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Solution Manual for Elementary Differential Equations with Boundary Value Problems, 2nd Edition

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    E LEMENTARY D IFFERENTIAL E QUATIONS WITH B OUNDARY V ALUE P ROBLEMS S ECOND E DITION Werner Kohler Virginia Tech Lee Johnson Virginia Tech ’ S S OLUTIONS M ANUAL J EREMY B OURDON McMaster University T ERRI B OURDON Virginia Tech

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    CONTENTS Chapter 1: Introduction to 1 Differential Equations Chapter 2: First Order 5 Differential Equations Chapter 3: Second and Higher Order 51 Linear Differential Equations Chapter 4: First Order Linear Systems 110 Chapter 5: Laplace Transforms 172 Chapter 6: Nonlinear Systems 212 Chapter 7: Numerical Methods 248 Chapter 8: Series Solutions of Linear 266 Differential Equations Chapter 9: Second Order Partial Differential 298 Equations and Fourier Series Chapter 10: First Order Partial Differential Equations 329 and the Method of Characteristics Chapter 11: Linear Two-Point Boundary 339 Value Problems

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    Chapter 1 Introduction to Differential Equations Section 1.2 1. This D.E. is of order two because the highest derivative in the equation is ′′ y . 2. Order is 1. 3. This D.E. is of order one because the highest derivative in the equation is ′ y . (Note: ( ) ′ ≠ ′′′ y y 3 ) 4. Order is 3. 5. Differentiating gives us ′ = y ke kt . Substitution yields ke e kt kt + = 2 0 . Therefore, k = − 2 . 6. ′′ − = ⇒ − = ⇒ = ± y y k e e k kt kt 0 0 1 2 . 7. Differentiating gives us ′ = − y k te k t 2 2 2 sin cos . Substitution yields − + = 2 2 2 0 2 2 k te te k t k t sin sin cos cos . Therefore, sin cos 2 2 1 0 2 te k k t − + ( ) = , and solving for k gives us k = 1 2 . 8. y ke y y ke ke t t t = ′ + = − + = − − − , , . 0 0 k can be any real number. 9 (a). y Ce t = 2 . Differentiating gives us ′ = ⋅ = y Ce t ty t 2 2 2 . Therefore, ′ − = y ty 2 0 for any value of C . 9 (b). Substituting into the differential equation yields y Ce Ce ( ) 1 1 2 = = . Using the initial condition, y Ce ( ) 1 2 = = . Solving for C , we find C e = − 2 1 . 10. ′′′ = ′′ = + = + + = + + + y y t c y t c t c y t c t c t c 2 2 3 2 1 1 2 1 2 3 1 2 2 3 . , , . Order = 3 3 arbitrary constants 11 (a). y C t C t = + 1 2 2 2 sin cos . Differentiating gives us ′ = − y C t C t 2 2 2 2 1 2 cos sin and ′′ = − − = − + = − y C t C t C t C t y 4 2 4 2 4 2 2 4 1 2 1 2 sin cos ( sin cos ) . Therefore, ′′ + = − + = y y y y 4 4 4 0 and thus y t C t C t ( ) sin cos = + 1 2 2 2 is a solution of the D.E. ′′ + = y y 4 0 . 11 (b). y C C C π 4 1 2 1 1 0 3 ( ) = + = = ( ) ( ) and ′( ) = − = − = − ⇒ = y C C C C π 4 1 2 2 2 2 0 2 1 2 2 1 ( ) ( ) . 12. y e y ky e ke k e t t t t = ′ + = − + = − = − − − − 2 8 2 2 4 0 4 4 4 4 . ( ) ∴ = = = ∴ = = . ( ) . , . k y y k y 4 0 2 4 2 0 0

