise 21 (Mixture Problem) ........................................................................................... 68 Solution of Exercise 22 (Reduction of Order) ....................................................................................... 70 Page 2 of 72
Solution of Exercise 1 (Linear and Nonlinear D.E) We will determine whether the equations under consideration is linear or nonlinear. To determine whether the equations under consideration is linear or nonlinear we should know that differential equation are said to be nonlinear if any product exist between the dependent variable and its derivatives, between the derivatives themselves, or the dependent variable is trancedental function. 1. π2 π¦ ππ₯ ππ¦ + 5 ππ₯ + 6π¦ = 0 (Linear differential equation) Since we see that the dependent variable of the differential equation above is π¦ and its various derivatives occur to the first degree only. 2. π4 π¦ ππ₯4 π3π¦ ππ¦ + π₯ 2 π π₯ 3 + π₯ 3 ππ₯ = π₯π π₯ ( Linear differential equation) Since we see that the dependent variable of the differential equation above is π¦ and its various derivatives occur to the first degree only. 3. π2 π¦ ππ₯2 ππ¦ + 5 ππ₯ + 6π¦ 2 = 0(Nonlinear differential equation) Since we see that the dependent variable of the differential equation above is π¦ and its various derivatives occur to the first degree only, but then, the differential equation above contains the product between the dependent variable themselves, that is in the term 6π¦ 2 . Therefore, the differential equation π2 π¦ ππ₯2 4. ππ¦ + 5 ππ₯ + 6π¦ 2 = 0 is nonlinear differential equation. π2 π¦ ππ¦ 3 ππ₯ ππ₯ 2 +5 + 6π¦ = 0 (Nonlinear differential equation) Since we see that the dependent variable of the differential equation above is π¦, but in the term 5 ππ¦ 3 ππ₯ involves the third degree of the first derivative or in the othere word there is product between three derivatives. Therefore, the π2 π¦ differential equation π π₯ 2 + 5 5. π2 π¦ ππ₯2 ππ¦ 3 ππ₯ + 6π¦ = 0 is nonlinear differential equation. ππ¦ + 5π¦ ππ₯ + 6π¦ = 0 (Nonlinear differential equation) Since we see that the dependent variable of the differential equation above is π¦, but there is product between the dependent variable and its derivative in the ππ¦ term 5π¦ ππ₯ . Therefore, the differential equation nonlinear differential equation. Page 3 of 72 π2 π¦ ππ₯2 +5 ππ¦ 3 ππ₯ + 6π¦ = 0 is
Solution of Exercise 2 (Ordinary and Partial D.E) We will classify of the following differential equations as ordinary or partial differential equations, state the order of each equation, and determine whether the equation under consideration is linear or nonlinear. Ordinary differential equation is the differential equation involving ordinary derivatives of one or more dependent variables with respect to a single independent variable. Partial differential equation is the differential equation involving ordinary derivatives of one or more dependent variables with respect to more than one independent variable. The order of a differential equation is equal to the order of the highest differential coefficient that it contains. The degree of a differential equation is the highest power of the highest order differential coefficient that the equation contains after it has been rationalized. 1. ππ¦ ππ₯ + π₯ 2 π¦ = π₯π π₯ . The differential equation ππ¦ ππ₯ + π₯ 2 π¦ = π₯π π₯ is ordinary differential equation (since it has only one independent variable, that is π₯), first order ordinary differential equation, ππ¦ ππ₯ , first degree ordinary differential equation, and linear differential equation (since no product between dependent variable (π¦) themselves, no product between π¦ and/or any of its derivatives, and its various derivatives occur to the first degree only). 2. π3 π¦ π2 π¦ ππ¦ + 4 π π₯ 2 β 5 ππ₯ + 3π¦ = sin π₯. ππ₯3 The differential equation π3π¦ ππ₯3 π2 π¦ ππ¦ + 4 π π₯ 2 β 5 ππ₯ + 3π¦ = sin π₯ is ordinary differential equation (since it has only one independent variable, that is π₯), third order ordinary differential equation, π3π¦ ππ₯3 , first degree ordinary differential equation, and linear differential equation (since no product between dependent variable (π¦) themselves, no product between π¦ and/or any of its derivatives, its various derivatives occur to the first degree only, and no trancendental function of π¦ and/ or its derivatives occur). 3. π2π’ ππ₯2 π2π’ + π π¦ 2 = 0. The differential equation π 2π’ ππ₯2 π 2π’ + ππ¦ 2 = 0 is partial differential equation (since it has more that one independent variable involved, that is π₯ and π¦), second order Page 4 of 72
partial differential equation, first degree partial differential equation, linear differential equation (since no product between dependent variable (π’) themselves, no product between π¦ and/or any of its derivatives, and no trancendental function of π’ and/ or its derivatives occur). 4. π₯ 2 ππ¦ + π¦ 2 ππ₯ = 0. π¦2 ππ¦ The differential equation is π₯ 2 ππ¦ + π¦ 2 ππ₯ = 0 β ππ₯ = β π₯ 2 ordinary differential equation (since it has only one independent variable involved, that is either π₯ or π¦), first order ordinary differential equation, first degree ordinary differential equation, nonlinear differential equation (since there is a product either between dependent variable π¦ themselves ( if π¦ is dependent variable) or between dependent variable π₯ themselves ( if π₯ is dependent variable) ). 5. π4 π¦ π2 π¦ ππ₯ ππ₯2 4 +3 5 + 5π¦ = 0. The differential equation π4π¦ +3 ππ₯4 π2π¦ ππ₯2 5 + 5π¦ = 0 is ordinary differential equation (since it has only one independent variable involved, that is π₯), fourth order ordinary differential equation, first degree ordinary differential equation, nonlinear differential equation (since in term 3 π2π¦ ππ₯2 5 involves the fifth power of the second derivatives ). 6. π4π’ ππ₯2 ππ¦2 π2π’ π 2π’ + π π₯ 2 + π π¦ 2 + π’ = 0. The differential equation π 4π’ ππ₯2 ππ¦ 2 π 2π’ ππ’ + π π₯ 2 + π π¦ 2 + π’ = 0 is partial differential equation (since it has more than one independent variable involved, that is π₯ and π4 π¦), fourth order partial differential equation (π π₯ 2 π π¦ 2 is the fourth derivatives of π’ π₯, π¦ ), first degree partial differential equation, linear differential equation (since since no product between dependent variable (π’) themselves, no product between π’ and/or any of its derivatives, and no trancendental function of π¦ and/ or its derivatives occur). 7. π2 π¦ ππ₯2 + π¦ sin π₯ = 0. The differential equation π2π¦ ππ₯2 + π¦ sin π₯ = 0 is ordinary differential equation (since it has only one independent variable, that is π₯), second order ordinary differential equation, π2π¦ ππ₯2 , first degree ordinary differential equation, and linear differential equation (since no product between dependent variable (π¦) themselves, no product between π¦ and/or any of its derivatives, and no trancendental function of π¦ and/ or its derivatives occur). Page 5 of 72
π2 π¦ 8. ππ₯2 + π₯ sin π¦ = 0. The differential equation π2π¦ ππ₯2 + π₯ sin π¦ = 0 is ordinary differential equation (since it has only one independent variable, that is π₯), second order ordinary differential equation, π2π¦ ππ₯2 , first degree ordinary differential equation, and nonlinear differential equation (since its dependent variable is π¦, but then, there is trancendental function of π¦ occur in term π₯ sin π¦). π6 π₯ 9. ππ‘6 + π4π₯ π3π₯ ππ‘ 4 ππ‘ 3 + π₯ = π‘. The differential equation π6π₯ ππ‘6 + π4π₯ π3π₯ ππ‘ 4 ππ‘ 3 + π₯ = π‘ is ordinary differential equation (since it has only one independent variable, that is π‘), sixth order ordinary differential equation, π6π₯ ππ‘6 , first degree ordinary differential equation, and nonlinear differential equation (since there is a product between the various derivatives of π₯ with respect to π‘ in the term 10. ππ 3 ππ = π2π π π 2 π4π₯ π3 π₯ ππ‘ 4 ππ‘3 ). +1 The differential equation ππ 3 ππ = π2π π π 2 + 1 is ordinary differential equation (since it has only one independent variable, that is π ), second order ordinary differential equation, , first degree ordinary differential equation, and nonlinear differential equation (since in term ππ 3 ππ involves the third power of the first derivatives). Solution of Exercise 3 (Solution of D.E) 1. a) We will show that π π₯ = π₯ + 3π βπ₯ is a solution of the differential equation that be ππ¦ defined as ππ₯ + π¦ = π₯ + 1 on every interval π < π₯ < π of the π₯ axis. To show this, we must show that π π₯ = π₯ + 3π βπ₯ satisfies the differential equation ππ¦ ππ₯ +π¦=π₯+1 ( ordinary linear nonhomogen differential equation). By differentiating π π₯ , we obtain π β² π₯ = 1 β 3π βπ₯ , βπ₯ β π, π . Afterwards, subtituting π(π₯) for π¦, π β² π₯ for ππ¦ ππ₯ in the mentioned differential equation. We obtain ππ¦ ππ₯ + π¦ = 1 β 3π βπ₯ + π₯ + 3π βπ₯ = 1 β 3π βπ₯ + π₯ + 3π βπ₯ = π₯ + 1, that is, π₯ + 1 = π₯ + 1. Therefore, the given differential equation is satisfied by π π₯ = π₯ + 3π βπ₯ . In the other word, π π₯ = π₯ + 3π βπ₯ is a solution of the differential ππ¦ equation that be defined as ππ₯ + π¦ = π₯ + 1 on every interval π < π₯ < π of the π₯ axisβ Page 6 of 72
b) We will show that π π₯ = 2π 3π₯ β 5π 4π₯ is a solution of the differential equation that be defined as π2π¦ ππ₯2 ππ¦ β 7 ππ₯ + 12π¦ = 0 on every interval π < π₯ < π of the π₯ axis. To show this, we must show that π π₯ = 2π 3π₯ β 5π 4π₯ , βπ₯ β π, π differential equation π2 π¦ ππ₯2 satisfies the ππ¦ β 7 ππ₯ + 12π¦ = 0. By differentiating π(π₯), we obtain π β² π₯ = 6π 3π₯ β 20π 4π₯ and π β²β² π₯ = 18π 3π₯ β 80π 4π₯ . Afterwards, subtituting π(π₯) for π¦, π β² (π₯) for ππ¦ , and π β²β² (π₯) for ππ₯ π2π¦ ππ₯2 in the mentioned differential equation. We obtain π2 π¦ ππ¦ β 7 + 12π¦ = 0 ππ₯ 2 ππ₯ β 18π 3π₯ β 80π 4π₯ β 7 6π 3π₯ β 20π 4π₯ + 12 2π 3π₯ β 5π 4π₯ = 0 β 18π 3π₯ β 80π 4π₯ β 42π 3π₯ + 140π 4 + 24π 3π₯ β 60π 4π₯ = 0 β 18 β 42 + 24 π 3π₯ + β80 + 140 β 60 π 4π₯ = 0 β 0. π 3π₯ + 0. π 4π₯ = 0 β 0 π 3π₯ + π 4π₯ = 0 β 0 = 0. Therefore, the given differential equation is satisfied by π π₯ = 2π 3π₯ β 5π 4π₯ . In the other word, π₯ = 2π 3π₯ β 5π 4π₯ is a solution of the differential equation that be π2 π¦ ππ¦ defined as π π₯ 2 β 7 ππ₯ + 12π¦ = 0 on every interval π < π₯ < π of the π₯ axisβ c) We will show that π π₯ = π π₯ + 2π₯ 2 + 6π₯ + 7 is a solution of the differential equation that be defined as π2π¦ ππ₯2 ππ¦ β 3 ππ₯ + 2π¦ = 4π₯ 2 on every interval π < π₯ < π of the π₯ axis. To show this, we must show that π π₯ = π π₯ + 2π₯ 2 + 6π₯ + 7, βπ₯ β π, π satisfies the differential equation π2 π¦ ππ₯2 ππ¦ β 3 ππ₯ + 2π¦ = 4π₯ 2 . By differentiating π(π₯), we obtain π β² π₯ = π π₯ + 4π₯ + 6 and π β²β² π₯ = π π₯ + 4. Afterwards, subtituting π(π₯) for π¦, ππ¦ π2π¦ π β² (π₯) for ππ₯ , and π β²β² (π₯) for π π₯ 2 in the mentioned differential equation. We obtain π2 π¦ ππ¦ β3 + 2π¦ = 4π₯ 2 2 ππ₯ ππ₯ β π π₯ + 4 β 3 π π₯ + 4π₯ + 6 + 2 π π₯ + 2π₯ 2 + 6π₯ + 7 = 4π₯ 2 β π π₯ + 4 β 3π π₯ β 12π₯ β 18 + 2π π₯ + 4π₯ 2 + 12π₯ + 14 = 4π₯ 2 β 1 β 3 + 2 π π₯ + 4π₯ 2 + β12 + 12 π₯ + 4 β 18 + 14 = 4π₯ 2 β 0. π π₯ + 4π₯ 2 + 0. π₯ + 0 = 4π₯ 2 β 0 + 4π₯ 2 + 0 + 0 = 4π₯ 2 Page 7 of 72
β 4π₯ 2 = 4π₯ 2 . Therefore, the given differential equation is satisfied by π π₯ = π π₯ + 2π₯ 2 + 6π₯ + 7. In the other word, π π₯ = π π₯ + 2π₯ 2 + 6π₯ + 7 is a solution of the differential equation π2π¦ ππ¦ that be defined as π π₯ 2 β 3 ππ₯ + 2π¦ = 4π₯ 2 on every interval π < π₯ < π of the π₯ axisβ 1 d) We will show that π π₯ = 1+π₯ 2 is a solution of the differential equation that be π2 π¦ defined as 1 + π₯ 2 ππ₯2 ππ¦ + 4π₯ ππ₯ + 2π¦ = 0 on every interval π < π₯ < π of the π₯ axis. To 1 show this, we must show that π π₯ = 1+π₯ 2 satisfies the differential equation (1 + π₯ 2 )π¦ β²β² + 4π₯π¦β² + 2π¦ = 0. By differentiating π(π₯), we obtain π β² π₯ = β π β²β² π₯ = β 2 1+π₯ 2 2 β2π₯ 2 1+π₯ 2 2π₯ 1+π₯ 2 4 1+π₯ 2 = 8π₯ 2 β2 1+π₯ 2 1+π₯ 2 4 = 6π₯ 2 β2 . 1+π₯ 2 3 2π₯ 1+π₯ 2 2 and Afterwards subtitut- π2 π¦ ππ¦ ing π(π₯) for π¦, π β² (π₯) for ππ₯ , and π β²β² (π₯) for π π₯ 2 in the mentioned differential equation. We obtain 1 + π₯ 2 π¦ β²β² + 4π₯π¦ β² + 2π¦ = 0 6π₯ 2 β 2 2π₯ + 4π₯ β 2 3 1+π₯ 1 + π₯2 β 1 + π₯2 6π₯ 2 β 2 8π₯ 2 β β 1 + π₯2 2 1 + π₯2 1 =0 1 + π₯2 +2 2 + 2π₯ 2 + =0 2 1 + π₯2 2 β 6 β 8 + 2 π₯ 2 + β2 + 2 =0 1 + π₯2 2 β 0. π₯ 2 + 0 =0 1 + π₯2 2 β 0 1 + π₯2 2 2 =0 β 0 = 0. 1 Therefore, the given differential equation is satisfied by π π₯ = 1+π₯ 2 . In the other 1 word, π π₯ = 1+π₯ 2 is a solution of the differential equation that be defined as 1 + π₯2 π2 π¦ ππ₯2 + 4π₯ ππ¦ ππ₯ + 2π¦ = 0 on every interval π < π₯ < π of the π₯ axisβ 2. a) We will show that π₯ 3 + 3π₯π¦ 2 = 1 β π¦ = of the differential equation 2π₯π¦ ππ¦ ππ₯ 1βπ₯ 3 3π₯ β π¦2 = 1βπ₯ 3 3π₯ is an implicit solution + π₯ 2 + π¦ 2 = 0 on the interval 0 < π₯ < 1. To show this, firstly, we differentiating π₯ 3 + 3π₯π¦ 2 = 1 implicitly with respect to π₯. We obtain Page 8 of 72