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    2 • Chapter 1 Introduction to Differential Equations 13. y ct = − 1 . Differentiating gives us ′ = − − y ct 2 . Thus ′ + = − + = − = − − − y y ct c t c c t 2 2 2 2 2 2 0 ( ) . Solving this for c , we find that c c c c 2 1 0 − = − = ( ) . Therefore, c = 0 1 , . 14. y e t t = − + − sin ′ + = = y y g t y y ( ), ( ) . 0 0 ′ = + − y e t t cos ′ + = + − + = ∴ = + = − = − − y y e t e t g g t t t y y t t cos sin ( ) cos sin , ( ) 0 1 0 15. y t r = . Differentiating gives us ′ = − y rt r 1 and ′′ = − − y r r t r ( ) 1 2 . Thus t y ty y r r t rt t r r r t r r r r 2 2 2 1 2 2 1 2 2 0 ′′ − ′ + = − − + = − − + = ( ) [ ( ) ] . Solving this for r , we find that r r r r r r r ( ) ( )( ) − − + = − + = − − = 1 2 2 3 2 2 1 0 2 . Therefore, r = 1 2 , . 16. y c e c e y c e c e y c e c e y t t t t t t = + ′ = − ′′ = + = − − − 1 2 2 2 1 2 2 2 1 2 2 2 2 2 4 4 4 . , ∴ ′′ − = . y y 4 0 17. From (16), y C e C e t t = + − 1 2 2 2 , which we differentiate to get ′ = − − y C e C e t t 2 2 1 2 2 2 . Using the initial conditions, y ( ) 0 2 = and ′ = y ( ) 0 0 , we have two equations containing C 1 and C 2 : C C 1 2 2 + = and 2 2 0 1 2 C C − = . Solving these simultaneous equations gives us C C 1 2 1 = = . Thus, the solution to the initial value problem is y e e t t t = + = − 2 2 2 2 cosh( ) . 18. y c c c c c c y t e t ( ) , , ( ) 0 1 2 2 2 1 0 1 2 1 2 1 2 2 = + = − = ∴ = = = . 19. From (16), y t C e C e t t ( ) = + − 1 2 2 2 . Using the initial condition y ( ) 0 3 = , we find that C C 1 2 3 + = . From the initial condition lim ( ) t y t →∞ = 0 and the equation for y t ( ) given to us in (16), we can conclude that C 1 0 = (if C 1 0 ≠ , then lim t →∞ = ±∞ ). Therefore, C 2 3 = and y t e t ( ) = − 3 2 . 20. c c y t c c y t e t t 1 2 2 1 2 10 0 0 10 10 + = = ⇒ = ∴ = = →−∞ lim ( ) ( ) and . 21. From the graph, we can see that ′ = − y 1 and that y ( ) 1 1 = . Thus m y = ′ − = − − = − 1 1 1 2 and y y 0 1 1 = = ( ) . 22. ′ = ⇒ = + = − = ∴ = y mt y m t c y t t . . 2 1 0 0 2 0 From the graph, only at Also From the graph . ( ) . c y m m = − = − ∴ − = − ⇒ = 1 1 0 5 1 2 2 1 1 . 23. We know that this is a freefall problem, so we can begin with the generic equation for freefall situations: y t g t v t y ( ) = − + + 2 2 0 0 . The object is released from rest, so v 0 0 = . The impact time corresponds to the time at which y = 0 , so we are left with the following equation for the impact time t : 0 2 2 0 = − + g t y . Solving this for t yields t y g = 2 0 . For the velocity at the time of impact: v y gt v gt gy = ′ = − + = − = − 0 0 2 .

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    Chapter 1 Introduction to Differential Equations • 3 24. ′′ = ′ = + = = ⇒ = + x a x at v v x x at , 0 0 0 2 0 2 0 . 88 8 11 8 11 64 2 352 2 = ⇒ = = =       = a a ft t x ft ( ) / sec . , . At Section 1.3 1 (a). The equation is autonomous because ′ y depends only on y . 1 (b). Setting ′ = y 0 , we have 0 1 = − + y . Solving this for y yields the equilibrium solution: y = 1 . 2 (a). not autonomous 2 (b). no equilibrium solutions, isoclines are t = const ant. 3 (a). The equation is autonomous because ′ y depends only on y . 3 (b). Setting ′ = y 0 , we have 0 = sin y . Solving this for y yields the equilibrium solutions: y n = ± π . 4 (a). autonomous 4 (b). y y y ( ) , , − = = 1 0 0 1 . 5 (a). The equation is autonomous because ′ y does not depend explicitly on t . 5 (b). There are no equilibrium solutions because there are no points at which ′ = y 0 . 6 (a). not autonomous 6 (b). y = 0 is equilibrium solution, isoclines are hyperbolas. 7 (a). c = − 1 : Setting c = − 1 gives us − + = − y 1 1 which, solved for y , reads y = 2 . This is the isocline for c = − 1 . c = 0 : Setting c = 0 gives us − + = y 1 0 which, solved for y , reads y = 1 . This is the isocline for c = 0 . c = 1: Setting c = 1 gives us − + = y 1 1 which, solved for y , reads y = 0 , the isocline for c = 1. 8 (a). − + = − ⇒ = + y t y t 1 1 − + = ⇒ = y t y t 0 − + = ⇒ = − y t y t 1 1 9 (a). c = − 1 : Setting c = − 1 gives us y t 2 2 1 − = − which can be simplified to t y 2 2 1 − = (a hyperbola). This is the isocline for c = − 1 . c = 0 : Setting c = 0 gives us y t 2 2 0 − = which can be simplified to y t = ± . This is the isocline for c = 0 . c = 1: Setting c = 1 gives us y t 2 2 1 − = (a hyperbola). This is the isocline for c = 1.

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    4 • Chapter 1 Introduction to Differential Equations 10. f f y y y ( ) ( ) ( ) 0 2 0 2 = = ′ = − ′ > < < ′ < − ∞ < < < < ∞ y y y y y 0 0 2 0 0 2 , for for and . 11. One example that would fit these criteria is ′ = − − y y ( ) 1 2 . For this autonomous D.E., ′ = y 0 at y = 1 and ′ < y 0 for −∞ < < y 1 and 1 < < ∞ y . 12. ′ = y 1 . 13. One example that would fit these criteria is ′ = y y sin( ) 2 π . For this autonomous D.E., ′ = y 0 at y n = 2 . 14. c. 15. f. 16. a. 17. b. 18. d. 19. e.

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    Chapter 2 First Order Differential Equations Section 2.1 1. This equation is linear because it can be written in the form ′ + = y p t y g t ( ) ( ) . It is nonhomogeneous because when it is put in this form, g t ( ) ≠ 0 . 2. nonlinear 3. This equation is nonlinear because it cannot be written in the form ′ + = y p t y g t ( ) ( ) . 4. nonlinear 5. This equation is nonlinear because it cannot be written in the form ′ + = y p t y g t ( ) ( ) . 6. linear, homogeneous 7. This equation is nonlinear because it cannot be written in the form ′ + = y p t y g t ( ) ( ) . 8. nonlinear 9. This equation is linear because it can be written in the form ′ + = y p t y g t ( ) ( ) . It is nonhomogeneous because when it is put in this form, g t ( ) ≠ 0 . 10. linear, homogeneous 11 (a). Theorem 2.1 guarantees a unique solution for the interval ( , ) −∞ ∞ , since t t 2 1 + and sin( ) t are both continuous for all t and − 2 is on this interval. 11 (b). Theorem 2.1 guarantees a unique solution for the interval ( , ) −∞ ∞ , since t t 2 1 + and sin( ) t are both continuous for all t and 0 is on this interval. 11 (c). Theorem 2.1 guarantees a unique solution for the interval ( , ) −∞ ∞ , since t t 2 1 + and sin( ) t are both continuous for all t and π is on this interval. 12 (a). 2 < < ∞ t 12 (b). − < < 2 2 t 12 (c). − < < 2 2 t 12 (d). −∞ < < − t 2