π π₯ 3 + 3π₯π¦ 2 π 1 = ππ₯ ππ₯ π π₯3 π 3π₯π¦ 2 π 1 β + = ππ₯ ππ₯ ππ₯ β 3π₯ 2 + 3 π π₯π¦ 2 =0 ππ₯ β 3π₯ 2 + 3 1. π¦ 2 + π₯ 2π¦ β 3π₯ 2 + 3π¦ 2 + 6π₯π¦ β =0 ππ¦ =0 ππ₯ ππ¦ 3(π₯ 2 + π¦ 2 ) =β βπ₯ β (0,1) β¦ β . ππ₯ 6π₯π¦ By subtituting (*) to 2π₯π¦ 2π₯π¦ ππ¦ ππ₯ ππ¦ ππ₯ + π₯ 2 + π¦ 2 = 0 we obtain ππ¦ + π₯2 + π¦2 = 0 ππ₯ 3(π₯ 2 + π¦ 2 ) β 2π₯π¦ β + π₯2 + π¦2 = 0 6π₯π¦ β β π₯2 + π¦2 + π₯2 + π¦2 = 0 β β π₯2 + 1 β π₯3 1 β π₯3 + π₯2 + = 0, βπ₯ β (0,1) 3π₯ 3π₯ β 0 = 0. Thus, we can conclude that π₯ 3 + 3π₯π¦ 2 = 1 is an implicit solution of the differential equation 2π₯π¦ ππ¦ + π₯ 2 + π¦ 2 = 0 on the interval 0 < π₯ < 1 β ππ₯ b) We will show that 5π₯ 2 π¦ 2 β 2π₯ 3 π¦ 2 = 1 is an implicit solution of the differential equation π₯ ππ¦ ππ₯ 5 + π¦ = π₯ 3 π¦ 3 on the interval 0 < π₯ < 2. To show this, firstly, we differentiating 5π₯ 2 π¦ 2 β 2π₯ 3 π¦ 2 = 1 implicitly with respect to π₯. We obtain π 5π₯ 2 π¦ 2 β 2π₯ 3 π¦ 2 π 1 = ππ₯ ππ₯ β 10π₯ 2 π¦ ππ¦ ππ¦ + 10π₯π¦ 2 β 4π₯ 3 π¦ β 6π₯ 2 π¦ 2 = 0 ππ₯ ππ₯ β 10π₯ 2 π¦ β 4π₯ 3 π¦ ππ¦ = 6π₯ 2 π¦ 2 β 10π₯π¦ 2 ππ₯ Page 9 of 72
ππ¦ 6π₯ 2 π¦ 2 β 10π₯π¦ 2 5 β = , βπ₯ β 0, β¦ β ππ₯ 10π₯ 2 π¦ β 4π₯ 3 π¦ 2 and 5π₯ 2 π¦ 2 β 2π₯ 3 π¦ 2 = 1 β 5π₯ 2 β 2π₯ 3 π¦ 2 = 1 βπ¦= 1 = 5π₯ 2 β 2π₯ 3 1 π₯ 5 β 2π₯ By subtituting (*) and (**) to π₯ π₯ , βπ₯ β 0, ππ¦ ππ₯ 5 β¦ ββ 2 + π¦ β π₯ 3 π¦ 3 = 0 we obtain 6π₯ 2 π¦ 2 β 10π₯π¦ 2 1 + = π₯ 3π¦3 2 3 2 3 10π₯ π¦ β 4π₯ π¦ 5π₯ β 2π₯ βπ₯ π₯π¦ 2 3π₯ β 5 1 + = π₯ 3π¦3 2 π₯ π¦ 5 β 2π₯ π₯ 5 β 2π₯ βπ¦ 3π₯ β 5 1 + = π₯ 3π¦3 5 β 2π₯ π₯ 5 β 2π₯ β β β β 3π₯ β 5 π₯ 5 β 2π₯ 5 β 2π₯ 3π₯ β 5 + 5 β 2π₯ π₯ 5 β 2π₯ 5 β 2π₯ π₯ π₯ 5 β 2π₯ 5 β 2π₯ 1 5 β 2π₯ β π₯3 βπ₯ 5 β 2π₯ + 1 π₯ 5 β 2π₯ = π₯3π¦3 = π₯3π¦3 = π₯ 3π¦3 1 π₯ 3 5 β 2π₯ 1 3 π₯ 5 β 2π₯ = π₯3π¦3 5 β 2π₯ = π₯ 3π¦3 3 = π₯ 3 π¦ 3 , βπ₯ β 0, 5 2 5 β π₯ 3 π¦ 3 = π₯ 3 π¦ 3 , βπ₯ β 0, 2 . Thus, we can conclude that 5π₯ 2 π¦ 2 β 2π₯ 3 π¦ 2 = 1 is an implicit solution of the differential equation π₯ 5 ππ¦ ππ₯ + π¦ = π₯3π¦3 β π₯ 0<π₯<2β Page 10 of 72 ππ¦ ππ₯ + π¦ β π₯ 3 π¦ 3 = 0 on the interval
Solution of Exercise 4 (Initial Value Problem) 1. Show that π¦ = 4π 2π₯ + 2π β3π₯ is a solution of initial- value problem π 2 π¦ ππ¦ + β 6π¦ = 0, ππ₯ 2 ππ₯ π¦ 0 = 6, π¦ β² 0 = 2. Is π¦ = 2π 2π₯ + 4π β3π₯ also a solution of this problem? Explain why and why not. Proof Assume that π¦ = 4π 2π₯ + 2π β3π₯ is a solution of given initial- value problem. Since π¦ = 4π 2π₯ + 2π β3π₯ , we obtain π¦ β² = 8π 2π₯ β 6π β3π₯ , π¦ β²β² = 16π 2π₯ + 18π β3π₯ , and π¦ β²β² + π¦ β 6 = 16π 2π₯ + 18π β3π₯ + 8π 2π₯ β 6π β3π₯ β 24π 2π₯ β 12π β3π₯ = 0 (satisfied). Moreover, by subtituting π₯ = 0 to π¦ = 4π 2π₯ + 2π β3π₯ and π¦β² = 8π 2π₯ β 6π 3π₯ we obtain π¦(0) = 4π 0 + 2π 0 = π 0 4 + 2 = 1.6 = 6 (the initial condition is satisfied) and π¦ β² 0 = 8π 0 β 6π 0 = π 0 8 β 6 = 1.2 = 2 (the second condition is satisfied). From this result we can say that our assumption is accepted. Therefore, we can conclude that π¦ = 4π 2π₯ + 2π β3π₯ is the solution of given initial- value problem. β Afterward, we will observe whether π¦ = 2π 2π₯ + 4π β3π₯ is also a solution of the given problem or not. We know that π¦ β² = 4π 2π₯ β 12π β3π₯ , π¦ β²β² = 8π 2π₯ + 36π β3π₯ and π¦ β²β² + π¦β² β 6π¦ = 8π 2π₯ + 36π β3π₯ + 4π 2π₯ β 12π β3π₯ + 12π 2π₯ + 24π β3π₯ = 0 (satisfied). Moreover, by subtituting π₯ = 0 to π¦ = 2π 2π₯ + 4π β3π₯ and π¦ β² = 4π 2π₯ β 12π β3π₯ , we obtain π¦ 0 = 2π 0 + 4π 0 = π 0 2 + 4 = 1.6 = 6 (satisfied) and π¦ β² 0 = 4π 0 β 12π 0 = π 0 4 β 12 = 1 β8 = 8 β 2 (unsatisfied). From this result, we can conclude that π¦ = 2π 2π₯ + 4π β3π₯ is not the solution of given problem since one of the initial conditions in initial- value problem, that is π¦ β² 0 = 2 not be satisfied when we subtituting π₯ = 0 to π¦ β² = 4π 2π₯ β 12π β3π₯ . 2. Every solution of the differential equation π2 π¦ ππ₯2 + π¦ = 0 may be written in the form π¦ = π1 sin π₯ + π2 cos π₯, for some choice of the arbitrary constants π1 and π2 . Using this information, show that boundary problem (a) and (b) possess solution but that (c) does not. π π2 π¦ π + π¦ = 0, π¦ 0 = 0, π¦ = 1. ππ₯ 2 2 Page 11 of 72
π2 π¦ π β² π + π¦ = 0, π¦ 0 = 1, π¦ = β1. ππ₯ 2 2 π2 π¦ π + π¦ = 0, π¦ 0 = 0, π¦ π = 1. ππ₯ 2 Solution π2 π¦ (a) Since every solution of differential equation π π₯ 2 + π¦ = 0 may be written in π form π¦ = π1 sin π₯ + π2 cos π₯, by subtituting π₯ = 0 and π₯ = 2 to π¦ = π1 sin π₯ + π2 cos π₯, we obtain π¦ 0 = π1 sin 0 + π2 cos 0 = π2 = 0 and π¦ π π π = π1 sin + π2 cos = π1 . 1 + π2 . 0 = π1 = 1. 2 2 2 Therefore, the given boundary problem possess solution and it particular solution is π¦ = sin π₯. π2 π¦ (b) Since every solution of differential equation π π₯ 2 + π¦ = 0 may be written in form π¦ = π1 sin π₯ + π2 cos π₯, by subtituting π₯ = 0 to π¦ = π1 sin π₯ + π2 cos π₯ we obtain that π¦ 0 = π1 sin 0 + π2 cos 0 = π1 . 0 + π2 . 1 = π2 = 1. Since π¦ = π1 sin π₯ + π2 cos π₯, we obtain that π¦ β² = π1 cos π₯ β π2 sin π₯. By π subtituting π₯ = to π¦ β² = π1 cos π₯ β π2 sin π₯, we obtain 2 π¦β² π π π = π1 cos β π2 sin = π1 . 0 β π2 . 1 = βπ2 = β1 β π2 = 1. 2 2 2 Since, the value of π2 consistent, that is π2 = 1, we can conclude that the given boundary problem possess solution. Moreover, the general solution for such boundary problem is π¦ = π1 sin(π₯) + cos(π₯). π2 π¦ (c) Since every solution of differential equation π π₯ 2 + π¦ = 0 may be written in form π¦ = π1 sin π₯ + π2 cos π₯, by subtituting π₯ = 0 and π₯ = π to π¦ = π1 sin π₯ + π2 cos π₯, we obtain π¦ 0 = π1 sin(0) + π2 cos(0) = π1 . 0 + π2 . 1 = π2 = 0 and π¦ π = π1 sin π + π2 cos π = π1 . 0 + π2 . β1 = βπ2 = 1 β π2 = 1. Since the value of π2 inconsistent, we can say that the given boundary problem has no solution. Page 12 of 72
*** Solution of Exercise 5 (Separable D.E) Determine whether each of the following differential equations is or is not separable. Note: An equation of the form πΉ π₯ πΊ π¦ ππ₯ + π π₯ π π¦ ππ¦ = 0 β¦ (β) is called an equation with variables searable or simply a separable equation. Equation (*) can be πΉ π₯ restated π 1. ππ¦ ππ₯ π₯ π π¦ ππ₯ + πΊ π¦ ππ¦ = 0. = 3π¦ 2 β π¦ 2 π Solution: Since ππ¦ = 3π¦ 2 β π¦ 2 sin π₯ ππ₯ ππ¦ β = π¦ 2 3 β sin π₯ ππ₯ 1 β 2 ππ¦ = 3 β sin π₯ ππ₯ π¦ 1 β 3 β sin π₯ ππ₯ β 2 ππ¦ = 0, π¦ we can conclude that the differential equation 2. ππ¦ ππ₯ ππ¦ ππ₯ = 3π¦ 2 β π¦ 2 sin π₯ is separable. = 3π₯ β π¦ sin π₯ Solution: Since ππ¦ = 3π₯ β π¦ sin π₯ ππ₯ β ππ¦ = 3π₯ β π¦ sin π₯ ππ₯ β π¦ sin π₯ β 3π₯ ππ₯ β ππ¦ = 0, ππ¦ we can conclude that the differential equation ππ₯ = 3π₯ β π¦ sin(π₯) is not separable. 3. π₯ ππ¦ ππ₯ = π₯βπ¦ 2 Solution: Since ππ¦ π₯ = π₯βπ¦ 2 ππ₯ β π₯ ππ¦ = π₯ β π¦ 2 ππ₯ 1 1 β ππ¦ = ππ₯ 2 π₯βπ¦ π₯ 1 1 β ππ¦ β ππ₯ = 0, 2 π₯βπ¦ π₯ we can say that the differential equation π₯ 4. ππ¦ ππ₯ ππ¦ ππ₯ = 1 + π₯2 Page 13 of 72 = π₯βπ¦ 2 is not separable.
Solution: Since ππ¦ = 1 + π₯2 ππ₯ β ππ¦ β 1 + π₯ 2 ππ₯ = 0, ππ¦ we can conclude that the differential equation ππ₯ = 1 + π₯ 2 is separable. 5. ππ¦ ππ₯ + 4π¦ = 8 Solution: Since ππ¦ + 4π¦ = 8 ππ₯ ππ¦ β = 8 β 4π¦ ππ₯ β ππ¦ = (8 β 4π¦)ππ₯ 1 β ππ¦ β ππ₯ = 0 8 β 4π¦ we can conclude that the differential equation 6. ππ¦ ππ₯ ππ¦ ππ₯ + 4π¦ = 8 is separable. + π₯π¦ = 4π₯ Solution: ππ¦ + π₯π¦ = 4π₯ ππ₯ ππ¦ β = 4π₯ β π₯π¦ ππ₯ ππ¦ β = π₯ 4βπ¦ ππ₯ 1 β ππ¦ = π₯ ππ₯ 4βπ¦ 1 β ππ¦ β π₯ππ₯ = 0, 4βπ¦ we can conclude that the differential equation 7. ππ¦ ππ₯ ππ¦ ππ₯ + π₯π¦ = 4π₯ is separable. + 4π¦ = π₯ 2 Solution: Since ππ¦ + 4π¦ = π₯ 2 ππ₯ ππ¦ β = π₯ 2 β 4π¦ ππ₯ β ππ¦ = π₯ 2 β 4π¦ ππ₯ β ππ¦ β π₯ 2 β 4π¦ ππ₯ = 0, ππ¦ we can say that the differential equation ππ₯ + 4π¦ = π₯ 2 is not separable. 8. ππ¦ ππ₯ = π₯π¦ β 3π₯ β 2π¦ + 6 Page 14 of 72
Solution: Since ππ¦ = π₯π¦ β 3π₯ β 2π¦ + 6 ππ₯ ππ¦ β = π₯β2 π¦β3 ππ₯ 1 β ππ¦ β π₯ β 2 ππ₯ = 0, π¦β3 ππ¦ we can say that the differential equation ππ₯ = π₯π¦ β 3π₯ β 2π¦ + 6 is separable. 9. ππ¦ ππ₯ = sin π₯ + π¦ Solution: Since ππ¦ = sin π₯ + π¦ ππ₯ β ππ¦ β sin π₯ + π¦ ππ₯ = 0, ππ¦ we can say that the differential equation ππ₯ = sin(π₯ + π¦) is not separable. ππ¦ 10. π¦ ππ₯ = π π₯β3π¦ 2 Solution: Since ππ¦ 2 π¦ = π π₯β3π¦ ππ₯ ππ¦ ππ₯ βπ¦ = ππ₯ π 3π¦ 2 2 β π¦π 3π¦ ππ¦ β π π₯ ππ₯ = 0, we can conclude that the differential equation π¦ ππ¦ ππ₯ 2 = π π₯β3π¦ is separable. *** Solution of Exercise 6 (General Solution of Separable D.E) Find the general solution for each of the following. Where possible, write your answer as an explicit solution. 1. ππ¦ ππ₯ = π₯π¦ β 4π₯ Solution: ππ¦ ππ¦ 1 Since ππ₯ = π₯π¦ β 4π₯ β ππ₯ = π₯ π¦ β 4 β π¦β4 ππ¦ β π₯ ππ₯ = 0, assume π¦ β 4, we obtain that 1 ππ¦ β π¦β4 π₯ ππ₯ = πΆ1 1 β ln π¦ β 4 β π₯ 2 = πΆ1 2 1 β ln π¦ β 4 = π₯ 2 + πΆ1 2 Page 15 of 72
1 2 +πΆ β π¦ β 4 = π 2π₯ 1 , πππ‘ πΆ = π πΆ1 1 2 β π¦ β 4 = πΆπ 2π₯ as the general solution for the differential equation ππ¦ ππ₯ = π₯π¦ β 4π₯. Note: 1 2 Since π¦ β 4 = πΆπ 2π₯ , 1 2 π¦β4 = 2. ππ¦ ππ₯ 1 2 πΆπ 2π₯ ππ π¦ β₯ 4 1 2 βπΆπ 2π₯ πΆπ 2π₯ + 4 β π¦ = 1 2 1 2 = πΎπ 2π₯ ππ π¦ < 4 πΎπ 2π₯ + 4 = 3π¦ 2 β π¦ 2 sin π₯ Solution: ππ¦ 1 Since ππ₯ = π¦ 2 3 β sin π₯ β π¦ 2 ππ¦ β 3 β sin π₯ ππ₯ = 0, we obtain that 1 ππ¦ β 3 β sin π₯ ππ₯ = πΆ1 π¦2 1 β β β 3π₯ + cos π₯ = πΆ1 π¦ 1 β β = 3π₯ + cos π₯ + πΆ1 π¦ 1 βπ¦=β , πππ‘ πΆ = βπΆ1 3π₯ + cos π₯ + πΆ1 1 βπ¦= πΆ β 3π₯ β cos π₯ ππ¦ as the general solution for the differential equation ππ₯ = 3π¦ 2 β π¦ 2 sin π₯ . 3. ππ¦ ππ₯ = π₯π¦ β 3π₯ β 2π¦ + 6 Solution: ππ¦ ππ¦ 1 Since ππ₯ = π₯π¦ β 3π₯ β 2π¦ + 6 β ππ₯ = π₯ β 2 π¦ β 3 β π¦β3 ππ¦ β π₯ β 2 ππ₯ = 0, we obtain 1 ππ¦ β π¦β3 (π₯ β 2) ππ₯ = πΆ1 1 β ln π¦ β 3 β π₯ 2 + 2π₯ = πΆ1 2 1 β ln π¦ β 3 = π₯ 2 β 2π₯ + πΆ1 2 1 2 β2π₯+πΆ1 β π¦ β 3 = π 2π₯ , πππ‘ πΆ = π πΆ1 1 2 β2π₯ β π¦ β 3 = πΆπ 2π₯ as the general solution for differential equation Note: 1 2 β2π₯ Since π¦ β 3 = πΆπ 2π₯ , Page 16 of 72 ππ¦ ππ₯ = π₯π¦ β 3π₯ β 2π¦ + 6.
1 2 β2π₯ π¦β3 = 4. ππ¦ ππ₯ πΆπ 2π₯ 1 2 β2π₯ βπΆπ 2π₯ 1 2 πΆπ 2π₯ β2π₯ + 3 β π¦ = 1 2 1 2 = πΎπ 2π₯ β2π₯ ππ π¦ < 3 πΎπ 2π₯ β2π₯ + 3 ππ π¦ β₯ 3 = tan π¦ Solution: ππ¦ 1 Since ππ₯ = tan(π¦) β tan π¦ ππ¦ β ππ₯ = 0, assume tan(π¦) β 0, we obtain 1 ππ¦ β ππ₯ = πΆ1 tan π¦ cos π¦ β ππ¦ β ππ₯ = πΆ1 sin π¦ 1 β π sin(π¦) β ππ₯ = πΆ1 sin(π¦) β ln | sin π¦ | β π₯ = πΆ1 β ln | sin π¦ | = π₯ + πΆ1 β sin π¦ = π π₯+πΆ1 , πππ‘ πΆ = π πΆ1 β sin π¦ = πΆπ π₯ ππ¦ as the general solution for ππ₯ = tan(π¦). Note: Since sin π¦ = πΆπ π₯ , πΆπ π₯ ππ0 β€ sin(π¦) β€ 1 arcsin(πΆπ π₯ ) sin(π¦) = β π¦ = βπΆπ π₯ = πΎπ π₯ ππ β1 β€ sin(π¦) < 0 arcsin(πΎπ π₯ ) 5. ππ¦ ππ₯ π¦ =π₯ Solution: ππ¦ π¦ 1 1 Since ππ₯ = π₯ β π¦ ππ¦ β π₯ ππ₯ = 0, we obtain 1 1 ππ¦ β ππ₯ = πΆ1 π¦ π₯ β ln π¦ β ln π₯ = πΆ1 β ln π¦ = ln π₯ + ln π πΆ1 β ln π¦ = ln(π πΆ1 |π₯|) , πππ‘ πΆ = π πΆ1 β π¦ = πΆ|π₯| ππ¦ π¦ as the general solution for the differential equation ππ₯ = π₯ . 6. ππ¦ ππ₯ 6π₯ 2 +4 = 3π¦ 2 β4π¦ Solution: ππ¦ 6π₯ 2 +4 Since ππ₯ = 3π¦ 2 β4π¦ β 3π¦ 2 β 4π¦ ππ¦ β 6π₯ 2 + 4 ππ₯ = 0, assume 3π¦ 2 β 4π¦ β 0 we obtain 3π¦ 2 β 4π¦ ππ¦ β 6π₯ 2 + 4 ππ₯ = πΆ β π¦ 3 β 2π¦ 2 β 2π₯ 3 β 4π₯ = πΆ Page 17 of 72
2π₯ 3 + 4π₯ + πΆ βπ¦= π¦ 2 β 2π¦ ππ¦ 6π₯ 2 +4 as the general solution for the diffeential equation ππ₯ = 3π¦ 2 β4π¦ . 7. π₯2 + 1 ππ¦ ππ₯ = π¦2 + 1 Solution: Since π₯ 2 + 1 ππ¦ ππ₯ 1 1 = π¦ 2 + 1 β 1+π¦ 2 ππ¦ β 1+π₯ 2 ππ₯ = 0, we obtain 1 1 ππ¦ β ππ₯ = πΆ 1 + π¦2 1 + π₯2 β πππ tan π¦ β arctan π₯ = πΆ β arctan π¦ = arctan π₯ + πΆ β π¦ = tan(arctan π₯ + πΆ) as the general solution for differential equation π₯ 2 + 1 8. π¦2 β 1 ππ¦ ππ₯ ππ¦ ππ₯ = π¦ 2 + 1. = 4π₯π¦ 2 Solution: Since π¦ 2 β 1 ππ¦ ππ₯ = 4π₯π¦ 2 β π¦ 2 β1 π¦2 ππ¦ β 4π₯ππ₯ = 0, assume π¦ β 0, we obtain π¦2 β 1 ππ¦ β 4π₯ ππ₯ = πΆ π¦2 1 β 1 β 2 ππ¦ β 4π₯ππ₯ = πΆ π¦ 1 β π¦ + β 2π₯ 2 = πΆ π¦ 2 π¦ +1 β = 2π₯ 2 + πΆ π¦ as the general solution for the differential equation π¦ 2 β 1 9. ππ¦ ππ₯ ππ¦ ππ₯ = 4π₯π¦ 2 . = π βπ¦ Solution: ππ¦ 1 Since ππ₯ = π βπ¦ β π βπ¦ ππ¦ β ππ₯ = 0 β π π¦ ππ¦ β ππ₯ = 0, we obtain π π¦ ππ¦ β ππ₯ = πΆ β ππ¦ β π₯ = πΆ β ππ¦ = π₯ + πΆ β π¦ = ln π₯ + πΆ ππ¦ as the general solution for differential equation ππ₯ = π βπ¦ . ππ¦ 10. ππ₯ = π βπ¦ + 1 Solution: ππ¦ 1 Since ππ₯ = π βπ¦ + 1 β π βπ¦ +1 ππ¦ β ππ₯ = 0, we obtain 1 ππ¦ β +1 π βπ¦ ππ₯ = πΆ1 Page 18 of 72