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    6 • Chapter 2 First Order Differential Equations 13 (a). For this equation, p t ( ) is continuous for all t ≠ − 2 2 , and g t ( ) is continuous for all t ≠ 3. Therefore, Theorem 2.1 guarantees a unique solution for ( , ) 3 ∞ , the largest interval that includes t = 5. 13 (b). For this equation, p t ( ) is continuous for all t ≠ − 2 2 , and g t ( ) is continuous for all t ≠ 3. Therefore, Theorem 2.1 guarantees a unique solution for ( , ) − 2 2 , the largest interval that includes t = − 3 2 . 13 (c). For this equation, p t ( ) is continuous for all t ≠ − 2 2 , and g t ( ) is continuous for all t ≠ 3. Therefore, Theorem 2.1 guarantees a unique solution for ( , ) − 2 2 , the largest interval that includes t = 0. 13 (d). For this equation, p t ( ) is continuous for all t ≠ − 2 2 , and g t ( ) is continuous for all t ≠ 3. Therefore, Theorem 2.1 guarantees a unique solution for ( , ) −∞ − 2 , the largest interval that includes t = − 5 . 13 (e). For this equation, p t ( ) is continuous for all t ≠ − 2 2 , and g t ( ) is continuous for all t ≠ 3. Therefore, Theorem 2.1 guarantees a unique solution for ( , ) − 2 2 , the largest interval that includes t = 3 2 . 14. ln ln t t t t t t + − = − − + 1 1 2 2 2 undefined at t = 0 2 , . 14 (a). 2 < < ∞ t . 14 (b). 0 2 < < t . 14 (c). −∞ < < t 0 . 14 (d). −∞ < < t 0 . 15. y t e t ( ) = 3 2 . Differentiating gives us ′ = = y e t ty t 3 2 2 2 ( ) . Substituting these values into the given equation yields 2 0 ty p t y + = ( ) . Solving this for p t ( ) , we find that p t t ( ) = − 2 . Putting t = 0 into the equation for y gives us y 0 3 = . 16(a). y Ct r = ′ = − y Crt r 1 2 6 0 ty y ′ − = ∴ − = − = ⇒ − = ⇒ − = ⇒ = ( ) ( ) 2 6 2 6 0 2 6 0 2 6 0 3 Crt Ct r Ct r y r r r r r y C C C C r ( ) ( ) ( ) − = − = ⇒ ≠ ∴ − = ⇒ = − 2 2 8 0 2 8 1 3 16 (b). −∞ < < t 0 since p t t ( ) = − 3 16 (c). y t t t ( ) , = − − ∞ < < ∞ 3 .

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    Chapter 2 First Order Differential Equations • 7 17. y t ( ) = 0 satisfies all of these conditions. Section 2.2 1 (a). First, we will integrate p t ( ) = 3 to find P t t ( ) = 3 . The general solution, then, is y t Ce Ce P t t ( ) ( ) = = − − 3 . 1 (b). y C ( ) 0 3 = = − . Therefore, the solution to the initial value problem is y e t = − − 3 3 . 2 (a). ′ − = y y 1 2 0 , ( ) , e y y Ce t t − ′ = = 2 2 0 . 2 (b). y Ce C e ( ) , − = = = − 1 2 2 1 2 1 2 y t e t ( ) ( ) = + 2 1 2 3 (a). We can rewrite this equation into the conventional form: ′ − = y ty 2 0 . Then we will integrate p t t ( ) = − 2 to find P t t ( ) = − 2 . The general solution, then, is y t Ce Ce P t t ( ) ( ) = = − 2 . 3 (b). y Ce ( ) 1 3 = = . Solving for C yields C e = − 3 1 . Therefore, the solution to the initial value problem is y t e e e t t ( ) ( ) = = − − 3 3 1 1 2 2 . 4 (a). ty y y t y ′ − = ⇒ ′ − = 4 0 4 0 . − = − = − ∴ = ∫ 4 4 1 4 4 t dt t t t ln ln( ) μ 1 4 0 4 5 4 t y t y t y ′ − = ′ = − ( ) y Ct = 4 . 4 (b). y C y t t ( ) ( ) 1 1 4 = = ∴ = . 5 (a). For this D.E., p t ( ) = − 3 . Integrating gives us P t t ( ) = − 3 . An integrating factor is, then, μ ( ) t e t = − 3 . Multiplying the D.E. by μ ( ) t , we obtain e y e y e y e t t t t − − − − ′ − = ′ = 3 3 3 3 3 6 ( ) . Integrating both sides yields e y e C t t − − = − + 3 3 2 . Solving for y gives us y Ce t = − + 2 3 . 5 (b). y C ( ) 0 1 2 = = − + . Solving for C yields C = 3 , and thus our final solution is y e t = − + 2 3 3 . 6 (a). ′ − = = y y e y t 2 0 3 3 , ( ) . ( ) e y e e y e C y e Ce t t t t t t − − ′ = ⇒ = + ⇒ = + 2 2 3 2 6 (b). y C C y e e t t ( ) , 0 1 3 2 2 3 2 = + = ⇒ = = + . 7 (a). Putting this D.E. in the conventional form, we have ′ + = y y e t 3 2 1 2 . For this D.E., p t ( ) = 3 2 . Integrating gives us P t t ( ) = 3 2 . An integrating factor is, then, μ ( ) t e t = 3 2 . Multiplying the D.E. by μ ( ) t , we obtain e y e y e y e t t t t 3 2 3 2 3 2 5 2 3 2 1 2 ′ + = ′ = ( ) . Integrating both sides yields e y e C t t 3 2 5 2 1 5 = + . Solving for y gives us y e Ce t t = + − 1 5 3 2 .

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    8 • Chapter 2 First Order Differential Equations 7 (b). y C ( ) 0 0 1 5 = = + . Solving for C yields C = − 1 5 , and thus our final solution is y e e t t = − − 1 5 1 5 3 2 . 8 (a). ′ + = + = ∴ ′ = + − y y e t y e y e t t t t 1 2 2 0 2 2 2 cos( ), ( ) ( ) cos π e y e t C y e t Ce t t t t = + + ⇒ = + + − − sin sin 2 1 2 . 8 (b). y Ce C e y e t e t t ( ) ; sin ( ) π π π π 2 1 0 1 2 2 2 2 = + = ⇒ = − = + − − − − − . 9 (a). Putting this D.E. in the conventional form, we have ′ + = − y t y t cos( ) cos( ) 2 3 2 . For this D.E., p t t ( ) cos( ) = 2 . Integrating gives us P t t ( ) sin( ) = 2 . An integrating factor is, then, μ ( ) sin( ) t e t = 2 . Multiplying the D.E. by μ ( ) t , we obtain e y t e y e y t e t t t t sin( ) sin( ) sin( ) sin( ) cos( ) ( ) cos( ) 2 2 2 2 2 3 2 ′ + = ′ = − . Integrating both sides yields e y e C t t sin( ) sin( ) 2 2 3 = − + . Solving for y gives us y Ce t = − + − 3 2 sin( ) . 9 (b). y C ( ) 0 4 3 = − = − + . Solving for C yields C = − 1 , and thus our final solution is y e t = − − − 3 2 sin( ) . 10 (a). ′ + = + + − = ′ = + + − y y e t y e e y e te e t t t t t 2 1 1 2 2 2 , ( ) , ( ) ye e te e e C y e t Ce t t t t t t t 2 2 2 2 2 1 2 1 4 1 2 2 1 4 = + − + + ⇒ = + + + − − . 10 (b). y e Ce e C e ( ) − = − + + = ⇒ = − 1 1 2 1 4 1 4 2 2 ∴ = + + + − − + ( ) y e t e t t 2 1 4 1 4 2 1 . 11. We can rewrite this equation into the conventional form: ′ + = y t y 4 0 . Then we will integrate p t t ( ) = 4 to find P t t t ( ) ln ln = = 4 4 . The general solution, then, is y t Ce Ce Ce Ct P t t t ( ) ( ) ln ln = = = = − − − − 4 4 4 . 12. μ = − ∴ = − − exp( cos ) ( ) ( cos ) t t y t Ce t t . 13. First, we will integrate p t t ( ) cos( ) = − 2 2 to find P t t ( ) sin( ) = − 2 . The general solution, then, is y t Ce Ce P t t ( ) ( ) sin( ) = = − 2 . 14. (( ) ) t y 2 1 0 + ′ = y C t = + 2 1 .