ππ¦ ππ¦ β ππ₯ = πΆ1 ππ¦ + 1 1 β π π π¦ β ππ₯ = πΆ1 π¦ π +1 β ln π π¦ + 1 β π₯ = πΆ1 β ln π π¦ + 1 = π₯ + πΆ1 β π π¦ = π π₯+πΆ1 β 1 , πππ‘ πΆ = π πΆ1 β π¦ = ln |πΆπ π₯ β 1| β ππ¦ as the general solution for differential equation ππ₯ = π βπ¦ + 1. ππ¦ 11. ππ₯ = 3π₯π¦ 3 Solution: ππ¦ 1 Since ππ₯ = 3π₯π¦ 3 β π¦ 3 ππ¦ β 3π₯ππ₯ = 0, assume π¦ β 0, we obtain 1 ππ¦ β 3π₯ππ₯ = πΆ1 π¦3 1 3 β β π₯ 2 = πΆ1 2 β2π¦ 2 1 β 2 = β3π₯ 2 β 2πΆ1 , πππ‘ πΆ = β2πΆ1 π¦ 1 β π¦2 = πΆ β 3π₯ 2 As the general solution for differential equation ππ¦ ππ¦ ππ₯ = 3π₯π¦ 3 . 2+ π₯ 12. ππ₯ = 2+ π¦ Solution: Since ππ¦ ππ₯ = 2+ π₯ 2+ π¦ β 2 + π¦ ππ¦ β 2 + π₯ ππ₯ = 0, we obtain 2 + π¦ ππ¦ β 2 + π₯ ππ₯ = πΆ2 2 2 β 2π¦ + π¦ π¦ β 2π₯ β π₯ π₯ = πΆ2 3 3 1 1 β 2π¦ 1 + π¦ = 2π₯ 1 + π₯ + πΆ2 3 3 3+ π¦ 3+ π₯ πΆ2 πΆ2 βπ¦ =π₯ + , πππ‘ πΆ1 = 3 3 2 2 π₯ βπ¦= βπ¦= 3+ π₯ + πΆ1 3 3+ π¦ 3 π₯ 3+ π₯ +πΆ , πππ‘ πΆ = 3πΆ1 3+ π¦ ππ¦ 2+ π₯ as the general solution for the differential equation ππ₯ = 2+ π¦ . Page 19 of 72
ππ¦ 13. ππ₯ β 3π₯ 2 π¦ 2 = β3π₯ 2 Solution: ππ¦ ππ¦ 1 Since ππ₯ β 3π₯ 2 π¦ 2 = β3π₯ 2 β ππ₯ β 3π₯ 2 π¦ 2 β 1 = 0 β π¦ 2 β1 ππ¦ β 3π₯ 2 ππ₯ = 0, we obtain 1 ππ¦ β 3π₯ 2 ππ₯ = πΆ 2 π¦ β1 1 1 2 β β 2 ππ¦ β π¦β1 π¦+1 3π₯ 2 ππ₯ = πΆ1 1 1 ln π¦ β 1 β ln π¦ + 1 β π₯ 3 = πΆ1 2 2 π¦β1 β ln = 2(π₯ 3 + πΆ1 ) π¦+1 π¦β1 3 β = π 2π₯ +πΆ , πππ‘ πΆ = 2πΆ1 π¦+1 β β π¦ β 1 = π¦ + 1 π 2π₯ 3 +πΆ as the general solution from the differential equation ππ¦ ππ¦ ππ₯ β 3π₯ 2 π¦ 2 = β3π₯ 2 . 14. ππ₯ β 3π₯ 2 π¦ 2 = 3π₯ 2 Solution: ππ¦ ππ¦ 1 Since ππ₯ β 3π₯ 2 π¦ 2 = 3π₯ 2 β ππ₯ = 3π₯ 2 1 + π¦ 2 β 1+π¦ 2 ππ¦ β 3π₯ 2 ππ₯ = 0, we obtain 1 ππ¦ β 3π₯ 2 ππ₯ = πΆ 1 + π¦2 β arctan π¦ β π₯ 3 = πΆ β π¦ = tan(π₯ 3 + πΆ) ππ¦ as the general solution of differential equation ππ₯ β 3π₯ 2 π¦ 2 = 3π₯ 2 . ππ¦ 15. ππ₯ = 200π¦ β 2π¦ 2 Solution: ππ¦ 1 1 Since ππ₯ = 200π¦ β 2π¦ 2 β 200π¦ β2π¦ 2 ππ¦ = ππ₯ β 200 π¦ β2π¦ 2 ππ¦ β ππ₯ = 0, assume π¦ β 100, we obtain 1 ππ¦ β ππ₯ = πΆ2 200π¦ β 2π¦ 2 1 β ππ¦ β ππ₯ = πΆ2 2π¦ 100 β π¦ 1 1 100 + 200 β ππ¦ β ππ₯ = πΆ2 2π¦ 100 β π¦ 1 1 ln π¦ β ln 100 β π¦ β π₯ = πΆ2 200 200 1 π¦ β ln = π₯ + πΆ2 200 100 β π¦ β Page 20 of 72
π¦ = 200 π₯ + πΆ2 , πππ‘ πΆ1 = 200πΆ2 100 β π¦ π¦ β = π 200π₯+πΆ1 , πππ‘ πΆ = π πΆ1 100 β π¦ β π¦ = 100 β π¦ πΆπ 200π₯ β ln as the general solution for the differential equation ππ¦ ππ₯ = 200π¦ β 2π¦ 2 . Solution of Quiz 1. Solve π π£ + 1 cos π’ ππ’ + π π£ sin π’ + 1 ππ£ = 0. Solution: From the differential equation π π£ + 1 cos π’ ππ’ + π π£ sin π’ + 1 ππ£ = 0, dividing both side by π π£ + 1 (sin π’ + 1), with the assumption that sin(π’) β β1 and we obviously π π£ β β1, we obtain cos π’ ππ£ ππ’ + π£ ππ£ = 0. sin π’ + 1 π +1 cos π’ ππ£ ππ’ + ππ£ = πΆ1 sin π’ + 1 ππ£ + 1 1 1 β π sin π’ + π π π£ = πΆ1 π£ sin π’ + 1 π +1 β ln sin π’ + 1 + ln π π£ + 1 = πΆ1 β¦ (β) β ln sin π’ + 1 = β ln π π£ + 1 + πΆ1 π£ β sin π’ = π β ln π +1 +πΆ1 β 1, πππ‘ πΆ = π πΆ1 π£ β1 β π’ = arcsin πΆπ ln π +1 β 1 β π’ = arcsin(πΆ π π£ + 1 β1 β 1) as the general solution of the separable differential equation π π£ + 1 cos π’ ππ’ + π π£ sin π’ + 1 ππ£ = 0. It is enough for us if we say that (*) is the general solution of given separable differential equation. 2. Solve π¦ + 2 ππ₯ + π¦ π₯ + 4 ππ¦ = 0, π¦ β3 = β1. Solution: From π¦ + 2 ππ₯ + π¦ π₯ + 4 ππ¦ = 0, dividing both side by π¦ + 2 (π₯ + 4), with the assumption that π¦ β β2 and π₯ β β4, we obtain 1 π¦ ππ₯ + ππ¦ = 0 π₯+4 π¦+2 1 π¦ ππ₯ + ππ¦ = πΆ π₯+4 π¦+2 1 π¦+2 β2 β ππ₯ + ππ¦ = πΆ π₯+4 π¦+2 1 2 β ππ₯ + 1β ππ¦ = πΆ π₯+4 π¦+2 Page 21 of 72
β ln π₯ + 4 + π¦ β 2 ln π¦ + 2 = πΆ β π¦ β 2 ln π¦ + 2 = β ln π₯ + 4 + πΆ β ln(π π¦ ) β ln π¦ + 2 2 = β ln π₯ + 4 + πΆ ππ¦ β ln = β ln π₯ + 4 + ln π πΆ π¦+2 2 ππ¦ ππΆ β ln = ln π¦+2 2 π₯+4 π¦ πΆ π π β = β¦ (β) π¦+2 2 π₯+4 Afterwards, we apply the initial condition to (*) and we obtain π β1 ππΆ = β πΆ = β1. β1 + 2 2 β3 + 4 Therefore, we can conclude that the solution of the initial- value problem under consideration is ππ¦ π β1 = π¦+2 2 π₯+4 or it can be written as π₯+4 π¦ π = π β1 . π¦+2 2 Solution of Exercise 7 (Homogeneous D.E) Definition An equation in differential form π π₯, π¦ ππ₯ + π π₯, π¦ ππ¦ = 0 is said to be homogeneous, ππ¦ if when written in derivative form ππ₯ = π π₯, π¦ = π π π₯, π¦ = π π¦ π₯ π¦ π₯ there exist a function π such that . Identify whether the following differential equations is homogeneous or not. 1. π₯ β π¦ ππ₯ + π₯ππ¦ = 0 Solution: Since π₯ β π¦ ππ₯ + π₯ππ¦ = 0 can be written as π¦ ππ¦ π₯βπ¦ π¦ π¦ =β = β1 + = β π₯ + π¦ π₯ ππ₯ π₯ π₯ π₯ π¦ we obtain that there exist a function π such that π π₯, π¦ = π π₯ . Therefore, we can conclude that the differential equation π₯ β π¦ ππ₯ + π₯ππ¦ = 0 is homogeneous. 2. π₯ 2 + π₯π¦ ππ₯ β π¦ 2 ππ¦ = 0 Page 22 of 72
Solution: Since π₯ 2 + π₯π¦ ππ₯ β π¦ 2 ππ¦ = 0 can be written as ππ¦ π₯ 2 + π₯π¦ π₯ 2 π₯ 1 1 = = + = + , π¦ 2 π¦ ππ₯ π¦2 π¦ π¦ π₯ π₯ we obtain that there exist a function π such that π π₯, π¦ = π π¦ π₯ . Thus, we can conclude that the differential equation π₯ 2 + π₯π¦ ππ₯ β π¦ 2 ππ¦ = 0 is homogeneous. 3. π₯ 2 + π₯π¦ + π¦ 2 ππ₯ β π¦ 2 + π₯ ππ¦ = 0 Solution: Since π₯ 2 + π₯π¦ + π¦ 2 ππ₯ β π¦ 2 + π₯ ππ¦ = 0 can not be written as a function π π¦ π₯ , that is π¦ π¦ π₯ π₯ ππ¦ π₯ 2 + π₯π¦ + π¦ 2 π₯π¦ π¦ + 1 + π₯ π¦+1+π₯ = = = π¦ 1 π¦ 1 ππ₯ π¦2 + π₯ π₯π¦ π₯ + π¦ π₯+π¦ can not be written as a function π π¦ π₯ , we say that the differential equation π₯ 2 + π₯π¦ + π¦ 2 ππ₯ β π¦ 2 + π₯ ππ¦ = 0 is not homogeneous. 4. π₯ 2 + π₯π¦ + π¦ 2 ππ₯ β π¦ 3 + π₯ 2 π¦ ππ¦ = 0 Solution: Since π₯ 2 + π₯π¦ + π¦ 2 ππ₯ β π¦ 3 + π₯ 2 π¦ ππ¦ = 0 can not be written as a function π π¦ π₯ , that is, π¦ π₯ ππ¦ π₯ 2 + π₯π¦ + π¦ 2 π₯π¦(π¦ + 1 + π₯ ) 1 = = = π¦2 ππ₯ π¦3 + π₯ 2π¦ π₯ π₯ 2π¦ 2 + 1 π₯ can not be written as a function π 2 2 3 π¦ π₯ π¦ 1 π¦ +1+π₯ π₯ π¦ 2 π₯ +1 , we can conclude that the differential equation π₯ + π₯π¦ + π¦ ππ₯ β π¦ + π₯ 2 π¦ ππ¦ = 0 is not homogeneous. *** Solution of Exercise 8 (Solution of Homogeneous D.E) Theorem If π π₯, π¦ ππ₯ + π π₯, π¦ ππ¦ = 0 is a homogeneous differential equation, then the change of variables π¦ = π£π₯ transform π π₯, π¦ ππ₯ + π π₯, π¦ ππ¦ = 0 into a separable equation in the variables π£ and π₯. Page 23 of 72
1. Find the general solution of ππ¦ ππ₯ = π₯π¦ +π¦ 2 π₯2 . Solution: The given differential equation in derivative form is homogeneous since ππ¦ π₯π¦ + π¦ 2 π¦ π¦ 2 π¦ = = + = π . ππ₯ π₯2 π₯ π₯ π₯ ππ¦ ππ£ Now, let π¦ = π£π₯. We obtain ππ¦ = π£ππ₯ + π₯ππ£ β ππ₯ = π£ + π₯ ππ₯ and βπ£+π₯ ππ£ = π£ + π£2 ππ₯ ππ£ = π£2 ππ₯ 1 1 β 2 ππ£ β ππ₯ = 0 π£ π₯ integrating 1 1 ππ£ β ππ₯ = πΆ 2 π£ π₯ 1 β β β ln π₯ = πΆ π£ π¦ substituting π£ = π₯ , we obtain π₯ π₯ β β ln π₯ = πΆ β π¦ = β , π¦ ln π₯ + πΆ βπ₯ as the general solution of of ππ¦ ππ¦ ππ₯ = π₯π¦ +π¦ 2 π₯2 . 2. Solve 2π₯π¦ ππ₯ = π₯ 2 + π¦ 2 given that π¦ = 0 at π₯ = 1. Solution: The given equation in derivative form is ππ¦ π₯ 2 + π¦ 2 π₯ π¦ 1 1 π¦ = = + = + ππ₯ 2π₯π¦ 2π¦ 2π₯ 2 π¦ 2 π₯ π₯ and from this form we obtain that the given differential equation is ππ¦ ππ£ homogeneous. Now, let π¦ = π£π₯. We obtain ππ¦ = π£ππ₯ + π₯ππ£ β ππ₯ = π£ + π₯ ππ₯ and ππ£ 1 1 = + π£ ππ₯ 2π£ 2 ππ£ 1 1 βπ₯ = β π£ ππ₯ 2π£ 2 ππ£ 1 β π£ 2 βπ₯ = ππ₯ 2π£ 2π£ 1 β ππ£ β ππ₯ = 0 1 β π£2 π₯ integrating 2π£ 1 ππ£ β ππ₯ = πΆ1 2 1βπ£ π₯ 1 1 2 ββ π βπ£ β ππ₯ = πΆ1 1 β π£2 π₯ β β ln 1 β π£ 2 β ln π₯ = πΆ1 π£+π₯ Page 24 of 72
β β ln 1 β π£ 2 π₯ = πΆ1 β ln 1 β π£ 2 π₯ = πΆ (let πΆ = βπΆ1 ) β 1 β π£2 π₯ = ππΆ . π¦ Substituting π£ = π₯ we obtain π¦2 π¦2 πΆ π₯ = π β π₯ β = ππΆ. π₯2 π₯ Using the initial condition π¦ = 0 at π₯ = 1, we obtain 1 β 0 = π πΆ β 1 = π πΆ . It means that 1 β π£2 π₯ = π πΆ β π₯β π¦2 1β = 1. From this result, we obtain π₯ 2 π¦ π¦2 = β1 β π¦ 2 = π₯ 2 + π₯ or π₯ β = 1 β π¦ 2 = π₯ 2 β π₯. π₯ π₯ as the solution for the given initial value problem. π₯β ππ¦ 3. Solve ππ₯ = π₯+π¦ π₯ and find the particular solution when π¦ 1 = 1. Solution: ππ¦ Since ππ₯ = π₯+π¦ π₯ π¦ =1+ π₯ =π π¦ π₯ , π₯ + π¦ ππ₯ β π₯ππ¦ = 0 is homogeneous. ππ¦ ππ£ Moreover, let π¦ = π£π₯ and we will obtain that ππ¦ = π£ππ₯ + π₯ππ£ β ππ₯ = π£ + π₯ ππ₯ and ππ£ ππ£ 1 =1+π£ βπ₯ = 1 β ππ£ β ππ₯ = 0 β¦ (β). ππ₯ ππ₯ π₯ Integrating (*), we obtain 1 ππ£ β ππ₯ = πΆ β π£ β ln |π₯| = πΆ β¦ ββ π₯ π¦ Substituting π£ = π₯ to (**) we obtain π¦ = π₯ ln π₯ + πΆπ₯. π£+π₯ Using initial condition, we obtain 1 = 1 ln 1 + πΆ β πΆ = 1. Therefore, we can conclude that π¦ = π₯ ln |π₯| + π₯ is the particular solution for the given initial value problem. ππ¦ 1 4. Solve π₯ ππ₯ = π₯ β π¦ and find the particular solution when π¦ 2 = 2. ππ¦ Since ππ₯ = π₯βπ¦ π₯ π¦ =1βπ₯ =π π¦ π₯ , π₯ β π¦ ππ₯ β π₯ππ¦ = 0 is homogeneous. ππ¦ ππ£ Moreover, let π¦ = π£π₯ and we will obtain that ππ¦ = π£ππ₯ + π₯ππ£ β ππ₯ = π£ + π₯ ππ₯ and ππ£ ππ£ 1 1 =1βπ£ βπ₯ = 1 β 2π£ β ππ£ β ππ₯ = 0 β¦ (β). ππ₯ ππ₯ 1 β 2π£ π₯ Integrating (*), we obtain 1 1 ππ£ β ππ₯ = πΆ1 1 β 2π£ π₯ 1 β β ln 1 β 2π£ β ln π₯ = πΆ1 2 β ln 1 β 2π£ + ln π₯ 2 = πΆ (let πΆ = β2πΆ1 ) β ln 1 β 2π£ π₯ 2 = πΆ β 1 β 2π£ π₯ 2 = π πΆ π£+π₯ Page 25 of 72
β π₯ 2 β 2π£π₯ 2 = π πΆ β¦ ββ π¦ Substituting π£ = π₯ to (**) we obtain π₯ 2 β 2π₯π¦ = π πΆ . Using initial condition, we obtain 4 β 2 = 2 = π πΆ . Therefore, we can conclude that π₯ 2 β 2π₯π¦ = 2 is the particular solution for the given initial value problem. ππ¦ 5. Solve ππ₯ = π₯β2π¦ π₯ and find the particular solution when π¦ 1 = β1. Solution: ππ¦ Since ππ₯ = π₯β2π¦ π₯ = 1β2 π¦ π₯ =π π¦ π₯ , π₯ β 2π¦ ππ₯ β π₯ππ¦ = 0 is homogeneous. ππ¦ ππ£ Moreover, let π¦ = π£π₯ and we will obtain that ππ¦ = π£ππ₯ + π₯ππ£ β ππ₯ = π£ + π₯ ππ₯ and ππ£ ππ£ 1 1 = 1 β 2π£ β π₯ = 1 β 3π£ β ππ£ β ππ₯ = 0 β¦ (β). ππ₯ ππ₯ 1 β 3π£ π₯ Integrating (*), we obtain 1 1 ππ£ β ππ₯ = πΆ1 1 β 3π£ π₯ 1 β β ln 1 β 3π£ β ln π₯ = πΆ1 3 β ln 1 β 3π£ + ln(π₯ 2 π₯ ) = πΆ (let πΆ = β3πΆ1 ) 2 β ln 1 β 3π£ π₯ π₯ = πΆ β 1 β 3π£ |π₯|π₯ 2 = π πΆ β π₯ 3 β 3π£π₯ 3 = π πΆ β¦ ββ π¦ Substituting π£ = π₯ to (**) we obtain π£+π₯ ππΆ π₯ β 3π₯ π¦ = π₯ π₯ β 3π¦ = π β π₯ β 3π¦ = 2 . π₯ 3 2 2 πΆ Using initial condition, we obtain 1 + 3 = ππΆ 1 4 β π πΆ = 4. Therefore, we can conclude that π₯ β 3π¦ = π₯ 2 = 2 2 π₯ is the particular solution for the given initial value problem. ππ¦ π₯+π¦ 6. Given that ππ₯ = π₯βπ¦ , prove that arctan π¦ π₯ 1 = 2 ln(π₯ 2 + π¦ 2 ) + π΄ where π΄ is an arbitrary constant. Proof: Since 1 π¦ +1 π¦ π₯ =π , π₯ + π¦ ππ₯ β (π₯ β π¦)ππ¦ = 0 1 π₯ β 1 π¦ π₯ is homogeneous. Moreover, let π¦ = π£π₯ and we will obtain that π₯ ππ¦ π₯ + π¦ π¦ + 1 = = = ππ₯ π₯ β π¦ π₯ β 1 π¦ ππ¦ ππ£ ππ¦ = π£ππ₯ + π₯ππ£ β ππ₯ = π£ + π₯ ππ₯ and Page 26 of 72
1 ππ£ π£ + 1 ππ£ 1 + π£ β π£ + π£ 2 1βπ£ 1 π£+π₯ = βπ₯ = β ππ£ β ππ₯ = 0 β¦ (β). ππ₯ 1 β 1 ππ₯ 1βπ£ 1 + π£2 π₯ π£ Integrating (*), we obtain 1βπ£ 1 ππ£ β ππ₯ = π΄ 2 1+π£ π₯ 1 1 2π£ 1 β ππ£ β ππ£ β ππ₯ = π΄ 1 + π£2 2 1 + π£2 π₯ 1 1 β arctan π£ β ln 1 + π£ 2 β ln(π₯ 2 ) = π΄ 2 2 1 β arctan π£ = ln 1 + π£ 2 π₯ 2 + π΄ β¦ (β) 2 π¦ Substituting π£ = π₯ to (*), we obtain π¦2 1 1 arctan(π£) = 2 ln 1 + π₯ 2 π₯ 2 + π΄ = 2 ln(π₯ 2 + π¦ 2 ) + π΄.β ππ¦ 7. Find the general solution of 2π₯ 2 ππ₯ = π₯ 2 + π¦ 2 . Solution: ππ¦ Since ππ₯ = π₯ 2 +π¦ 2 2π₯ 2 1 π¦ 2 1 =2+2 π₯ =π π¦ π₯ , π₯ 2 + π¦ 2 ππ₯ β 2π₯ 2 ππ¦ = 0 is homogeneous. Moreover, let π¦ = π£π₯ and we will obtain that ππ¦ = π£ππ₯ + π₯ππ£ β ππ¦ ππ₯ =π£+π₯ ππ£ ππ₯ and ππ£ 1 + π£ 2 ππ£ π£β1 2 2 1 = βπ₯ = β ππ£ β ππ₯ = 0 β¦ (β). ππ₯ 2 ππ₯ 2 π£β1 2 π₯ Integrating (*), we obtain 2 1 2 2 ππ£ β ππ₯ = πΆ β β β ln |π₯| = πΆ β = ln |π₯| + πΆ β¦ ββ 2 π£β1 π₯ π£β1 1βπ£ π¦ Substituting π£ = π₯ to (**) we obtain 2π₯ 2π₯ = ln π₯ + πΆ β π¦ = π₯ β π₯βπ¦ ln π₯ + πΆ π£+π₯ ππ¦ as the general solution of 2π₯ 2 ππ₯ = π₯ 2 + π¦ 2 . 8. Find the general solution of 2π₯ β π¦ ππ¦ ππ₯ = 2π¦ β π₯ . Solution: Since 1 2β π¦ ππ¦ 2π¦ β π₯ π₯ = π π¦ , (2π¦ β π₯)ππ₯ β (2π₯ β π¦)ππ¦ = 0 = = π₯ = 2 ππ₯ 2π₯ β π¦ 2 π₯ π¦ β1 π¦ β1 π₯ is homogeneous. Moreover, let π¦ = π£π₯ and we will obtain that π₯ 2βπ¦ ππ¦ ππ£ ππ¦ = π£ππ₯ + π₯ππ£ β ππ₯ = π£ + π₯ ππ₯ and Page 27 of 72
1 ππ£ 2 β π£ ππ£ 2π£ β 1 β π£(2 β π£) π£β2 1 π£+π₯ = βπ₯ = ββ 2 ππ£ β ππ₯ = 0 β¦ (β). ππ₯ 2 β 1 ππ₯ 2βπ£ π£ β1 π₯ π£ Integrating (*), we obtain π£β2 1 β 2 ππ£ β ππ₯ = πΆ1 π£ β1 π₯ 1 2π£ 2 1 ββ ππ£ + ππ£ β ππ₯ = πΆ1 2 π£2 β 1 π£2 β 1 π₯ 1 1 1 1 1 ββ π π£2 + β ππ£ β ππ₯ = πΆ1 2 2 π£ β1 π£β1 π£+1 π₯ 1 β β ln π£ 2 β 1 + ln π£ β 1 β ln π£ + 1 β ln π₯ = πΆ1 2 β ln π£ 2 β 1 β ln π£ β 1 2 + ln π£ + 1 2 + ln(π₯ 2 ) = πΆ2 (let πΆ2 = β2πΆ1 ) 2 2 2 π£ β1 π£+1 π₯ β ln = πΆ2 π£β1 2 π£2 β 1 π£ + 1 2 π₯ 2 β = π πΆ2 = πΆ β¦ (ββ) let πΆ = π πΆ2 . π£β1 2 π¦ Substituting π£ = π₯ to (**), we obtain 2 2 π¦2 π¦ 2 2 2 π¦ β 1 + 1 π₯ π¦ β π₯ + 1 2 π¦+π₯ π₯ π₯ π₯ = = π¦+π₯ 2 2 π¦ π¦βπ₯ π¦βπ₯ β 1 π₯ π₯2 ππ¦ as the general solution of 2π₯ β π¦ ππ₯ = 2π¦ β π₯ . 2 =πΆ *** Solution of Exercise 9 (Non Homogeneous D.E) Note: If π π π π = π β π , suppose π π π = π = π, then the differential equation ππ₯ + ππ¦ + π ππ₯ + ππ₯ + ππ¦ + π ππ¦ = 0 can be written as ππ₯ + ππ¦ + π ππ₯ + π ππ₯ + ππ¦ + π ππ¦ = 0. To solve this, let π’ = ππ₯ + ππ¦. Find the solution of π₯ + π¦ + 1 ππ₯ + 2π₯ + 2π¦ + 1 ππ¦ = 0. Solution: Since π₯ + π¦ + 1 ππ₯ + 2π₯ + 2π¦ + 1 ππ¦ = 0 β π₯ + π¦ + 1 ππ₯ + 2 π₯ + π¦ + 1 ππ¦ = 0, let π’ = π₯ + π¦. We obtain ππ’ = ππ₯ + ππ¦ β ππ₯ = ππ’ β ππ¦ and π’ + 1 ππ₯ + 2π’ + 1 ππ¦ = 0 β π’ + 1 ππ’ β ππ¦ + 2π’ + 1 ππ¦ = 0 β π’ + 1 ππ’ β π’ + 1 ππ¦ + 2π’ + 1 ππ¦ = 0 β π’ + 1 ππ’ + βπ’ β 1 + 2π’ + 1 ππ¦ = 0 β π’ + 1 ππ’ + π’ππ¦ = 0. Dividing both sides of π’ + 1 ππ’ + π’ππ¦ = 0 by π’, we obtain Page 28 of 72
π’+1 ππ’ + ππ¦ = 0. π’ Furthermore, we obtain π’+1 ππ’ + ππ¦ = πΆ π’ 1 β 1 + ππ’ + ππ¦ = πΆ π’ β π’ + ln π’ + π¦ = πΆ β¦ (β) Substituting π’ = π₯ + π¦ to (*), we obtain β π₯ + π¦ + ln π₯ + π¦ + π¦ = πΆ β π₯ + 2π¦ + ln |π₯ + π¦| = πΆ as the general solution of π₯ + π¦ + 1 ππ₯ + 2π₯ + 2π¦ + 1 ππ¦ = 0. *** Solution of Exercise 10 (Solution of Non Homogeneous D.E) 1. Solve 3π₯ β π¦ + 1 ππ₯ β 6π₯ β 2π¦ β 3 ππ¦ = 0. Solution: Since 3π₯ β π¦ + 1 ππ₯ β 6π₯ β 2π¦ β 3 ππ¦ = 0 β 3π₯ β π¦ + 1 ππ₯ β 2 3π₯ β π¦ β 3 ππ¦ = 0, 1 1 3 3 let π’ = 3π₯ β π¦. We obtain ππ’ = 3ππ₯ β ππ¦ β ππ₯ = ππ’ + ππ¦ and π’ + 1 ππ₯ β 2π’ β 3 ππ¦ = 0 1 1 β π’+1 ππ’ + ππ¦ β 2π’ β 3 ππ¦ = 0 3 3 1 1 β π’ + 1 ππ’ + π’ + 1 ππ¦ β 2π’ β 3 ππ¦ = 0 3 3 1 1 5 10 β π’ + ππ’ + β π’ + ππ¦ = 0. 3 3 3 3 1 1 5 10 5 10 Dividing both sides of 3 π’ + 3 ππ’ + β 3 π’ + 3 ππ¦ = 0 by β 3 π’ + 3 , we obtain π’+1 ππ’ + ππ¦ = 0. β5π’ + 10 Furthermore, we obtain π’+1 ππ’ + ππ¦ = πΆ2 β5π’ + 10 β5 π’ + 1 β ππ’ + ππ¦ = πΆ2 β5 β5π’ + 10 1 β5π’ + 10 β15 β β + ππ’ + ππ¦ = πΆ2 5 β5π’ + 10 β5π’ + 10 1 3 β5 ββ ππ’ β ππ’ + ππ¦ = πΆ2 5 5 β5π’ + 10 1 3 1 ββ ππ’ β π β5π’ + 10 + ππ¦ = πΆ2 5 5 β5π’ + 10 1 3 β β π’ β ln β5π’ + 10 + π¦ = πΆ2 β¦ (β) 5 5 Page 29 of 72
Substituting π’ = 3π₯ β π¦ to (*), we obtain 1 3 β 3π₯ β π¦ β ln β5 3π₯ β π¦ + 10 + π¦ = πΆ2 5 5 β 3π₯ β π¦ + 3 ln β15π₯ + 5π¦ + 10 β 5π¦ = β5πΆ2 β 3π₯ β 6π¦ + 3 ln 5 β3π₯ + π¦ + 2 = πΆ1 πΆ1 β π₯ β 2π¦ + ln 5 + ln β3π₯ + π¦ + 2 = 3 πΆ1 β π₯ β 2π¦ + ln β3π₯ + π¦ + 2 = β ln 5 3 β π₯ β 2π¦ + ln | β 3π₯ + π¦ + 2| = πΆ as the general solution of 3π₯ β π¦ + 1 ππ₯ β 6π₯ β 2π¦ β 3 ππ¦ = 0. 2. Solve the initial value problem of 2π₯ + 3π¦ + 1 ππ₯ + 4π₯ + 6π¦ + 1 ππ¦ = 0; π¦ β2 = 2. Solution: Since 2π₯ + 3π¦ + 1 ππ₯ + 4π₯ + 6π¦ + 1 ππ¦ = 0 β 2π₯ + 3π¦ + 1 ππ₯ + 2 2π₯ + 3π¦ + 1 ππ¦ = 0, 1 3 let π’ = 2π₯ + 3π¦. We obtain ππ’ = 2ππ₯ + 3ππ¦ β ππ₯ = 2 ππ’ β 2 ππ¦ and π’ + 1 ππ₯ + 2π’ + 1 ππ¦ = 0 1 3 β π’+1 ππ’ β ππ¦ + 2π’ + 1 ππ¦ = 0 2 2 π’+1 3π’ 3 β ππ’ β ππ¦ β ππ¦ + 2π’ππ¦ + ππ¦ = 0 2 2 2 π’+1 π’β1 β ππ’ + ππ¦ = 0 2 2 π’+1 π’ β1 Dividing both sides of 2 ππ’ + 2 ππ¦ = 0 by π’β1 2 , we obtain π’+1 ππ’ + ππ¦ = 0 π’β1 π’β1 +2 β ππ’ + ππ¦ = 0 π’β1 2 β 1+ ππ’ + ππ¦ = 0. π’β1 Furthermore, we obtain 2 1+ ππ’ + ππ¦ = πΆ1 π’β1 β π’ + 2 ln |π’ β 1| + π¦ = πΆ1 β¦ (β) Substituting π’ = 2π₯ + 3π¦ to (*), we obtain 2π₯ + 3π¦ + 2 ln 2π₯ + 3π¦ β 1 + π¦ = πΆ1 β 2π₯ + 4π¦ + 2 ln 2π₯ + 3π¦ β 1 = πΆ1 πΆ1 β π₯ + 2π¦ + ln 2π₯ + 3π¦ β 1 = 2 β π₯ + 2π¦ + ln |2π₯ + 3π¦ β 1| = πΆ β¦ (ββ) Afterwards, we substituting π₯ = β2 and π¦ = 2 (initial condition) to (**), we obtain β2 + 4 + ln | β 4 + 6 β 1| = πΆ β 2 + ln(1) = πΆ β 2 = πΆ. Page 30 of 72
Therefore, π₯ + 2π¦ + ln |2π₯ + 3π¦ β 1| = 2 is the solution for the given initial value problem. *** Part 3 Solve the following differential equations. 1. ππ¦ ππ₯ 2π₯ β7π¦ = 3π¦ β8π₯ Solution: π¦ π₯ 2β7 ππ¦ 2π₯ β 7π¦ π₯ = = ππ₯ 3π¦ β 8π₯ 1 π¦ 3β8 π¦ π₯ 1 = π¦ π₯ π¦ π₯ 8 3β π¦ π₯ 2β7 π¦ Let π¦ = π£π₯ β π£ = π₯ . We obtain, ππ¦ = π₯ππ£ + π£ππ₯ β consequence, we obtain ππ£ 1 2 β 7π£ = ππ₯ π£ 3 β 8 π£ ππ£ 2 β 7π£ β π£+π₯ = ππ₯ 3π£ β 8 ππ£ β3π£ 2 + π£ + 2 βπ₯ = ππ₯ 3π£ β 8 3π£ β 8 1 ββ 2 ππ£ β ππ₯ = 0 3π£ β π£ β 2 π₯ 6 1 1 ββ β ππ£ β ππ₯ = 0 3π£ + 2 π£ β 1 π₯ 2.3 1 1 β β ππ£ + ππ₯ = 0. 3π£ + 2 π£ β 1 π₯ Furthermore, we obtain 2.3 1 1 β ππ£ + ππ₯ = πΆ 3π£ + 2 π£ β 1 π₯ β 2 ln 3π£ + 2 β ln π£ β 1 + ln π₯ = πΆ 3π¦ π¦ β 2 ln + 2 β ln β 1 + ln |π₯| = πΆ π₯ π₯ ππ¦ 2π₯β7π¦ as the general solusion of ππ₯ = 3π¦ β8π₯ . β π£+π₯ 2. π₯ 3 + 1 ππ¦ ππ₯ + 6π₯ 2 π¦ = 6π₯ 2 Solution: ππ¦ + 6π₯ 2 π¦ = 6π₯ 2 ππ₯ ππ¦ β π₯3 + 1 = 6π₯ 2 1 β π¦ ππ₯ 1 6π₯ 2 β ππ¦ β 3 ππ₯ = 0. 1βπ¦ π₯ +1 π₯3 + 1 Page 31 of 72 ππ¦ ππ₯ ππ£ = π£ + π₯ ππ₯ and as a
Afterwards, we obtain 1 3π₯ 2 ππ¦ β 2 ππ₯ = πΆ1 1βπ¦ π₯3 + 1 1 1 ββ ππ¦ β 2 π π₯ 3 = πΆ1 π¦β1 π₯3 + 1 β β ln π¦ β 1 β 2 ln π₯ 3 + 1 = πΆ1 β ln π¦ β 1 + ln π₯ 3 + 1 2 = βπΆ1 = πΆ as the general solution of π₯ 3 + 1 3. ππ¦ ππ₯ 2π₯ 2 +π¦ 2 ππ¦ ππ₯ + 6π₯ 2 π¦ = 6π₯ 2 . = 2π₯π¦ βπ₯ 2 Solution: π¦ 2 2 π₯ 2 + ππ¦ 2π₯ + π¦ π₯ = = 2 π₯ π₯ ππ₯ 2π₯π¦ β π₯ π¦2 2 π¦ β π¦ π¦ 2 2 + 1 π₯ . 2 = π¦ 2 2 1 β π¦ π¦ 2 π₯ π₯ π₯ π¦ ππ¦ ππ£ Let π¦ = π£π₯ β π£ = π₯ . We obtain, ππ¦ = π₯ππ£ + π£ππ₯ β ππ₯ = π£ + π₯ ππ₯ and as a 2 2 consequence, we obtain ππ£ 1 2 + π£2 π£+π₯ = ππ₯ π£ 2 2 β 1 π£ π£2 ππ£ 2 + π£ 2 β π£+π₯ = ππ₯ 2π£ β 1 ππ£ β(π£ 2 β π£ β 2) βπ₯ = ππ₯ 2π£ β 1 1 β 2π£ 1 β 2 ππ£ β ππ₯ = 0 π£ βπ£β2 π₯ β1 1 1 β β ππ£ β ππ₯ = 0 π£β2 π£+1 π₯ Moreover, β1 1 1 β ππ£ β ππ₯ = πΆ1 π£β2 π£+1 π₯ β β ln |π£ β 2| β ln |π£ + 1| β ln |π₯| = πΆ1 β ln |π£ β 2| + ln |π£ + 1| + ln |π₯| = βπΆ1 β ln π£ β 2 π£ + 1 π₯ = πΆ π¦ π¦ β ln β2 +1 π₯ =πΆ π₯ π₯ π¦2 π¦ β ln β β2 π₯ =πΆ π₯2 π₯ π¦2 β ln β π¦ β 2π₯ = πΆ π₯ ππ¦ 2π₯ 2 +π¦ 2 as the general solution of ππ₯ = 2π₯π¦ βπ₯ 2 . 4. 3π₯ β 5π¦ ππ₯ + π₯ + π¦ ππ¦ = 0 Page 32 of 72
Solution: 3π₯ β 5π¦ ππ₯ + π₯ + π¦ ππ¦ = 0 ππ¦ β 3π₯ β 5π¦ β = ππ₯ π₯+π¦ β ππ¦ π¦ =β ππ₯ π₯ 1 3 π¦ β5 π₯ . π¦ 1+π₯ π¦ ππ¦ ππ£ Let π¦ = π£π₯ β π£ = π₯ . We obtain, ππ¦ = π₯ππ£ + π£ππ₯ β ππ₯ = π£ + π₯ ππ₯ and as a consequence, we obtain 3 β5 ππ£ π£+π₯ = βπ£ π£ ππ₯ 1+π£ ππ£ β3 + 5π£ = ππ₯ 1+π£ ππ£ βπ£ 2 + 4π£ β 3 βπ₯ = ππ₯ 1+π£ 1+π£ 1 ββ 2 ππ£ β ππ₯ = 0 π£ β 4π£ + 3 π₯ 2 1 1 ββ β ππ£ β ππ₯ = 0 π£β3 π£β1 π₯ 1 2 1 β β ππ£ β ππ₯ = 0. π£β1 π£β3 π₯ Furthermore, we obtain 1 2 1 β ππ£ β ππ₯ = πΆ π£β1 π£β3 π₯ β ln π£ β 1 β 2 ln π£ β 3 β ln π₯ = πΆ π¦ π¦ β ln β 1 β 2 ln β 3 β ln π₯ = πΆ π₯ π₯ as the general solution of 3π₯ β 5π¦ ππ₯ + π₯ + π¦ ππ¦ = 0. 5. π₯ 2 + π¦ 2 ππ₯ β 2π₯π¦ππ¦ = 0; π¦ 1 = 0 Solution: ππ¦ π₯ 2 + π¦ 2 1 1 1 π¦ π₯ 2 + π¦ 2 ππ₯ β 2π₯π¦ππ¦ = 0 β = = .π¦ + . . ππ₯ 2π₯π¦ 2 2 π₯ π₯ π¦ ππ¦ ππ£ Let π¦ = π£π₯ β π£ = π₯ . We obtain, ππ¦ = π₯ππ£ + π£ππ₯ β ππ₯ = π£ + π₯ ππ₯ and as a β π£+π₯ consequence, we obtain ππ£ 1 π£ π£+π₯ = + ππ₯ 2π£ 2 ππ£ 1 + π£ 2 β π£+π₯ = ππ₯ 2π£ 2 ππ£ 1 β π£ βπ₯ = ππ₯ 2π£ Page 33 of 72
2π£ 1 ππ£ β ππ₯ = 0. 2 1βπ£ π₯ Furthermore, we obtain β2π£ 1 β ππ£ β ππ₯ = πΆ1 1 β π£2 π₯ 1 1 ββ π βπ£ 2 β ππ₯ = πΆ1 2 1βπ£ π₯ β β ln 1 β π£ 2 β ln π₯ = πΆ1 β ln 1 β π£ 2 + ln π₯ = πΆ π¦2 β ln 1 β 2 + ln |π₯| = πΆ β¦ (β) π₯ Substituting π₯ = 1 and π¦ = 0 (initial condition), we obtain ln 1 β 0 + ln 1 = πΆ β 0 = πΆ. β π¦2 Therefore, ln 1 β π₯ 2 + ln |π₯| = 0 is the solution of the given inital value problem. 6. 5π₯ + 2π¦ + 1 ππ₯ + 2π₯ + π¦ + 1 ππ¦ = 0 Solution: Let π’ = 5π₯ + 2π¦ + 1 and π£ = 2π₯ + π¦ + 1. We obtain ππ’ = 5ππ₯ + 2ππ¦ and ππ£ = 2ππ₯ + ππ¦. Using elimination, we obtain ππ₯ = ππ’ β 2ππ£ and ππ¦ = 5ππ£ β 2ππ’. Afterwards, we obtain π’ππ₯ + π£ππ¦ = 0 β π’ ππ’ β 2ππ£ + π£ 5ππ£ β 2ππ’ = 0 β π’ππ’ β 2π’ππ£ + 5π£ππ£ β 2π£ππ’ = 0 β π’ β 2π£ ππ’ + β2π’ + 5π£ ππ£ = 0 π£ π£ β 1 β 2 ππ’ + β2 + 5 ππ£ = 0. π’ π’ π£ Let π€ = β π£ = π’π€. As a consequence, we obtain ππ£ = π€ππ’ + π’ππ€ and π’ 1 β 2π€ ππ’ + β2 + 5π€ π€ππ’ + π’ππ€ = 0 β ππ’ β 2π€ππ’ β 2π€ππ’ β 2π’ππ€ + 5π€ 2 ππ’ + 5π€π’ππ€ = 0 β 1 β 4π€ + 5π€ 2 ππ’ + β2 + 5π€ π’ππ€ = 0 1 β2 + 5π€ β ππ’ + ππ€ = 0. π’ 1 β 4π€ + 5π€ 2 Moreover, we obtain 1 β2 + 5π€ ππ’ + ππ€ = πΆ π’ 1 β 4π€ + 5π€ 2 1 1 β4 + 10π€ β ππ’ + ππ€ = πΆ π’ 2 1 β 4π€ + 5π€ 2 1 1 1 β ππ’ + π 1 β 4π€ + 5π€ 2 = πΆ π’ 2 1 β 4π€ + 5π€ 2 1 β ln π’ + ln 1 β 4π€ + 5π€ 2 = πΆ 2 1 4 2π₯ + π¦ + 1 2π₯ + π¦ + 1 2 β ln |5π₯ + 2π¦ + 1| + ln 1 β +5 =πΆ 2 5π₯ + 2π¦ + 1 5π₯ + 2π¦ + 1 as the general solution of 5π₯ + 2π¦ + 1 ππ₯ + 2π₯ + π¦ + 1 ππ¦ = 0. Page 34 of 72
Solution of Exercise 11 (Exact and Non Exact D.E) ( In exercises 1-10 determine whether or not each of the given equation is exact; solve those that are exact). 1. 3π₯ + 2π¦ ππ₯ + 2π₯ + π¦ ππ¦ = 0. Solution: Our first duty is to determine whether or not the equation is exact or not. Here π π₯, π¦ = 3π₯ + 2π¦, π π₯, π¦ = 2π₯ + π¦, ππ π₯, π¦ ππ π₯, π¦ = 2, = 2. ππ¦ ππ₯ Since ππ π₯, π¦ ππ π₯, π¦ = =2 ππ¦ ππ₯ we can conclude that the differential equation 3π₯ + 2π¦ ππ₯ + 2π₯ + π¦ ππ¦ = 0 is exact differential equation. Furthermore, we must find πΉ such that ππΉ π₯, π¦ ππΉ π₯, π¦ = π π₯, π¦ = 3π₯ + 2π¦ and = π π₯, π¦ = 2π₯ + π¦. ππ₯ ππ¦ From the first of these, 3 πΉ π₯, π¦ = π π₯, π¦ ππ₯ + π π¦ = 3π₯ + 2π¦ ππ₯ + π π¦ = π₯ 2 + 2π₯π¦ + π π¦ . 2 Then ππΉ π₯, π¦ ππ π¦ = 2π₯ + ππ¦ ππ¦ But we must have ππΉ π₯, π¦ = π π₯, π¦ = 2π₯ + π¦. ππ¦ Thus ππ π¦ ππ π¦ 2π₯ + π¦ = 2π₯ + βπ¦= . ππ¦ ππ¦ 1 Thus π π¦ = 2 π¦ 2 + πΆ0 , where πΆ0 is an arbitrary constant, and so 3 1 πΉ π₯, π¦ = π₯ 2 + 2π₯π¦ + π¦ 2 + πΆ0 . 2 2 Hence a one- parameter family of solution is πΉ π₯, π¦ = πΆ1 , or 3 2 1 π₯ + 2π₯π¦ + π¦ 2 + πΆ0 = πΆ1 2 2 Combining the constsnts πΆ0 and πΆ1 we may write this solution as 3 2 1 π₯ + 2π₯π¦ + π¦ 2 = πΆ 2 2 where πΆ = πΆ1 β πΆ0 is an arbitrary constant. So, we conclude that the general solution of the exact differential equation 3 1 3π₯ + 2π¦ ππ₯ + 2π₯ + π¦ ππ¦ = 0 is 2 π₯ 2 + 2π₯π¦ + 2 π¦ 2 = πΆ. 2. (π¦ 2 + 3)ππ₯ + 2π₯π¦ β 4 ππ¦ = 0. Solution: Our first duty is to determine whether or not the equation is exact or not. Here π π₯, π¦ = π¦ 2 + 3, π π₯, π¦ = 2π₯π¦ β 4, Page 35 of 72
ππ π₯, π¦ ππ π₯, π¦ = 2π¦, = 2π¦. ππ¦ ππ₯ Since ππ π₯, π¦ ππ π₯, π¦ = = 2π¦ ππ¦ ππ₯ we can conclude that the differential equation (π¦ 2 + 3)ππ₯ + 2π₯π¦ β 4 ππ¦ = 0 is exact differential equation. Afterwards, we must find πΉ such that ππΉ π₯, π¦ ππΉ π₯, π¦ = π π₯, π¦ = π¦ 2 + 3 and = π π₯, π¦ = 2π₯π¦ β 4. ππ₯ ππ¦ From the first of these, πΉ π₯, π¦ = π π₯, π¦ ππ₯ + π π¦ = π¦ 2 + 3 ππ₯ + π π¦ = π₯π¦ 2 + 3π₯ + π π¦ . Then ππΉ π₯, π¦ ππ π¦ = 2π₯π¦ + ππ¦ ππ¦ But we must have ππΉ π₯, π¦ = π π₯, π¦ = 2π₯π¦ β 4. ππ¦ Thus ππ π¦ ππ π¦ β β4 = . ππ¦ ππ¦ Thus π π¦ = β4π¦ + πΆ0 , where πΆ0 is an arbitrary constant, and so πΉ π₯, π¦ = π₯π¦ 2 + 3π₯ + β4π¦ + πΆ0 . Hence a one- parameter family of solution is πΉ π₯, π¦ = πΆ1 , or π₯π¦ 2 + 3π₯ + β4π¦ + πΆ0 = πΆ1 Combining the constsnts πΆ0 and πΆ1 we may write this solution as π₯π¦ 2 + 3π₯ + β4π¦ = πΆ where πΆ = πΆ1 β πΆ0 is an arbitrary constant. So, we conclude that the general solution of the exact differential equation (π¦ 2 + 3)ππ₯ + 2π₯π¦ β 4 ππ¦ = 0 is π₯π¦ 2 + 3π₯ + β4π¦ = πΆ. 3. 2π₯π¦ + 1 ππ₯ + π₯ 2 + 4π¦ ππ¦ = 0. Solution: Our first duty is to determine whether or not the equation is exact or not. Here π π₯, π¦ = 2π₯π¦ + 1, π π₯, π¦ = π₯ 2 + 4π¦, ππ π₯, π¦ ππ π₯, π¦ = 2π₯, = 2π₯. ππ¦ ππ₯ Since ππ π₯, π¦ ππ π₯, π¦ = = 2π₯ ππ¦ ππ₯ we can conclude that the differential equation 2π₯π¦ + 1 ππ₯ + π₯ 2 + 4π¦ ππ¦ = 0 is exact differential equation. Afterwards, we must find πΉ such that ππΉ π₯, π¦ ππΉ π₯, π¦ = π π₯, π¦ = 2π₯π¦ + 1 and = π π₯, π¦ = π₯ 2 + 4π¦. ππ₯ ππ¦ From the first of these, 2π₯π¦ β 4 = 2π₯π¦ + Page 36 of 72