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    Chapter 2 First Order Differential Equations • 9 15. We can rewrite this equation into the conventional form: ′ − + = y t y 3 1 0 2 ( ) . Then we will integrate p t t ( ) ( ) = − + 3 1 2 to find P t t t ( ) = − − 3 3 . The general solution, then, is y t Ce Ce P t t t ( ) ( ) = = − + 3 3 . 16. ′ + = ∴ = − − − − ∫ y e y e dt e t t t 0 ( ) − ′ = − e y e t 0 y Ce e t = − . 17. For this D.E., p t ( ) = 2 . Integrating gives us P t t ( ) = 2 . An integrating factor is, then, μ ( ) t e t = 2 . Multiplying the D.E. by μ ( ) t , we obtain e y e y e y e t t t t 2 2 2 2 2 ′ + = ′ = ( ) . Integrating both sides yields e y e C t t 2 2 1 2 = + . Therefore, the general solution is y t Ce t ( ) = + − 1 2 2 . 18. ′ + = ⇒ ′ = ⇒ = + ⇒ = + − − − y y e e y e e y e C y e Ce t t t t t t t 2 2 2 2 ( ) . 19. For this D.E., p t ( ) = 2 . Integrating gives us P t t ( ) = 2 . An integrating factor is, then, μ ( ) t e t = 2 . Multiplying the D.E. by μ ( ) t , we obtain e y e y e y t t t 2 2 2 2 1 ′ + = ′ = ( ) . Integrating both sides yields e y t C t 2 = + . Therefore, the general solution is y t te Ce t t ( ) = + − − 2 2 . 20. ′ + = ⇒ ′ = ⇒ = + ⇒ = + − y ty t e y te e y e C y Ce t t t t t 2 1 2 1 2 2 2 2 2 2 ( ) . 21. Putting this equation into the conventional form gives us ′ + = y t y t 2 . For this D.E., p t t ( ) = 2 . Integrating gives us P t t ( ) ln = 2 . An integrating factor is, then, μ ( ) ln t e t t = = 2 2 . Multiplying the D.E. by μ ( ) t , we obtain t y ty t y t 2 2 3 2 ′ + = ′ = ( ) . Integrating both sides yields t y t C 2 4 1 4 = + . Therefore, the general solution is y t t Ct ( ) = + − 1 4 2 2 . 22. ( ) ( ) , ln( ) t y ty t t y t t y t e t t 2 2 2 2 2 4 2 4 2 4 2 4 4 2 + ′ + = + ⇒ ′ + + = = = + + μ ∴ + ′ = + = + ⇒ + = + + (( ) ) ( ) ( ) t y t t t t t y t t C 2 2 2 4 2 2 5 3 4 4 4 4 5 4 3 y C t t t = + + + 5 3 5 4 3 2 4 ( ) . 23. For this D.E., p t ( ) = 1 . Integrating gives us P t t ( ) = . An integrating factor is, then, μ ( ) t e t = . Multiplying the D.E. by μ ( ) t , we obtain e y e y e y te t t t t ′ + = ′ = ( ) . Integrating both sides yields e y te e C t t t = − + . Therefore, the general solution is y t t Ce t ( ) = − + − 1 . 24. ′ + = ⇒ ′ = y y t e y e t t t 2 3 3 2 2 cos ( ) cos u e t = 2 dv tdt = cos 3 du e dt t = 2 2 v t = 1 3 3 sin e tdt e t e tdt t t t 2 2 2 3 3 3 2 3 3 cos sin sin = − ∫ ∫