πΉ π₯, π¦ = π π₯, π¦ ππ₯ + π π¦ = 2π₯π¦ + 1 ππ₯ + π π¦ = π₯ 2 π¦ + π₯ + π π¦ . Then ππΉ π₯, π¦ ππ π¦ = π₯2 + ππ¦ ππ¦ But we must have ππΉ π₯, π¦ = π π₯, π¦ = π₯ 2 + 4π¦. ππ¦ Thus ππ π¦ ππ π¦ β 4π¦ = . ππ¦ ππ¦ Thus π π¦ = 2π¦ 2 + πΆ0 , where πΆ0 is an arbitrary constant, and so πΉ π₯, π¦ = π₯ 2 π¦ + π₯ + 2π¦ 2 + πΆ0 . Hence a one- parameter family of solution is πΉ π₯, π¦ = πΆ1 , or π₯ 2 π¦ + π₯ + 2π¦ 2 + πΆ0 = πΆ1 Combining the constsnts πΆ0 and πΆ1 we may write this solution as π₯ 2 π¦ + π₯ + 2π¦ 2 = πΆ where πΆ = πΆ1 β πΆ0 is an arbitrary constant. So, we conclude that the general solution of the exact differential equation 2π₯π¦ + 1 ππ₯ + π₯ 2 + 4π¦ ππ¦ = 0 is π₯ 2 π¦ + π₯ + 2π¦ 2 = πΆ. 4. 3π₯ 2 π¦ + 2 ππ₯ β π₯ 3 + π¦ ππ¦ = 0. Solution: Our first duty is to determine whether or not the equation is exact or not. Here π π₯, π¦ = 3π₯ 2 π¦, π π₯, π¦ = βπ₯ 3 β π¦, ππ π₯, π¦ ππ π₯, π¦ = 3π₯ 2 , = β3π₯ 2 . ππ¦ ππ₯ Since ππ π₯, π¦ ππ π₯, π¦ = 3π₯ 2 β = β3π₯ 2 ππ¦ ππ₯ we can conclude that the differential equation 3π₯ 2 π¦ + 2 ππ₯ β π₯ 3 + π¦ ππ¦ = 0 is not exact (non- exact) differential equation. 5. 6π₯π¦ + 2π¦ 2 β 5 ππ₯ + 3π₯ 2 + 4π₯π¦ β 6 ππ¦ = 0. Solution: Our first duty is to determine whether or not the equation is exact or not. Here π π₯, π¦ = 6π₯π¦ + 2π¦ 2 β 5, π π₯, π¦ = 3π₯ 2 + 4π₯π¦ β 6, ππ π₯, π¦ ππ π₯, π¦ = 6π₯ + 4π¦, = 6π₯ + 4π¦. ππ¦ ππ₯ Since ππ π₯, π¦ ππ π₯, π¦ = = 6π₯ + 4π¦ ππ¦ ππ₯ we can conclude that the differential equation 6π₯π¦ + 2π¦ 2 β 5 ππ₯ + 3π₯ 2 + 4π₯π¦ β 6 ππ¦ = 0 is exact differential equation. Moreover, we must find πΉ such that ππΉ π₯, π¦ ππΉ π₯, π¦ = π π₯, π¦ = 6π₯π¦ + 2π¦ 2 β 5 and = π π₯, π¦ = 3π₯ 2 + 4π₯π¦ β 6. ππ₯ ππ¦ π₯ 2 + 4π¦ = π₯ 2 + Page 37 of 72
From the first of these, πΉ π₯, π¦ = = π π₯, π¦ ππ₯ + π π¦ 6π₯π¦ + 2π¦ 2 β 5 ππ₯ + π π¦ = 3π₯ 2 π¦ + 2π₯π¦ 2 β 5π₯ + π π¦ Then ππΉ π₯, π¦ ππ π¦ = 3π₯ 2 + 4π₯π¦ + ππ¦ ππ¦ But we must have ππΉ π₯, π¦ = π π₯, π¦ = 3π₯ 2 + 4π₯π¦ β 6. ππ¦ Thus ππ π¦ ππ π¦ β β6 = . ππ¦ ππ¦ Thus π π¦ = β6π¦ + πΆ0 , where πΆ0 is an arbitrary constant, and so πΉ π₯, π¦ = 3π₯ 2 π¦ + 2π₯π¦ 2 β 5π₯ β 6π¦ + πΆ0 . Hence a one- parameter family of solution is πΉ π₯, π¦ = πΆ1 , or 3π₯ 2 π¦ + 2π₯π¦ 2 β 5π₯ β 6π¦ + πΆ0 = πΆ1 Combining the constsnts πΆ0 and πΆ1 we may write this solution as 3π₯ 2 π¦ + 2π₯π¦ 2 β 5π₯ β 6π¦ = πΆ where πΆ = πΆ1 β πΆ0 is an arbitrary constant. So, we conclude that the general solution of the exact differential equation 6π₯π¦ + 2π¦ 2 β 5 ππ₯ + 3π₯ 2 + 4π₯π¦ β 6 ππ¦ = 0 is 3π₯ 2 π¦ + 2π₯π¦ 2 β 5π₯ β 6π¦ = πΆ. 6. π 2 + 1 cos π ππ + 2π sin π ππ = 0. Solution: Our first duty is to determine whether or not the equation is exact or not. Here π π, π = π 2 + 1 cos π , π π, π = 2π sin π, ππ π, π ππ π, π = 2π cos π , = 2π cos π. ππ ππ Since ππ π, π ππ π, π = = 2π cos π ππ ππ we can conclude that the differential equation π 2 + 1 cos π ππ + 2π sin π ππ = 0 is exact differential equation. Moreover, we must find πΉ such that ππΉ π, π ππΉ π, π = π π, π = π 2 + 1 cos π and = π π, π = 2π sin π ππ ππ From the first of these, 3π₯ 2 + 4π₯π¦ β 6 = 3π₯ 2 + 4π₯π¦ + πΉ π, π = = π π, π ππ + π π π 2 + 1 cos π ππ + π π = π 2 + 1 sin π + π π Then Page 38 of 72
But we must have ππΉ π, π ππ π = 2π sin π + ππ ππ ππΉ π, π = π π, π = 2π sin π. ππ Thus ππ π ππ π β0= . ππ ππ Thus π π = πΆ0 , where πΆ0 is an arbitrary constant, and so πΉ π, π = π 2 + 1 sin π + πΆ0 . Hence a one- parameter family of solution is πΉ π, π = πΆ1 , or π 2 + 1 sin π + πΆ0 = πΆ1 Combining the constsnts πΆ0 and πΆ1 we may write this solution as π 2 + 1 sin π = πΆ where πΆ = πΆ1 β πΆ0 is an arbitrary constant. So, we conclude that the general solution of the exact differential equation π 2 + 1 cos π ππ + 2π sin π ππ = 0 is π 2 + 1 sin π = πΆ. 7. π¦ sec 2 π₯ + sec π₯ tan π₯ ππ₯ + tan π₯ + 2π¦ ππ¦ = 0. Solution: Our first duty is to determine whether or not the equation is exact or not. Here π π₯, π¦ = π¦ sec 2 π₯ + sec π₯ tan π₯ , π π₯, π¦ = tan π₯ + 2π¦, ππ π₯, π¦ ππ π₯, π¦ = sec 2 π₯ , = sec 2 π₯. ππ¦ ππ₯ Since ππ π₯, π¦ ππ π₯, π¦ = = sec 2 π₯ ππ¦ ππ₯ we can conclude that the equation π¦ sec 2 π₯ + sec π₯ tan π₯ ππ₯ + tan π₯ + 2π¦ ππ¦ = 0 is exact differential equation. Furthermore, we must find πΉ such that ππΉ π₯, π¦ ππΉ π₯, π¦ = π π₯, π¦ = π¦ sec 2 π₯ + sec π₯ tan π₯ and = π π₯, π¦ = tan π₯ + 2π¦. ππ₯ ππ¦ From the first of these, 2π sin π = 2π sin π + πΉ π₯, π¦ = = π π₯, π¦ ππ₯ + π π¦ π¦ sec 2 π₯ + sec π₯ tan π₯ ππ₯ + π π¦ = π¦ tan π₯ + sec π₯ + π π¦ . Then ππΉ π₯, π¦ ππ π¦ = tan π₯ + ππ¦ ππ¦ But we must have ππΉ π₯, π¦ = π π₯, π¦ = tan π₯ + 2π¦. ππ¦ Thus Page 39 of 72
ππ π¦ ππ π¦ β 2π¦ = . ππ¦ ππ¦ Thus π π¦ = π¦ 2 + πΆ0 , where πΆ0 is an arbitrary constant, and so πΉ π₯, π¦ = π¦ tan π₯ + sec π₯ + π¦ 2 + πΆ0 . Hence a one- parameter family of solution is πΉ π₯, π¦ = πΆ1 , or π¦ tan π₯ + sec π₯ + π¦ 2 + πΆ0 = πΆ1 Combining the constsnts πΆ0 and πΆ1 we may write this solution as π¦ tan π₯ + sec π₯ + π¦ 2 = πΆ where πΆ = πΆ1 β πΆ0 is an arbitrary constant. So, we conclude that the general solution of the exact differential equation π¦ sec 2 π₯ + sec π₯ tan π₯ ππ₯ + tan π₯ + 2π¦ ππ¦ = 0 is π¦ tan π₯ + sec π₯ + π¦ 2 = πΆ. tan π₯ + 2π¦ = tan π₯ + 8. π₯ π¦2 + π₯ ππ₯ + π₯2 π¦3 + π¦ ππ¦ = 0. Solution: Our first duty is to determine whether or not the equation is exact or not. Here π₯ π₯2 π π₯, π¦ = 2 + π₯, π π₯, π¦ = 3 + π¦, π¦ π¦ ππ π₯, π¦ 2π₯ ππ π₯, π¦ 2π₯ =β 3, = 3. ππ¦ π¦ ππ₯ π¦ Since ππ π₯, π¦ 2π₯ ππ π₯, π¦ 2π₯ =β 3β = 3 ππ¦ π¦ ππ₯ π¦ we can conclude that the differential equation π₯ π¦2 + π₯ ππ₯ + π₯2 π¦3 + π¦ ππ¦ = 0 is not exact (non- exact) differential equation. 9. 2π β1 π‘ ππ + π βπ 2 π‘2 ππ‘ = 0. Solution: Our first duty is to determine whether or not the equation is exact or not. Here 2π β 1 π β π 2 π π , π‘ = , π π , π‘ = , π‘ π‘2 ππ π , π‘ 1 β 2π ππ π , π‘ 1 β 2π = , = . 2 ππ‘ π‘ ππ π‘2 Since ππ π , π‘ ππ π , π‘ 1 β 2π = = ππ‘ ππ π‘2 2π β1 π βπ 2 we can conclude that the differential equation π‘ ππ + π‘ 2 ππ‘ = 0 is exact differential equation. Furthermore, we must find πΉ such that ππΉ π , π‘ 2π β 1 ππΉ π , π‘ π β π 2 = π π , π‘ = and = π π , π‘ = . ππ π‘ ππ‘ π‘2 From the first of these, πΉ π , π‘ = = π π , π‘ ππ₯ + π π‘ 2π β 1 ππ + π π‘ π‘ Page 40 of 72
Then π 2 β π = +π π‘ . π‘ ππΉ π , π‘ π 2 π ππ π‘ π β π 2 ππ π‘ =β 2+ 2+ = + ππ‘ π‘ π‘ ππ‘ π‘2 ππ‘ But we must have ππΉ π , π‘ π β π 2 = π π , π‘ = . ππ‘ π‘2 Thus π β π 2 π β π 2 ππ π‘ ππ π‘ = + β0= . 2 2 π‘ π‘ ππ‘ ππ‘ Thus π π‘ = πΆ0 , where πΆ0 is an arbitrary constant, and so π 2 β π πΉ π , π‘ = + πΆ0 . π‘ Hence a one- parameter family of solution is πΉ π₯, π¦ = πΆ1 , or π 2 β π + πΆ0 = πΆ1 π‘ Combining the constsnts πΆ0 and πΆ1 we may write this solution as π 2 β π =πΆ π‘ or we can write this as π 2 β π = πΆπ‘, where πΆ = πΆ1 β πΆ0 is an arbitrary constant. So, we conclude that the general solution of the exact differential equation 2π β1 π‘ 10. 3 2π¦ 2 +1 1 π₯2 ππ + π βπ 2 ππ‘ = 0 is π 2 β π = πΆπ‘. π‘2 1 1 ππ₯ + 3π₯ 2 π¦ 2 β 1 ππ¦ = 0. Solution: Our first duty is to determine whether or not the equation is exact or not. Here 3 π π₯, π¦ = Since 2π¦ 2 + 1 1 1 1 , π π₯, π¦ = 3π₯ 2 π¦ 2 β 1, π₯2 1 1 ππ π₯, π¦ π¦ 2 ππ π₯, π¦ 3 β1 1 3 π¦ 2 =3 1, = π₯ 2π¦2 = . ππ¦ ππ₯ 2 2 12 2 π₯ π₯ 1 1 ππ π₯, π¦ π¦ 2 ππ π₯, π¦ 3 π¦2 =3 1β = ππ¦ ππ₯ 2 12 π₯2 π₯ 3 we can conclude that the differential equation is not exact (non- exact) differential equation. Page 41 of 72 2π¦ 2 +1 1 π₯2 1 1 ππ₯ + 3π₯ 2 π¦ 2 β 1 ππ¦ = 0
Solution of Exercise 12 (Integrating Factor) ( In each of the following equations determine the constant π΄ such that the equation is exact, and solve the resulting exact equation). a. π₯ 2 + 3π₯π¦ ππ₯ + π΄π₯ 2 + 4π¦ ππ¦ = 0. Solution: Suppose π π₯, π¦ = π₯ 2 + 3π₯π¦ and π π₯, π¦ = π΄π₯ 2 + 4π¦. Then, we obtain ππ π₯, π¦ ππ π₯, π¦ = 3π₯ and = 2π΄π₯. ππ¦ ππ₯ In order to make the differential equation become exact differential equation, it must be ππ π₯, π¦ ππ π₯, π¦ = = 3π₯. ππ¦ ππ₯ Thus 3 2π΄π₯ = 3π₯ β π΄ = . 2 Therefore, we obtain that the differential equation 3 π₯ 2 + 3π₯π¦ ππ₯ + 2 π₯ 2 + 4π¦ ππ¦ = 0 is exact differential equation. Furthermore, we must find πΉ such that ππΉ π₯, π¦ ππΉ π₯, π¦ 3 = π π₯, π¦ = π₯ 2 + 3π₯π¦ and = π π₯, π¦ = π₯ 2 + 4π¦. ππ₯ ππ¦ 2 From the first of these, 1 3 πΉ π₯, π¦ = π π₯, π¦ ππ₯ + π π¦ = π₯ 2 + 3π₯π¦ ππ₯ + π π¦ = π₯ 3 + π₯ 2 π¦ + π π¦ . 3 2 Then ππΉ π₯, π¦ 3 ππ π¦ = π₯2 + . ππ¦ 2 ππ¦ But we must have ππΉ π₯, π¦ 3 = π π₯, π¦ = π₯ 2 + 4π¦. ππ¦ 2 Thus 3 2 3 ππ π¦ ππ π¦ π₯ + 4π¦ = π₯ 2 + β 4π¦ = . 2 2 ππ¦ ππ¦ Thus π π¦ = 2π¦ 2 + πΆ0 , where πΆ0 is an arbitrary constant, and so 1 3 πΉ π₯, π¦ = π₯ 3 + π₯ 2 π¦ + πΆ0 . 3 2 Hence a one- parameter family of solution is πΉ π₯, π¦ = πΆ1 , or 1 3 3 2 π₯ + π₯ π¦ + 2π¦ 2 + πΆ0 = πΆ1 3 2 Combining the constsnts πΆ0 and πΆ1 we may write this solution as 1 3 3 2 π₯ + π₯ π¦ + 2π¦ 2 = πΆ 3 2 where πΆ = πΆ1 β πΆ0 is an arbitrary constant. Page 42 of 72
So, we conclude that the general solution of the exact differential equation π₯ 2 + 3π₯π¦ ππ₯ + b. 1 π₯2 1 + π¦ 2 ππ₯ + 3 π₯ 2 + 4π¦ ππ¦ = 0 is 2 π΄π₯ +1 π¦3 1 3 3 π₯ 3 + 2 π₯ 2 π¦ + 2π¦ 2 = πΆ. ππ¦ = 0. Solution: Suppose π π₯, π¦ = 1 1 π΄π₯ + 1 + 2 and π π₯, π¦ = . 2 π₯ π¦ π¦3 Then, we obtain ππ π₯, π¦ 2 ππ π₯, π¦ π΄ = β 3 and = 3. ππ¦ π¦ ππ₯ π¦ In order to make the differential equation become exact differential equation, it must be ππ π₯, π¦ ππ π₯, π¦ 2 = = β 3. ππ¦ ππ₯ π¦ Thus π΄ 2 = β 3 β π΄ = β2. 3 π¦ π¦ Therefore, we obtain that the differential equation 1 π₯2 1 + π¦ 2 ππ₯ + β2π₯+1 π¦3 ππ¦ = 0 is exact differential equation. Furthermore, we must find πΉ such that ππΉ π₯, π¦ 1 1 ππΉ π₯, π¦ β2π₯ + 1 = π π₯, π¦ = 2 + 2 and = π π₯, π¦ = . ππ₯ π₯ π¦ ππ¦ π¦3 From the first of these, 1 1 1 π₯ πΉ π₯, π¦ = π π₯, π¦ ππ₯ + π π¦ = + 2 ππ₯ + π π¦ = β + 2 + π π¦ . 2 π₯ π¦ π₯ π¦ Then ππΉ π₯, π¦ 2π₯ ππ π¦ =β 3+ . ππ¦ π¦ ππ¦ But we must have ππΉ π₯, π¦ β2π₯ + 1 = π π₯, π¦ = . ππ¦ π¦3 Thus β2π₯ + 1 2π₯ ππ π¦ 1 ππ π¦ =β 3+ β 3= . 3 π¦ π¦ ππ¦ π¦ ππ¦ 1 Thus π π¦ = β 2π¦ 2 + πΆ0 , where πΆ0 is an arbitrary constant, and so 1 π₯ 1 πΉ π₯, π¦ = β + 2 β 2 + πΆ0 . π₯ π¦ 2π¦ Hence a one- parameter family of solution is πΉ π₯, π¦ = πΆ1 , or 1 π₯ 1 β + 2 β 2 + πΆ0 = πΆ1 π₯ π¦ 2π¦ Combining the constsnts πΆ0 and πΆ1 we may write this solution as Page 43 of 72
1 π₯ 1 β + 2β 2=πΆ π₯ π¦ 2π¦ where πΆ = πΆ1 β πΆ0 is an arbitrary constant. So, we conclude that the general solution of the exact differential equation 1 π₯2 1 + π¦ 2 ππ₯ + β2π₯+1 π¦3 1 π₯ 1 ππ¦ = 0 is β π₯ + π¦ 2 β 2π¦ 2 = πΆ. Solution of Exercise 13 (Grouping Method) ( Solve using grouping method) 1. 1 π₯ π¦ ππ¦ β π₯ 2 ππ₯ = 0. Solution: 1 π¦ From π₯ ππ¦ β π₯ 2 ππ₯ = 0 we obtain π So π¦ =π πΆ . π₯ π¦ =πΆ π₯ 1 π¦ is the general solution of the differential equation π₯ ππ¦ β π₯ 2 ππ₯ = 0. ππ¦ 2. 2π₯π¦ ππ₯ + π¦ 2 β 2π₯ = 0. Solution: ππ¦ 2π₯π¦ + π¦ 2 β 2π₯ = 0 ππ₯ ππ¦ β 2π₯π¦ = 2π₯ β π¦ 2 ππ₯ β 2π₯ β π¦ 2 ππ₯ β 2π₯π¦ππ¦ = 0 From 2π₯ β π¦ 2 ππ₯ β 2π₯π¦ππ¦ = 0, we group the term as follows 2π₯ππ₯ β (π¦ 2 ππ₯ + 2π₯π¦ππ¦) = 0. Thus π π₯ 2 β π π₯π¦ 2 = π πΆ . So, π₯ 2 β π₯π¦ = πΆ ππ¦ is the general solution of the differential equation 2π₯π¦ ππ₯ + π¦ 2 β 2π₯ = 0. 3. 2 π¦ + 1 π π₯ ππ₯ + 2 π π₯ β 2π¦ ππ¦ = 0. Solution: From 2 π¦ + 1 π π₯ ππ₯ + 2 π π₯ β 2π¦ ππ¦ = 0, we group the term as follows 2π¦π π₯ ππ₯ + 2π π₯ ππ¦ + 2π π₯ ππ₯ β 4π¦ππ¦ = 0. Thus π 2π¦π π₯ + π 2π π₯ β π 2π¦ 2 = π πΆ . Page 44 of 72
Therefore, 2π¦π π₯ + 2π π₯ β 2π¦ 2 = πΆ is the general solution of the differential equation 2 π¦ + 1 π π₯ ππ₯ + 2 π π₯ β 2π¦ ππ¦ = 0. 4. 2π₯π¦ + 6π₯ ππ₯ + π₯ 2 + 4π¦ 3 ππ¦ = 0. Solution: From 2π₯π¦ + 6π₯ ππ₯ + π₯ 2 + 4π¦ 3 ππ¦ = 0, we group the term as follows 2π₯π¦ππ₯ + π₯ 2 ππ¦ + 6π₯ππ₯ + 4π¦ 3 ππ¦ = 0. Thus π π₯ 2 π¦ + π 3π₯ 2 + π π¦ 4 = π πΆ . So, π₯ 2 π¦ + 3π₯ 2 + π¦ 4 = πΆ is the general solution of the differential equation 2π₯π¦ + 6π₯ ππ₯ + π₯ 2 + 4π¦ 3 ππ¦ = 0. Solution of Quiz 1. Which of the following differential equations can be made exact by multiplying by π₯ 2 ? ππ¦ 2 (a) ππ₯ + π₯ π¦ = 4. Solution: ππ¦ 2 ππ¦ 2 2 + π¦=4β = 4 β π¦ β 4 β π¦ ππ₯ β ππ¦ = 0 ππ₯ π₯ ππ₯ π₯ π₯ By multiplying both sides by π₯ 2 , we obtain 4π₯ 2 β 2π₯π¦ ππ₯ + βπ₯ 2 ππ¦ = 0. Here π π₯, π¦ = 4π₯ 2 β 2π₯π¦, π π₯, π¦ = βπ₯ 2 , ππ π₯, π¦ ππ π₯, π¦ = β2π₯, = β2π₯ ππ¦ ππ₯ Since ππ π₯, π¦ ππ π₯, π¦ = = β2π₯ ππ¦ ππ₯ we can conclude that the differential equation 3π₯ + 2π¦ ππ₯ + 2π₯ + π¦ ππ¦ = 0 is exact differential equation. In the other word, the differential equation ππ¦ 2 + π¦ = 4 can be made exact by multiplying by π₯ 2 . ππ₯ π₯ ππ¦ (b) π₯ ππ₯ + 3π¦ = π₯ 2 . Solution: ππ¦ ππ¦ π₯ + 3π¦ = π₯ 2 β π₯ = π₯ 2 β 3π¦ β π₯ 2 β 3π¦ ππ₯ + βπ₯ ππ¦ = 0. ππ₯ ππ₯ By multiplying both sides by π₯ 2 , we obtain π₯ 4 β 3π₯ 2 π¦ ππ₯ + βπ₯ 3 ππ¦ = 0. Page 45 of 72
Here π π₯, π¦ = π₯ 4 β 3π₯ 2 π¦, π π₯, π¦ = βπ₯ 3 , ππ π₯, π¦ ππ π₯, π¦ = β3π₯ 2 , = β3π₯ 2 . ππ¦ ππ₯ Since ππ π₯, π¦ ππ π₯, π¦ = = β3π₯ 2 ππ¦ ππ₯ we can conclude that the differential equation π₯ 4 β 3π₯ 2 π¦ ππ₯ + βπ₯ 3 ππ¦ = 0 is exact differential equation. In the other ππ¦ word, the differential equation π₯ ππ₯ + 3π¦ = π₯ 2 can be made exact by multiplying by π₯ 2 . 1 ππ¦ 1 (c) π₯ ππ₯ β π₯ 2 π¦ = π₯. Solution: 1 ππ¦ 1 1 ππ¦ 1 1 1 β 2π¦ =π₯ β = π₯ + 2 π¦ β π₯ + 2 π¦ ππ₯ + β ππ¦ = 0. π₯ ππ₯ π₯ π₯ ππ₯ π₯ π₯ π₯ 2 By multiplying both sides by π₯ , we obtain π₯ 3 + π¦ ππ₯ + βπ₯ ππ¦ = 0. Here π π₯, π¦ = π₯ 3 + π¦, π π₯, π¦ = βπ₯, ππ π₯, π¦ ππ π₯, π¦ = 1, = β1. ππ¦ ππ₯ Since ππ π₯, π¦ ππ π₯, π¦ =1β = β1 ππ¦ ππ₯ we can conclude that the differential equation π₯ 3 + π¦ ππ₯ + βπ₯ ππ¦ = 0 is not exact differential equation. In the other 1 ππ¦ 1 word, the differential equation π₯ ππ₯ β π₯ 2 π¦ = π₯ can not be made exact by multiplying by π₯ 2 . 1 ππ¦ 1 (d) π₯ ππ₯ + π₯ 2 π¦ = 3. Solution: 1 ππ¦ 1 1 ππ¦ 1 1 1 + 2π¦ =3β = 3 β 2 π¦ β 3 β 2 π¦ ππ₯ + β ππ¦ = 0. π₯ ππ₯ π₯ π₯ ππ₯ π₯ π₯ π₯ By multiplying both sides by π₯ 2 , we obtain 3π₯ 2 β π¦ ππ₯ + βπ₯ ππ¦ = 0. Here π π₯, π¦ = 3π₯ 2 β π¦, π π₯, π¦ = βπ₯, ππ π₯, π¦ ππ π₯, π¦ = β1, = β1. ππ¦ ππ₯ Since ππ π₯, π¦ ππ π₯, π¦ = = β1 ππ¦ ππ₯ we can conclude that the differential equation Page 46 of 72
3π₯ 2 β π¦ ππ₯ + βπ₯ ππ¦ = 0 is exact differential equation. In the other word, 1 ππ¦ 1 the differential equation π₯ ππ₯ + π₯ 2 π¦ = 3 can be made exact by multiplying by π₯ 2. 2. Consider the differential equation 4π₯ + 3π¦ 2 ππ₯ + 2π₯π¦ππ¦ = 0. (a) Show that this equation is not exact. Proof. Here π π₯, π¦ = 4π₯ + 3π¦ 2 , π π₯, π¦ = 2π₯π¦, ππ π₯, π¦ ππ π₯, π¦ = 6π¦, = 2π¦. ππ¦ ππ₯ Since ππ π₯, π¦ ππ π₯, π¦ = 6π¦ β = 2π¦ ππ¦ ππ₯ we can conclude that the differential equation 4π₯ + 3π¦ 2 ππ₯ + 2π₯π¦ππ¦ = 0 is is not exact differential equation. β (b) Find an integrating factor of the form π₯ π , where n is a positive integer. Solution: From (a) we know that the differential equation 4π₯ + 3π¦ 2 ππ₯ + 2π₯π¦ππ¦ = 0 is not exact. But then, we can find an integrating factor π’ π₯, π¦ = π₯ π , where π is a positive integer such that the differential equation π₯ π 4π₯ + 3π¦ 2 ππ₯ + π₯ π 2π₯π¦ ππ¦ = 0 is exact. Assume that 4π₯ + 3π¦ 2 ππ₯ + π₯ π 2π₯π¦ ππ¦ = 0 is exact differential equation. Here π₯ π 4π₯ + 3π¦ 2 ππ₯ + π₯ π 2π₯π¦ ππ¦ = 0 β 4π₯ π+1 + 3π₯ π π¦ 2 ππ₯ + 2π₯ π+1 π¦ ππ¦ = 0. Let π π₯, π¦ = 4π₯ π+1 + 3π₯ π π¦ 2 and π π₯, π¦ = 2π₯ π+1 π¦. Then, we obtain ππ(π₯, π¦) ππ π₯, π¦ = 6π₯ π π¦ and = 2π + 2 π₯ π π¦ ππ¦ ππ₯ Since 4π₯ + 3π¦ 2 ππ₯ + π₯ π 2π₯π¦ ππ¦ = 0 is exact differential equation, we must obtain ππ(π₯, π¦) ππ π₯, π¦ 6π₯ π π¦ = = = 2π + 2 π₯ π π¦, ππ¦ ππ₯ that is 6π₯ π π¦ = 2π + 2 π₯ π π¦. Hence π = 2. Thus, the integrating factor of the form π₯ π , where π is a positive integer such that the differential equation π₯ π 4π₯ + 3π¦ 2 ππ₯ + π₯ π 2π₯π¦ ππ¦ = 0 is exact is π₯ 2 . (c) Multiplying the given equation through by the integrating factor found in (b) and solve the resulting exact equation. Solution: Page 47 of 72
From (b) we know that the differential equation 4π₯ 3 + 3π₯ 2 π¦ 2 ππ₯ + 2π₯ 3 π¦ ππ¦ = 0 is exact. Now, we will find the solution of this exact differential equation or in the other word we must find πΉ such that ππΉ π₯, π¦ ππΉ π₯, π¦ = π π₯, π¦ = 4π₯ 3 + 3π₯ 2 π¦ 2 and = π π₯, π¦ = 2π₯ 3 π¦. ππ₯ ππ¦ From the first of these, πΉ π₯, π¦ = = π π₯, π¦ ππ₯ + π π¦ 4π₯ 3 + 3π₯ 2 π¦ 2 ππ₯ + π π¦ = π₯4 + π₯3π¦2 + π π¦ . Then ππΉ π₯, π¦ ππ π¦ = 2π₯ 3 π¦ + ππ¦ ππ¦ But we must have ππΉ π₯, π¦ = π π₯, π¦ = 2π₯ 3 π¦. ππ¦ Thus ππ π¦ ππ π¦ β0= . ππ¦ ππ¦ Thus π π¦ = πΆ0 , where πΆ0 is an arbitrary constant, and so πΉ π₯, π¦ = π₯ 4 + π₯ 3 π¦ 2 + πΆ0 . Hence a one- parameter family of solution is πΉ π₯, π¦ = πΆ1 , or π₯ 4 + π₯ 3 π¦ 2 + πΆ0 = πΆ1 Combining the constsnts πΆ0 and πΆ1 we may write this solution as π₯ 4 + π₯ 3π¦2 = πΆ where πΆ = πΆ1 β πΆ0 is an arbitrary constant. So, we conclude that the general solution of the exact differential equation 4π₯ 3 + 3π₯ 2 π¦ 2 ππ₯ + 2π₯ 3 π¦ ππ¦ = 0 is π₯ 4 + π₯ 3 π¦ 2 = πΆ. 2π₯ 3 π¦ = 2π₯ 3 π¦ + Solution of Exercise 14 (Linear D.E) Solve the given differential equations. ππ¦ 3π¦ 1. ππ₯ + π₯ = 6π₯ 2 . Solution: Here π π₯ = and hence Therefore, we obtain π π π₯ ππ₯ π¦= π π π π₯ ππ₯ π π₯ ππ₯ =π 3 ππ₯ π₯ 3 , π π₯ = 6π₯ 2 π₯ 3 = π 3 ln π₯ = π ln π₯ = π₯ 3 . π π₯ ππ₯ Page 48 of 72
β π₯ 3π¦ = 2. π₯ 3 6π₯ 2 ππ₯ β π₯ 3π¦ = π₯ 6 + πΆ 1 β π¦ = π₯3 + 3 πΆ π₯ as the genaral solution of the given differential equation, where πΆ is an arbitrary constant. ππ¦ π₯ 4 ππ₯ + 2π₯ 3 π¦ = 1. Solution: ππ¦ ππ¦ 2π₯ 3 1 ππ¦ 2 1 π₯4 + 2π₯ 3 π¦ = 1 β + 4 π¦= 4β + π¦ = 4. ππ₯ ππ₯ π₯ π₯ ππ₯ π₯ π₯ Here 2 1 π π₯ = ,π π₯ = 4 π₯ π₯ and hence Therefore, we obtain π π π₯ ππ₯ π¦= β π₯ 2π¦ = π π₯2 π π₯ ππ₯ π π π₯ ππ₯ =π 2 ππ₯ π₯ 2 = π 2 ln π₯ = π ln π₯ = π₯ 2 . π π₯ ππ₯ 1 ππ₯ π₯4 1 β π₯ 2π¦ = β + πΆ π₯ 1 1 βπ¦ =β 3+ 2πΆ π₯ π₯ as the genaral solution of the given differential equation, where πΆ is an arbitrary constant. 3. ππ¦ + 3π¦ = 3π₯ 2 π β3π₯ . Solution: Here ππ₯ and hence Therefore, we obtain π π π₯ ππ₯ π¦= β π 3π₯ π¦ = 4. π π π₯ ππ₯ π π₯ = 3, π π₯ = 3π₯ 2 π β3π₯ π π π₯ ππ₯ =π 3ππ₯ = π 3π₯ . π π₯ ππ₯ π 3π₯ (3π₯ 2 π β3π₯ )ππ₯ β π 3π₯ π¦ = π₯ 3 + πΆ π₯3 1 β π¦ = 3π₯ + 3π₯ πΆ π π as the genaral solution of the given differential equation, where πΆ is an arbitrary constant. ππ¦ + 4π₯π¦ = 8π₯. ππ₯ Solution: Page 49 of 72
Here π π₯ = 4π₯, π π₯ = 8π₯ and hence π Therefore, we obtain π π π₯ ππ₯ π¦= 2 β π 2π₯ π¦ = 2 β π 2π₯ π¦ = 2 2 β π 2π₯ π¦ = 2 2 π π₯ ππ₯ π π π₯ ππ₯ =π 4π₯ππ₯ 2 = π 2π₯ . π π₯ ππ₯ 2 π 2π₯ (8π₯)ππ₯ π 2π₯ 2 (4π₯)ππ₯ 2 π 2π₯ π 2π₯ 2 2 β π 2π₯ π¦ = 2π 2π₯ + πΆ 1 β π¦ = 2 + 2π₯ 2 πΆ π as the genaral solution of the given differential equation, where πΆ is an arbitrary constant. Solution of Exercise 15 (Solution of Linear D.E) Consider the differential equation ππ¦ + π π₯ π¦ = 0. ππ₯ (a) Show that if π and π are two solutions of this equation and π1 and π2 are arbitrary constants, then π1 π + π2 π is also a solution of this equation. Proof: Here ππ¦ ππ¦ ππ¦ +π π₯ π¦=0β = βπ π₯ π¦ β + π π₯ ππ₯ = 0 β¦ β . ππ₯ ππ₯ π¦ By integrating both sides of (*), we obtain ππ¦ + π π₯ ππ₯ = πΆ0 π¦ β ln π¦ = β π π₯ ππ₯ + πΆ0 β |π¦| = π β π π₯ ππ₯ π πΆ0 β π¦ = πΆπ β π π₯ ππ₯ , where πΆ0 and πΆ are constants, as the general solution of the differential equati on ππ¦ + π π₯ π¦ = 0. Now, let π and π are two solutions of that equation, π and π are ππ₯ given as follow π = πΆπ π β π π₯ ππ₯ and π = πΆπ π β π π₯ ππ₯ , where πΆπ and πΆπ are constants. For π1 and π2 are arbitrary constants, we have π1 π + π2 π = π1 πΆπ π β π π₯ ππ₯ + π2 πΆπ π β π π₯ ππ₯ = π1 πΆπ + π2 πΆπ π β π π₯ ππ₯ Page 50 of 72
= πΆπ π β π π₯ ππ₯ β¦ (β), where πΆπ is constant. From (*) we can conclude that π1 π + π2 π also a solution of the ππ¦ differential equation ππ₯ + π π₯ π¦ = 0 β . (b) Extending the result of (a), show that if π1 , π2 , β¦ , ππ are π solutions of this equation and π1 , π2 , β¦ , ππ are π arbitrary constants, then π ππ ππ π=1 is also a solution of this equation. Proof: By extending the result of (a), let π1 , π2 , β¦ , ππ are π solutions of the differential ππ¦ equation ππ₯ + π π₯ π¦ = 0. Then π1 = πΆπ1 π β π π₯ ππ₯ , π2 = πΆπ2 π β π π₯ ππ₯ , β¦ , ππ = πΆππ π β where πΆπ1 , πΆπ2 , β¦ , and πΆππ are constants. For π1 , π2 , β¦ , ππ are π arbitrary constants, we obtain π π₯ ππ₯ , π ππ ππ = π1 πΆπ1 π β π π₯ ππ₯ + π2 πΆπ2 π β π π₯ ππ₯ + β― + ππ πΆππ π β π π₯ ππ₯ π=1 = π1 πΆπ1 + π2 πΆπ2 + β― + ππ πΆππ π β π π₯ ππ₯ = πΆπ π β π π₯ ππ₯ β¦ (β), where πΆπ is constant. From (*) we can conclude that the differential equation ππ¦ + π π₯ π¦ = 0 β . ππ₯ π π =1 ππ ππ is also a solution of Solution of Exercise 16 (Properties of Linear D.E) (a) Let π1 be a solution of and π2 be a solution of ππ¦ + π π₯ π¦ = π1 π₯ ππ₯ ππ¦ + π π₯ π¦ = π2 π₯ , ππ₯ Where π, π1 , and π2 are all defined on the same real interval πΌ. Prove that π1 + π2 is a solution of ππ¦ + π π₯ π¦ = π1 π₯ + π2 π₯ ππ₯ On πΌ. Proof: Since π1 is a solution of ππ¦ + π π₯ π¦ = π1 π₯ ππ₯ and π2 is a solution of ππ¦ + π π₯ π¦ = π2 π₯ , ππ₯ we have Page 51 of 72
π1 = π¦1 = π π π₯ ππ₯ π1 π₯ ππ₯ π π π₯ ππ₯ π2 π₯ π π₯ ππ₯ ππ₯ . π π Since π, π1 , and π2 are all defined on the same real interval πΌ, we know that the solution of ππ¦ + π π₯ π¦ = π1 π₯ + π2 π₯ , ππ₯ is π π π₯ ππ₯ π1 π₯ + π2 π₯ ππ₯ π¦= π π π₯ ππ₯ π π₯ ππ₯ π π1 (π₯) + π π π₯ ππ₯ π2 (π₯)ππ₯ = π π π₯ ππ₯ π π π₯ ππ₯ π1 π₯ ππ₯ π π π₯ ππ₯ π1 π₯ ππ₯ = + π π π₯ ππ₯ π π π₯ ππ₯ = π1 + π2 . β (b) Use the result of (a) to solve the equation ππ¦ + π¦ = 2 sin π₯ + 5 sin 2π₯. ππ₯ Solution: Let π π₯ = 1, π1 π₯ = 2 sin π₯, and π2 π₯ = 5 sin 2π₯. Since π π π₯ ππ₯ π1 π₯ ππ₯ π¦1 = π π π₯ ππ₯ π₯ 2π sin π₯ ππ₯ = ππ₯ π π₯ sin π₯ β π π₯ cos π₯ + πΆ1 = ππ₯ πΆ1 = sin π₯ β cos π₯ + π₯ π ππ¦ is the solution of ππ₯ + π¦ = 2 sin π₯, and π π π₯ ππ₯ π2 π₯ ππ₯ π¦2 = π π π₯ ππ₯ π₯ 5π sin 2π₯ ππ₯ = ππ₯ π₯ π sin 2π₯ β 2π π₯ cos 2π₯ + πΆ2 = ππ₯ πΆ2 = sin 2π₯ β 2 cos 2π₯ + π₯ π ππ¦ is the solution of ππ₯ + π¦ = 2 sin π₯, using the result (a), we obtain π¦ = π¦1 + π¦2 πΆ πΆ = sin π₯ β cos π₯ + π 1π₯ + sin 2π₯ β 2 cos 2π₯ + π 2π₯ πΆ1 + πΆ2 = sin π₯ β cos π₯ + sin 2π₯ β 2 cos 2π₯ + ππ₯ = sin π₯ β cos π₯ + sin 2π₯ β 2 cos 2π₯ + πΆπ βπ₯ ; πΆ = πΆ1 + πΆ2 as the solution of π π₯ ππ₯ and π2 = π¦2 = Page 52 of 72
ππ¦ + π¦ = 2 sin π₯ + 5 sin 2π₯. ππ₯ Notes: 2π π₯ sin π₯ ππ₯ = 2π π₯ sin π₯ β 2π π₯ cos π₯ ππ₯ = 2π π₯ sin π₯ β 2π π₯ cos π₯ + 2π π₯ sin π₯ ππ₯ 2π π₯ sin π₯ β 2π π₯ cos π₯ + πΆ0 2 = π π₯ sin π₯ β π π₯ cos π₯ + πΆ1 . = 5π π₯ sin 2π₯ ππ₯ = 5π π₯ sin 2π₯ β 2 5π π₯ cos 2π₯ ππ₯ = 5π π₯ sin 2π₯ β 10π π₯ cos 2π₯ + 4 5π π₯ sin 2π₯ ππ₯ 5π π₯ sin 2π₯ β 10π π₯ cos 2π₯ + πΆ0 = 5 = π π₯ sin 2π₯ β 2π π₯ cos 2π₯ + πΆ2 . Solution of Exercise 17 (Integrating Factor of Linear D.E) Solve each differential equation by first finding an integrating factor. 1. 5π₯π¦ + 4π¦ 2 + 1 ππ₯ + π₯ 2 + 2π₯π¦ ππ¦ = 0. Solution: Here π π₯, π¦ = 5π₯π¦ + 4π¦ 2 + 1 and π π₯, π¦ = π₯ 2 + 2π₯π¦ Since ππ π₯, π¦ π 5π₯π¦ + 4π¦ 2 + 1 ππ π₯, π¦ π π₯ 2 + 2π₯π¦ = = 5π₯ + 8π¦ β = = 2π₯ + 2π¦, ππ¦ ππ¦ ππ₯ ππ₯ we can say that the differential equation 5π₯π¦ + 4π¦ 2 + 1 ππ₯ + π₯ 2 + 2π₯π¦ ππ¦ = 0 is not exact (non- exact) differential equation. Furthermore, we will find an integrating factor π’(π₯, π¦) such that the differential equation π’ π₯, π¦ 5π₯π¦ + 4π¦ 2 + 1 ππ₯ + π’(π₯, π¦) π₯ 2 + 2π₯π¦ ππ¦ = 0 is exact differential equation. Let π’(π₯) is the integrating factor depends only upon π₯. Then we have Since we obtain π’ π₯ =π ππ π₯,π¦ ππ π₯,π¦ β ππ¦ ππ₯ ππ₯ π π₯,π¦ . ππ π₯, π¦ ππ π₯, π¦ β ππ₯ 5π₯ + 8π¦ β 2π₯ β 2π¦ 3(π₯ + 2π¦) 3 ππ¦ = = = , π π₯, π¦ π₯ 2 + 2π₯π¦ π₯(π₯ + 2π¦) π₯ 3 3 π’ π₯ = π π₯ ππ₯ = π ln π₯ = π₯ 3 . Multiplying the differential equation by this integrating factor, we obtain exact differential equation 5π₯ 4 π¦ + 4π₯ 3 π¦ 2 + π₯ 3 ππ₯ + π₯ 5 + 2π₯ 4 π¦ ππ¦ = 0. Page 53 of 72
Moreover, we will find a solution of this exact differential equation by using grouping method. From 5π₯ 4 π¦ + 4π₯ 3 π¦ 2 + π₯ 3 ππ₯ + π₯ 5 + 2π₯ 4 π¦ ππ¦ = 0, we group the terms as follows 5π₯ 4 π¦ππ₯ + π₯ 5 ππ¦ + 4π₯ 3 π¦ 2 ππ₯ + 2π₯ 4 π¦ππ¦ + π₯ 3 ππ₯ = 0. Thus 1 π π₯ 5π¦ + π π₯ 4π¦2 + π π₯ 4 = π πΆ . 4 So, 1 π₯ 5π¦ + π₯ 4π¦2 + π₯ 4 = πΆ 4 is the general solution of exact differential equation 5π₯ 4 π¦ + 4π₯ 3 π¦ 2 + π₯ 3 ππ₯ + π₯ 5 + 2π₯ 4 π¦ ππ¦ = 0. 2. 2π₯ + tan π¦ ππ₯ + π₯ β π₯ 2 tan π¦ ππ¦ = 0. Solution: Here π π₯, π¦ = 2π₯ + tan π¦ and π π₯, π¦ = π₯ β π₯ 2 tan π¦. Since ππ π₯, π¦ π 2π₯ + tan π¦ ππ π₯, π¦ π π₯ β π₯ 2 tan π¦ = = sec 2 π¦ β = = 1 β 2π₯ tan π¦, ππ¦ ππ¦ ππ₯ ππ₯ we can say that the differential equation 2π₯ + tan π¦ ππ₯ + π₯ β π₯ 2 tan π¦ ππ¦ = 0 is not exact (non- exact) differential equation. Furthermore, we will find an integrating factor π’(π₯, π¦) such that the differential equation π’ π₯, π¦ 2π₯ + tan π¦ ππ₯ + π’(π₯, π¦) π₯ β π₯ 2 tan π¦ ππ¦ = 0 is exact differential equation. Let π’(π¦) is the integrating factor depends only upon π¦. Then we have π’ π¦ =π β ππ π₯,π¦ ππ π₯,π¦ β ππ¦ ππ₯ ππ¦ π π₯,π¦ . Since ππ π₯, π¦ ππ π₯, π¦ β ππ₯ sec 2 π¦ β 1 + 2π₯ tan π¦ tan π¦ tan π¦ + 2π₯ ππ¦ = = = tan π¦, π π₯, π¦ 2π₯ + tan π¦ 2π₯ + tan π¦ we obtain β tan π¦ ππ¦ βsin π¦ ππ¦ cos π¦ 1 π cos π¦ cos π¦ π’ π¦ =π =π =π = π ln |cos π¦| = cos π¦. Multiplying the differential equation by this integrating factor, we obtain exact differential equation 2π₯ cos π¦ + sin π¦ ππ₯ + π₯ cos π¦ β π₯ 2 sin π¦ ππ¦ = 0. Moreover, we will find a solution of this exact differential equation by using grouping method. From 2π₯ cos π¦ + sin π¦ ππ₯ + π₯ cos π¦ β π₯ 2 sin π¦ ππ¦ = 0, we group the terms as follows 2π₯ cos π¦ ππ₯ β π₯ 2 sin π¦ ππ¦ + sin π¦ ππ₯ + π₯ cos π¦ ππ¦ = 0. Thus π π₯ 2 cos π¦ + π π₯ sin π¦ = π πΆ . So, π₯ 2 cos π¦ + π₯ sin π¦ = πΆ is the general solution of the exact differential equation 2π₯ cos π¦ + sin π¦ ππ₯ + π₯ cos π¦ β π₯ 2 sin π¦ ππ¦ = 0. 3. π¦ 2 π₯ + 1 + π¦ ππ₯ + 2π₯π¦ + 1 ππ¦ = 0. Solution: Here Page 54 of 72
π π₯, π¦ = π¦ 2 π₯ + 1 + π¦ = π₯π¦ 2 + π¦ 2 + π¦ and π π₯, π¦ = 2π₯π¦ + 1. Since ππ π₯, π¦ π(π₯π¦ 2 + π¦ 2 + π¦) ππ π₯, π¦ π(2π₯π¦ + 1) = = 2π₯π¦ + 2π¦ + 1 β = = 2π¦, ππ¦ ππ¦ ππ₯ ππ₯ we can say that the differential equation π¦ 2 π₯ + 1 + π¦ ππ₯ + 2π₯π¦ + 1 ππ¦ = 0 is not exact (non- exact) differential equation. Furthermore, we will find an integrating factor π’(π₯, π¦) such that the differential equation π’ π₯, π¦ π¦ 2 π₯ + 1 + π¦ ππ₯ + π’(π₯, π¦) 2π₯π¦ + 1 ππ¦ = 0 is exact differential equation. Let π’(π₯) is the integrating factor depends only upon π₯. Then we have Since π’ π₯ =π ππ π₯,π¦ ππ π₯,π¦ β ππ¦ ππ₯ ππ₯ π π₯,π¦ . ππ π₯, π¦ ππ π₯, π¦ β 2π₯π¦ + 2π¦ + 1 β 2π¦ ππ¦ ππ₯ = = 1, π π₯, π¦ 2π₯π¦ + 1 we obtain π’ π₯ = π ππ₯ = π π₯ . Multiplying the differential equation by this integrating factor, we obtain exact differential equation [π¦ 2 π π₯ (π₯ + 1) + π¦π π₯ ]ππ₯ + 2π₯π¦π π₯ + π π₯ ππ¦ = 0. Moreover, we will find a solution of this exact differential equation by using grouping method. From [π¦ 2 π π₯ (π₯ + 1) + π¦π π₯ ]ππ₯ + 2π₯π¦π π₯ + π π₯ ππ¦ = 0, we group the terms as follows [π¦ 2 π π₯ π₯ + 1 ππ₯ + 2π₯π¦π π₯ ππ¦] + (π¦π π₯ ππ₯ + π π₯ ππ¦) = 0. Thus π π₯π¦ 2 π π₯ + π π¦π π₯ = π πΆ . So, π₯π¦ 2 π π₯ + π¦π π₯ = πΆ is the general solution of exact differential equation [π¦ 2 π π₯ (π₯ + 1) + π¦π π₯ ]ππ₯ + 2π₯π¦π π₯ + π π₯ ππ¦ = 0. 4. 2π₯π¦ 2 + π¦ ππ₯ + 2π¦ 3 β π₯ ππ¦ = 0. Solution: Here π π₯, π¦ = 2π₯π¦ 2 + π¦ dan π π₯, π¦ = 2π¦ 3 β π₯. Since ππ π₯, π¦ π(2π₯π¦ 2 + π¦) ππ π₯, π¦ π(2π¦ 3 β π₯) = = 4π₯π¦ + 1 β = = β1, ππ¦ ππ¦ ππ₯ ππ₯ we can say that the differential equation 2π₯π¦ 2 + π¦ ππ₯ + 2π¦ 3 β π₯ ππ¦ = 0 is not exact (non- exact) differential equation. Furthermore, we will find an integrating factor π’(π₯, π¦) such that the differential equation π’ π₯, π¦ 2π₯π¦ 2 + π¦ ππ₯ + π’(π₯, π¦) 2π¦ 3 β π₯ ππ¦ = 0 is exact differential equation. Let π’(π¦) is the integrating factor depends only upon π¦. Then we have π’ π¦ = Since ππ π₯,π¦ ππ π₯,π¦ β ππ¦ ππ₯ β ππ¦ π π₯,π¦ π . ππ π₯, π¦ ππ π₯, π¦ β 4π₯π¦ + 2 2 2π₯π¦ + 1 2 ππ¦ ππ₯ = = = , π π₯, π¦ 2π₯π¦ 2 + π¦ π¦ 2π₯π¦ + 1 π¦ Page 55 of 72
we obtain 1 π¦2 1 . π¦2 Multiplying the differential equation by this integrating factor, we obtain exact differential equation 1 π₯ 2π₯ + ππ₯ + 2π¦ β 2 ππ¦ = 0. π¦ π¦ Moreover, we will find a solution of this exact differential equation by using 1 π₯ grouping method. From 2π₯ + π¦ ππ₯ + 2π¦ β π¦ 2 ππ¦ = 0, we group the terms as π’ π¦ =π β follows 2π₯ππ₯ + 2 ππ¦ π¦ = π β2 ln |π¦| = π ln = 1 π₯ ππ₯ β 2 ππ¦ + 2π¦ππ¦ = 0. π¦ π¦ Thus π π₯2 + π π₯ + π π¦2 = π πΆ . π¦ So, π₯ + π¦2 = πΆ π¦ is the general solution of the exact differential equation 1 π₯ 2π₯ + ππ₯ + 2π¦ β 2 ππ¦ = 0. π¦ π¦ π₯2 + Find an integrating factor of the form π₯ π π¦ π and solve. 4π₯π¦ 2 + 6π¦ ππ₯ + 5π₯ 2 π¦ + 8π₯ ππ¦ = 0. Solution: Here π π₯, π¦ = 4π₯π¦ 2 + 6π¦ and π π₯, π¦ = 5π₯ 2 π¦ + 8π₯. Since ππ π₯, π¦ π(4π₯π¦ 2 + 6π¦) ππ π₯, π¦ π 5π₯ 2 π¦ + 8π₯ = = 8π₯π¦ + 6 β = = 10π₯π¦ + 8, ππ¦ ππ¦ ππ₯ ππ₯ we can say that the differential equation 4π₯π¦ 2 + 6π¦ ππ₯ + 5π₯ 2 π¦ + 8π₯ ππ¦ = 0 is not exact (non- exact) differential equation. Furthermore, we will find an integrating factor of the form π₯ π π¦ π such that the differential equation 4π₯ π+1 π¦ π+2 + 6π₯ π π¦ π+1 ππ₯ + 5π₯ π+2 π¦ π+1 + 8π₯ π+1 π¦ π ππ¦ = 0 is exact differential equation. Since 4π₯ π+1 π¦ π+2 + 6π₯ π π¦ π+1 ππ₯ + 5π₯ π+2 π¦ π+1 + 8π₯ π+1 π¦ π ππ¦ = 0 is exact differential equation, π 4π₯ π+1 π¦ π+2 + 6π₯ π π¦ π+1 π 5π₯ π+2 π¦ π+1 + 8π₯ π+1 π¦ π = ππ¦ ππ₯ π+1 π+1 π π β 4 π+2 π₯ π¦ + 6 π + 1 π₯ π¦ = 5 π + 2 π₯ π+1 π¦ π+1 + 8 π + 1 π₯ π π¦ π β¦ β . From (*) we have 4 π + 2 = 5 π + 2 β 5π β 4π = β2 β¦ ββ and 6 π + 1 = 8 π + 1 β 8π β 6π = β2 β 4π β 3π = β1 β¦ βββ . Page 56 of 72
From (**) and (***) we obtain π = 2 and π = 3. Hence, π₯ 2 π¦ 3 is the integrating factor as desired. Moreover, we will find the solution of the exact differential equation 4π₯ 3 π¦ 5 + 6π₯ 2 π¦ 4 ππ₯ + 5π₯ 4 π¦ 4 + 8π₯ 3 π¦ 3 ππ¦ = 0 by using grouping method. From 4π₯ 3 π¦ 5 + 6π₯ 2 π¦ 4 ππ₯ + 5π₯ 4 π¦ 4 + 8π₯ 3 π¦ 3 ππ¦ = 0, we group the terms as follows 4π₯ 3 π¦ 5 ππ₯ + 5π₯ 4 π¦ 4 ππ¦ + 6π₯ 2 π¦ 4 ππ₯ + 8π₯ 3 π¦ 3 ππ¦ = 0. Thus π π₯ 4 π¦ 5 + π 2π₯ 3 π¦ 4 = π πΆ . So, π₯ 4 π¦ 5 + 2π₯ 3 π¦ 4 = πΆ is the general solution of the exact differential equation 4π₯ 3 π¦ 5 + 6π₯ 2 π¦ 4 ππ₯ + 5π₯ 4 π¦ 4 + 8π₯ 3 π¦ 3 ππ¦ = 0. Solution of Exercise 18 (Orthogonal and Oblique Trajectories) In exercises 1 β 4 find the orthogonal trajectories of each give family of curves. In each case sketch several numbers of the family and several of the orthogonal trajectories on the same set of axes. 1. π¦ = ππ₯ 3 . Solution: Step 1. We first find the differential equation of the given family π¦ = ππ₯ 3 β¦ i . Differentiating, we obtain ππ¦ = 3ππ₯ 2 β¦ ii . ππ₯ Eliminating the parameter π between Equations (i) and (ii), we obtain the differential equation of the family (i) in the form ππ¦ 3π¦ = β¦ iii . ππ₯ π₯ Step 2. We now find the differential equation of the orthogonal trajectories by replacing 3π¦/π₯ in iii by its negative reciprocal, obtaining ππ¦ π₯ = β β¦ iv . ππ₯ 3π¦ Step 3. We now solve the differential equation iv . Separating variables, we have 3π¦ππ¦ = βπ₯ππ₯. Integrating, we obtain the one- parameter family of solutions of i in the form 3 2 1 2 π¦ + π₯ = πΆ12 or π₯ 2 + 3π¦ 2 = πΆ 2 , 2 2 where πΆ is an arbitrary constant. Page 57 of 72
2. π¦ 2 = ππ₯. Solution: Step 1. We first find the differential equation of the given family π¦ 2 = ππ₯ β¦ i . Differentiating, we obtain ππ¦ 2π¦ = π β¦ ii . ππ₯ Eliminating the parameter π between Equations (i) and (ii), we obtain the differential equation of the family (i) in the form ππ¦ π¦ = β¦ iii . ππ₯ 2π₯ Step 2. We now find the differential equation of the orthogonal trajectories by replacing π¦/2π₯ in iii by its negative reciprocal, obtaining ππ¦ 2π₯ = β β¦ iv . ππ₯ π¦ Step 3. We now solve the differential equation iv . Separating variables, we have π¦ππ¦ = β2π₯ππ₯. Integrating, we obtain the one- parameter family of solutions of i in the form 1 2 π¦ + π₯ 2 = πΆ12 or 2π₯ 2 + π¦ 2 = πΆ 2 , 2 where πΆ is an arbitrary constant. Page 58 of 72
3. ππ₯ 2 + π¦ 2 = 1. Step 1. We first find the differential equation of the given family ππ₯ 2 + π¦ 2 = 1 β¦ i . Differentiating, we obtain ππ¦ ππ¦ ππ₯ 2ππ₯ + 2π¦ = 0 or =β β¦ ii . ππ₯ ππ₯ π¦ Eliminating the parameter π between Equations (i) and (ii), we obtain the differential equation of the family (i) in the form ππ¦ 1 β π¦2 =β β¦ iii . ππ₯ π₯π¦ Step 2. We now find the differential equation of the orthogonal trajectories by replacing β(1 β π¦ 2 )/π₯π¦ in iii by its negative reciprocal, obtaining ππ¦ π₯π¦ = β¦ iv . ππ₯ 1 β π¦ 2 Step 3. We now solve the differential equation iv . Separating variables, we have 1 β π¦ ππ¦ = π₯ππ₯. π¦ Integrating, we obtain the one- parameter family of solutions of i in the form 1 1 ln |π¦| β π¦ 2 β π₯ 2 = πΆ1 or π₯ 2 + π¦ 2 β ln(π¦ 2 ) = πΆ 2 2 where πΆ is an arbitrary constant. Page 59 of 72
4. π¦ = π ππ₯ . Step 1. We first find the differential equation of the given family π¦ = π ππ₯ β¦ i . Differentiating, we obtain ππ¦ = ππ ππ₯ β¦ ii . ππ₯ Eliminating the parameter π between Equations (i) and (ii), we obtain the differential equation of the family (i) in the form ππ¦ ln π¦ ln π¦ ππ¦ π¦ ln π¦ = π or = β¦ iii . ππ₯ π₯ ππ₯ π₯ Step 2. We now find the differential equation of the orthogonal trajectories by replacing π¦ ln |π¦| /π₯ in iii by its negative reciprocal, obtaining ππ¦ π₯ =β β¦ iv . ππ₯ π¦ ln π¦ Step 3. We now solve the differential equation iv . Separating variables, we have π¦ ln |π¦| ππ¦ = βπ₯ππ₯. Integrating, we obtain the one- parameter family of solutions of i in the form 1 2 1 1 π¦ ln |π¦| β π¦ 2 + π₯ 2 = πΆ1 or 2π¦ 2 ln |π¦| β π¦ 2 + 2π₯ 2 = πΆ, 2 4 2 where πΆ is an arbitrary constant. Page 60 of 72
*** 12. Find the value of πΎ such that the parabolas π¦ = π1 π₯ 2 + πΎ are the orthogonal trajectories of the family of ellipses π₯ 2 + 2π¦ 2 β π¦ = π2 . Solution: Step 1. We first find the differential equation of the given family π₯ 2 + 2π¦ 2 β π¦ = π2 β¦ i . Differentiating, we obtain ππ¦ ππ¦ ππ¦ 2π₯ 2π₯ + 4π¦ β = 0 or =β β¦ ii . ππ₯ ππ₯ ππ₯ 4π¦ β 1 Step 2. We now find the differential equation of the orthogonal trajectories by replacing β2π₯/(4π¦ β 1) in ii by its negative reciprocal, obtaining ππ¦ 4π¦ β 1 = β¦ iv . ππ₯ 2π₯ Step 3. We now solve the differential equation iv . Separating variables, we have 1 1 ππ¦ = ππ₯. 4π¦ β 1 2π₯ Integrating, we obtain the one- parameter family of solutions of i in the form 1 1 1 ln |4π¦ β 1| β ln |π₯| = π2 or π¦ = π1 π₯ 2 + , 4 2 4 where π2 and π1 are arbitrary constants. Hence, the value of πΎ that we want is 1/4, πΎ = 1/4. *** 13. Find the value of π such that the curves π₯ π + π¦ π = π1 are orthogonal trajectories of the family π₯ π¦= . 1 β π2 π₯ Solution: Page 61 of 72
Step 1. We first find the differential equation of the given family π¦= π₯ β¦ i . 1 β π2 π₯ Differentiating, we obtain ππ¦ 1 = β¦ ii . ππ₯ 1 β π2 π₯ 2 Eliminating the parameter π2 between Equations (i) and (ii), we obtain the differential equation of the family (i) in the form ππ¦ 1 ππ¦ π¦ 2 = or = β¦ iii . ππ₯ ππ₯ π₯ 2 π₯ 2 1β 1βπ¦ Step 2. We now find the differential equation of the orthogonal trajectories by replacing π¦ 2 /π₯ 2 in iii by its negative reciprocal, obtaining ππ¦ π₯2 = β 2 β¦ iv . ππ₯ π¦ Step 3. We now solve the differential equation iv . Separating variables, we have π¦ 2 ππ¦ = βπ₯ 2 ππ₯. Integrating, we obtain the one- parameter family of solutions of i in the form 1 3 1 3 π¦ + π₯ = π2 or π₯ 3 + π¦ 3 = π1 , 3 3 where π2 and π1 are arbitrary constant. Therefore, the value of π that we want is 3, π = 3. *** 15. Find a family of oblique trajectories that intersect the family of circles π₯ 2 + π¦ 2 = π 2 at angle 45o . Solution: Step 1. We first find the differential equation of the given family π₯ 2 + π¦2 = π2. Differentiating, we obtain ππ¦ ππ¦ π₯ 2π₯ + 2π¦ = 0 or =β β¦ i . ππ₯ ππ₯ π¦ Step 2. We replace π π₯, π¦ = βπ₯/π¦ in Equation (i) by π₯ βπ¦ +1 π¦ β π₯ π π₯, π¦ + tan πΌ = π₯ = π¦ + π₯. 1 β π π₯, π¦ tan πΌ 1+π¦ Thus the differential equation of the desired oblique trajectories is ππ¦ π¦ β π₯ = β¦ ii . ππ₯ π¦ + π₯ Step 3. We now solve the differential equation (ii). Observing that it is a homogeneous differential equation, we let π¦ = π£π₯ to obtain ππ£ π£ β 1 π£+π₯ = . ππ₯ π£ + 1 Page 62 of 72
After simplifications this becomes π£+1 ππ₯ ππ£ = β . 2 π£ +1 π₯ Integrating we obtain or 1 ln π£ 2 + 1 + arctan(π£) = β ln π₯ + πΆ1 2 ln π₯ 2 π£ 2 + 1 + 2 arctan π£ = πΆ. Replacing π£ by π¦/π₯, we obtain the family of oblique trajectories in the form π¦ ln π¦ 2 + π₯ 2 + 2 arctan = πΆ, π₯ where πΆ is an arbitrary constant. *** 16. Find a family of oblique trajectories that intersect the family of parabolas π¦ 2 = ππ₯ at angle 60o . Solution: Step 1. We first find the differential equation of the given family π¦ 2 = ππ₯ β¦ i . Differentiating, we obtain ππ¦ ππ¦ 1 2π¦ = π or = π β¦ ii . ππ₯ ππ₯ 2π¦ Eliminating the parameter π between Equation (i) and (ii), we obtain the differential equation ππ¦ π¦ = β¦ iii . ππ₯ 2π₯ of the given family of parabolas. Step 2. We replace π π₯, π¦ = π¦/2π₯ in Equation (iii) by π¦ + 3 π π₯, π¦ + tan πΌ 2π₯ = . 1 β π π₯, π¦ tan πΌ π¦ 3 1 β 2π₯ Thus the differential equation of the desired oblique trajectories is π¦ ππ¦ 2π₯ + 3 = β¦ iv . ππ₯ π¦ 3 1 β 2π₯ Step 3. We now solve the differential equation (iv). Observing that it is a homogeneous differential equation, we let π¦ = π£π₯ to obtain 1 π£+ 3 ππ£ π£+2 3 π£+π₯ = 2 = . ππ₯ 1 β 1 π£ 3 2 β π£ 3 2 After simplifications this becomes Page 63 of 72
1 3 2 2π£ 3 β 1 β 2 ππ£ = β ππ₯ . π₯ π£2 3 β π£ + 2 3 Integrating 1 1 3 ππ£ = β ππ₯ π£ π₯ π£2 π£2 β 3 3 + 2 1 2π£ 3 β 1 3 1 β ππ£ β ππ£ = β 2 π£2 3 β π£ + 2 3 2 3 3 69 π£ 2 β 2 6 π£ + 36 + 36 1 2 β β 2π£ 3 β 1 3 ππ£ β 2 3βπ£+2 3 1 2 1 2 2π£ 3 β 1 π£2 3 β π£ + 2 3 2π£ 3 β 1 π£2 3 β π£ + 2 3 ππ£ β ππ£ β 3 2 1 3 π£β 6 3 36 69 2 69 6 2 + 2 ππ£ 69 6 6 69 6π£ β 3 69 =β 1 ππ₯ π₯ ππ£ = β 2 1 ππ₯ π₯ +1 1 ππ₯ π₯ 1 3 6π£ β 3 ln π£ 2 β π£ + 2 3 β 3 arctan + ln π₯ = πΆ1 2 23 69 6 6π£ β 3 β ln |π£ 2 β π£ + 2 3| β 3 arctan + ln π₯ 2 = πΆ. 23 69 Substituting π£ = π¦/π₯, we obtain π¦ 6π₯ β 3 6 2 2 ln π¦ β π₯π¦ β 2π₯ 3 β 3 arctan = πΆ. 23 69 β Therefore, the family of oblique trajectories in the form ln π¦ 2 β π₯π¦ β 2π₯ 2 π¦ 6π₯ β 3 6 3 β 3 arctan = πΆ. 23 69 *** Solution of Exercise 19 (Problem in Mechanics (Frictional Forces)) A man is pushing a loaded sled across a level field of ice at the constant speed of 10 ft/sec. When the man in halfway across the ice field, he stops pushing and lets the loaded sled continue on. The combined weight of the sheld and its load is 80 lb 3 (π€ = 80 lb); the air resistence (in pounds) is numerically equal to 4 π£, where π£ is the velocity of the sheld (in feet per second); and the coefficient of fraction of the runners on the ice is 0.04 (π = 0.04). How far will the sheld continue to move after the man stops pushing? Solution: Page 64 of 72
3 5 ππ£ 0.04 80 β π£ = 4 2 ππ‘ 16 3 5 ππ£ β β π£= 5 4 2 ππ‘ 1 1 β ππ‘ = ππ£ 32 3 5 β π£ 5 2 1 1 β β ππ‘ = ππ£ 3 32 5 2π£ β 5 Integrating 1 1 β ππ‘ = ππ£ 3 32 5 2π£ β 5 1 2 1 3 ββ ππ‘ = π ππ£ 5 3 3 π£ β 32 2 2 5 1 2 3 32 β β π‘ = ln π£ β + πΆ1 5 3 2 5 3 3 32 β β π‘ + πΆ2 = ln π£ β 10 2 5 3 64 β πΆπ β10 π‘ + = π£(π‘) 15 We know that π£ 0 = 10 feet/sec. Thus 64 150 86 πΆ+ = βπΆ= . 15 15 15 Therefore 86 β 3 π‘ 64 π 10 + =π£ π‘ . 15 15 The shled will stop when π£ π‘ = 0. 3 32 10 32 π β10 π‘ = β β β ln β =π‘ 43 3 43 Integrating this we obtain 172 β 3 π‘ 64 π₯ π‘ =β π 10 + π‘ + πΆ4 9 15 We know that π₯ 0 = 0. Thus 172 = πΆ4 . 9 Therefore 172 β 3 π‘ 64 172 π₯ π‘ =β π 10 + π‘ + 9 15 9 and 10 32 π₯ π‘ = β ln β β 9.09 feet. 3 43 Therefore, in 9.09 feet the sheld will continue to move after the man stops pushing. *** Page 65 of 72
Solution of Exercise 20 (Rate Problems (Rate of Growth and Decay and Population) Growth) 5. Assume that the population of a certain city increases at a rate proportional to the number of its inhabitants at any time. If the population doubles in 40 years, in how many years will it triple? Solution: Let π₯ be the number of individuals in at time π‘. We know that the population of a certain city increases at a rate proportional to the number of its inhabitants at any time. Hence, we are led to the differential equation ππ₯ = ππ₯ β¦ i , ππ‘ where π is a constant of proportionality. The population π₯ is positive and is increasing and hence ππ₯/ππ‘ > 0. Therefore, from (i), we must have π > 0. Now, suppose that at timeπ‘0 = 0 the population is π₯0 . Then, in addition to the differential equation (i), we have the initial condition π₯ π‘0 = π₯(0) = π₯0 β¦ ii . The differential equation (i) is separable. Separating variables, integrating, and simplifying, we obtain π₯ = πΆπ ππ‘ . Applying the initial condition, π₯ = π₯0 at π‘ = π‘0 = 0, to this, we have π₯0 = πΆπ ππ‘0 = πΆ From this we at once find πΆ = π₯0 π βππ‘0 and hence we obtain the unique solution π₯ = π₯0 π π π‘βπ‘0 of the differential equation (i), which satisfies the initial condition (ii). Now, when π‘ = 40, we have π₯ = 2π₯0 . Hence, we obtain ln 2 2π₯0 = π₯0 π 40π β 2 = π 40π β = π. 40 If we let π₯ = 3π₯0 , then we obtain ln 2 ln 2 ln 2 ln 3 3π₯0 = π₯0 π 40 π‘ β 3 = π 40 π‘ β ln 3 = π‘ β π‘ = 40 β 63.40 40 ln 2 Therefore, the population will triple in about 63.40 years. 6. The population of the city of Bingville increases at a rate proportional to the numbers of its inhabitants present ant any time π‘. If the population of Bingville was 30,000 in 1970 and 35,000 in 1980, what will be the population of Bingville in 1990? Solution: According to the formula in the exercise 5, we have π₯ = π₯0 π π π‘βπ‘0 . Hence we obtain 35,000 1 7 π₯ 1980 = 30,000π π 1980 β1970 β = π 10π β ln = π. 30,000 10 6 Therefore, the population of Bingville in 1990 is Page 66 of 72
π₯ 1990 = 30,000π 2 ln *** 7 6 β 40,833. 9. The human population π₯ of a certain island satisfies the logistic law ππ₯ = ππ₯ β ππ₯ 2 β¦ (i) ππ‘ with π = 0.03 = 3 10 β2 , π = 3 10 β8 , and time π‘ measured in years. (a) If the population in 1980 is 200,000, find a formula for the population in future years. Solution: We must solve the separable differential equation (i) subject to the initial solution π₯ 1980 = 200,000 β¦ ii . Separating variable in (ii), we obtain ππ₯ ππ₯ = 0.03π₯ β 3 10 β8 π₯ 2 β = ππ‘ β2 ππ‘ 3 10 π₯ β 3 10 β8 π₯ 2 and hence ππ₯ = ππ‘. 3 10 β2 π₯ 1 β (10 β6 π₯] Using partial fractions, this becomes 100 1 10 β6 + = ππ‘. 3 π₯ 1 β 10 β6 π₯ Integrating, assuming 0 < π₯ < 106 , we obtain 100 ln π₯ β ln 1 β 10 β6 = π‘ + πΆ1 3 and hence π₯ 3 ln = π‘ + πΆ2 . 1 β 10 β6 100 Thus we find 3π‘ π₯ = πΆπ 100 . β6 1 β 10 Solving this for π₯, we finally obtain 3π‘ π₯= πΆπ 100 3π‘ β6 πΆπ 100 β¦ iii . 1 + 10 Now, applying the initial conditions (ii) to this, we have πΆπ 59.4 2 10 5 = , 1 + 10 β6 πΆπ 59.4 from which we obtain 2 10 5 10 6 πΆ = 59.4 = . π 1 β 2 10 5 10 β6 4π 59.4 Substituting this value for π back into (iii) and simplifying, we obtain the solution in the form Page 67 of 72
10 π₯= 6 3π‘ β¦ iv . 1 + 4π 59.4β100 This gives the population π₯ as a function of time for π‘ > 1980. Therefore, we can conclude that the formula for the population in future years is given by 10 6 π₯= 3π‘ . 59.4β 100 1 + 4π (b) According to the formula of part (a), what will be the population in the year 2000? Solution: Let π‘ = 2000 in (iv) and we obtain 10 6 π₯= β 312,965. 1 + 4π β0.6 Therefore, the population in the year 2000 is 312,965 people. (c) What is the limiting value of the population as π‘ β β? Solution: 10 lim π₯ = lim π‘ββ π‘ββ? 1+ 6 3π‘ 4π 59.4β100 = 10 6 = 1,000,000. *** Solution of Exercise 21 (Mixture Problem) 18. A large tank initially contains 200 gal of brine in which 15lb of salt is dissolved. Starting at π‘ = 0, brine containing 4lb of salt per gallon flows into the tank at the rate of 3.5 gal/min. The mixture is kept uniform by stirring and the well- stirred mixture leaves the tank at the rate of 4 gal/min. (a) How much salt is in the tank at the end of one hour? Solution: Let π₯ denotes the amount of salt in the tank at time π‘. We apply the basic equation ππ₯ = πΌπ β πππ. ππ‘ The brine flows in at the rate of 3.5 gal/min, and each gallon contains 4 lb of salt. Thus πΌπ =(4 lb/gal)(3.5 gal/min)=14 lb/min, and πππ = (πΆlb/gal)(4 gal/min)=4πΆ lb/min, where πΆ lb/gal denotes the concentration. But here, since the rate of outflow is different from that of inflow, the concentration is not quite so simple. At time π‘ = 0, the tank contains 200 gal of brine. Since brine flows in at the rate of 3.5 gal/min but flows out at the faster rate of 4 gal/min, there is a gross gain of Page 68 of 72
3.5 β 4 = β0.5 gal/min of brine in the tank. Thus at the end of π‘ minutes the amount of brine in the tank is 200 β 0.5π‘ gal. Hence the concentration at time π‘ minutes is π₯ lb/gal, 200 β 0.5π‘ and so 4π₯ πππ = lb/min. 200 β 0.5π‘ Thus the differential equation becomes ππ₯ 4π₯ 8π₯ = 14 β = 14 β β¦ i . ππ‘ 200 β 0.5π‘ 400 β π‘ Since there was initially 15 lb of salt in the tank, we have the initial condition π₯ 0 = 15 β¦ ii . The differential equation (i) is not separable but it is linear. Putting it in standard form, we obtain ππ₯ 8 + π₯ = 14, ππ‘ 400 β π‘ we find the integrating factor 8 1 exp ππ‘ = 400 β π‘ β8 = . 400 β π‘ 400 β π‘ 8 Multiplying through by this we have 1 ππ₯ 8 14 + π₯= 8 9 400 β π‘ ππ‘ 400 β π‘ 400 β π‘ 8 or π 1 14 π₯ = . 8 ππ‘ 400 β π‘ 400 β π‘ 8 Thus 1 2 π₯= +πΆ 8 400 β π‘ 400 β π‘ 7 or π₯ = 400 β π‘ + 400 β π‘ 8 πΆ. Applying condition (ii), π₯ = 15 at π‘ = 0, we obtain 15 = 400 + 4008 πΆ or 375 πΆ=β . 4008 Thus the amount of salt at any time π‘ > 0 is given by 400 β π‘ 8 π₯ = 400 β π‘ β 375 400 Therefore, the amount of salt in the tank at the end of one hour (π‘ = 60) is 340 8 π₯ 60 = 340 β 375 = 340 β 375 β 238 lb. 400 (b) How much salt is in the tank when the tank contains only 50 gal of brine? Page 69 of 72
Solution: According to part (a), we know that at the end of π‘ minutes the amount of brine in the tank is 200 β 0.5π‘ gal. Thus we obtain 200 β 0.5π‘ = 50 and hence π‘ = 300. Therefore, the amount of salt in the tank when the tank contains only 50 gal of brine is 100 8 π₯ 300 = 100 β 375 β 100 lb. 400 Solution of Exercise 22 (Reduction of Order) 1. Prove that the differential equation π 2 π¦ ππ¦ β β 2π¦ = 0 ππ₯ 2 ππ₯ has two solutions of the form π¦ = π ππ₯ , with π is a constant. Proof: Let π¦ = π ππ₯ be the solution of the given differential equation. Then we obtain π 2 π ππ₯ π π ππ₯ β β 2π ππ₯ = 0 ππ₯ 2 ππ₯ π ππ ππ₯ β β ππ ππ₯ β 2π ππ₯ = 0 ππ₯ β π2 π ππ₯ β ππ ππ₯ β 2π ππ₯ = 0 β π ππ₯ π2 β π β 2 = 0 β π ππ₯ π + 1 π β 2 = 0 β π ππ₯ = 0 βπ + 1 = 0βπ β 2 = 0. Hence there are two values for π, i.e., π = β1 or π = 2. Therefore, we can conclude that the given differential equation has two solutions of the form π¦ = π ππ₯ , i.e., π¦ = π βπ₯ or π¦ = π 2π₯ . β 2. If a differential equation has general solution of the form π¦ = π1 π 2π₯ + π2 π β3π₯ , then determine that differential equation. Solution: Consider the second order homogeneous differential equation with constant coefficient π2 π¦ ππ¦ + π + ππ¦ = 0. ππ₯ 2 ππ₯ ππ₯ Suppose that π¦ = π , π β β is the solution of that differential equation (where π will be determined). Then we will have the characteristic equation π 2 + ππ + π = 0. Page 70 of 72
Thus, in order to π¦ = π ππ₯ becomes the solution for that differential equation, π has to satisfies that characteristic equation. One of the possibilities for the roots of the characteristic equation above is if the characteristic equation has two different roots, say π1 , π2 β β, then the general solution for the differential equation above is π¦ π₯ = π1 π π1 π₯ + π2 π π2 π₯ . Since we have that the general solution of a differential equation is π¦ = π1 π 2π₯ + π2 π β3π₯ we obtai that the roots of a characteristic equation are π1 = 2 or π2 = β3. Moreover, the characteristic equation that we want is π β 2 π + 3 = 0 β π 2 + π β 6 = 0. Hence we obtain π = 1 and π = β6. Therefore, the diferential equation that has the general solution π¦ = π1 π 2π₯ + π2 π β3π₯ is π 2 π¦ ππ¦ + β 6π¦ = 0. ππ₯ 2 ππ₯ 3. Given that π¦1 π₯ = π₯ 2 is one of the solutions of the differential equation π 2 π¦ 1 ππ¦ 4 + β π¦ = 0 β¦ (i) ππ₯ 2 π₯ ππ₯ π₯ 2 Determine a general solution of that differential equation. Solution: Observe that π¦1 π₯ = π₯ 2 does satisfy the given differential equation, i.e., π2 π₯ 2 1 π π₯2 4 1 + β 2 π₯ 2 = 2 + 2π₯ β 4 = 0. 2 ππ₯ π₯ ππ₯ π₯ π₯ Let π¦2 (π₯) = π₯ 2 π£(π₯). We have π π¦2 π₯ π π£ π₯ = π₯2 + 2π₯π£ π₯ ππ₯ ππ₯ and π 2 π¦2 π₯ π2 π£ π₯ π π£ π₯ 2 =π₯ + 4π₯ + 2π£ π₯ . 2 2 ππ₯ ππ₯ ππ₯ Substituting the expression for π¦, ππ¦/ππ₯, and π 2 π¦/ππ₯ 2 into Equation (i), we obtain π2 π£ π₯ π π£ π₯ 1 2π π£ π₯ 4 2 π₯ + 4π₯ + 2π£ π₯ + π₯ + 2π₯π£ π₯ β 2 π₯ 2 π£ π₯ = 0 2 ππ₯ ππ₯ π₯ ππ₯ π₯ or π2 π£ π₯ π π£ π₯ 2 π₯ + 5π₯ = 0. ππ₯ 2 ππ₯ Letting π€ = ππ£/ππ₯ we obtain the first- order homogeneous linear equation ππ€ π₯2 + 5π₯π€ = 0. ππ₯ Treating this as a separable equation, we obtain 1 5 1 1 ππ€ = β ππ₯ β ππ€ + ππ₯ = 0. π€ π₯ 5π€ π₯ Integrating this, we obtain the general solution 1 1 π55 π4 ln |π€| + ln |π₯| = π6 β ln π€ 5 π₯ = ln π5 β π€ = 5 = 5 5 π₯ π₯ where π6 , π5 , and π4 are constants. Page 71 of 72
Afterwards, we choose π4 = 1, we recall that ππ£/ππ₯ = π€ and integrate to obtain the function π£ given by 1 π£ π₯ = β 4. 4π₯ Now forming π¦2 π₯ = π¦1 π₯ π£(π₯), where π¦1 (π₯) denotes the known solution, we obtain the function π¦2 defined by 1 1 1 π¦2 π₯ = π₯ 2 β 4 = β 2 = β π₯ β2 . 4π₯ 4π₯ 4 Therefore, the general solution of Equation (i) is 1 π¦(π₯) = π1 π₯ 2 + π3 β π₯ β2 = π1 π₯ 2 + π2 π₯ β2 4 where π1 , π2 , and π3 are constants. Page 72 of 72 View publication stats