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    10 • Chapter 2 First Order Differential Equations u e t = 2 dv tdt = sin 3 du e dt t = 2 2 v t = − 1 3 3 cos e tdt e t e tdt t t t 2 2 2 3 3 3 2 3 3 sin cos cos = − + ∫ ∫ ∴ = − − +       ⇒ + = + sin cos ( ) (sin cos ) I e t e t I I e t t t t t 2 2 2 3 3 2 3 3 3 2 3 1 4 9 3 3 2 3 ∴ = + (sin cos ) I e t t t 3 13 3 2 3 2 ∴ = + + ⇒ = + + − (sin cos ) (sin cos ) e y e t t C y t t Ce t t t 2 2 2 3 13 3 2 3 3 13 3 2 3 25 (a). #2 25 (b). #3 25 (c). #1 26. y t y e t ( ) = − 0 α 4 1 0 0 3 = = − − y e y e α α , Divide: 4 1 2 4 2 2 = ⇒ = = e α α ln ln and y e e e y t e t 0 3 4 8 2 3 2 8 8 = = = = ∴ = − α ln ln( ) (ln ) . ( ) . 27. First, we should put the equation into our conventional form: ′ − = y t y α 0. Integrating p t t ( ) = − α gives us P t t t ( ) ln ln = − = − α α . The general solution, then, is y t Ce Ce Ce Ct P t t t ( ) ( ) ln ln = = = = − − − α α α . Using the general solution and the point ( , ) 2 1 , we can solve for C in terms of α : y C ( ) 2 1 2 = = ⋅ α ; C = − 2 α . We can then substitute this value for C into the general solution at the point ( , ) 4 4 : y ( ) / / 4 4 2 4 4 4 4 2 2 = = ⋅ = ⋅ = − − α α α α α . Setting the exponents equal to each other yields 1 2 2 = = α α ; . Finally, solving for y 0 , y y 0 2 2 1 2 1 1 4 = = ⋅ = − ( ) . 28 (a). The general solution is y ce t = − 2 , so the corresponding graph is graph 2. y 0 2 ( ) = . 28 (b). The general solution is y ce t = −( ) sin 2 2 , so the corresponding graph is graph 4. y 0 3 ( ) = . 28 (c). The general solution is y ce t t = −       0 1 2 2 . sin , so the corresponding graph is graph 1. y 0 1 ( ) = . 28 (d). The general solution is y ce t = 10 , so the corresponding graph is graph 3. y 0 2 ( ) = . 29 (a). B c A c A dB dc dA dc kB ( ) = ( ) − = = − * and differentiating gives us , B A ( ) * 0 = − .

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    Chapter 2 First Order Differential Equations • 11 29 (b). B c A e A c A A c A e kc kc ( ) ( ) ( ) ( ) * * * = − = − = − − − , and substitution gives us 1 . The activity does not ever exceed A * because A c A c ( ) * only approaches the value of as approaches ∞ . Alternatively, the value of 1 − ( ) − e kc is never higher than 1. 29 (c). Substituting the condition into our given equation, we have 0 95 1 . ( ) * * A A e kc = − − . Simplification gives us − = − ⇒ − = = − − 0 05 20 1 20 . ln( ) ln( ) e kc kc , and solving for c yields ln( ) . c k 0 95 1 20 = . 30. ′ + = = y t y t t 4 4 α μ , t y t y t t y t y t C y t Ct 4 3 5 4 4 6 2 4 4 6 6 ′ + = = ′ ⇒ = + ⇒ = + − α α α ( ) y C C y t ( ) , 1 1 3 6 1 3 6 0 2 3 2 = − = + ⇒ = − − ≡ ⇒ = − = − α α α . 31. Multiplying both sides of the equation by the integrating factor, μ ( ) t e t = 2 , we have e y e Ce t e t C t t t t 2 2 2 2 1 1 = + + = + + − ( ) ( ) . Differentiating gives us ( ) ( ) ( ) ( ) e y e e t e t t t t t 2 2 2 2 1 2 1 2 3 ′ = + + = + . Therefore, ( ) ( ( ) ) ( ) ( ) ( ) ( ) e y t y t g t e t g t t t t 2 2 2 3 2 3 ′ = ′ = ⋅ = + ⇒ = + μ μ and μ ( ) ( ) ( ) ( ) t e e P t t p t t P t = = ⇒ = ⇒ = 2 2 2 . 32. 2 0 2 2 2 tCe pCe p t t t t + = ⇒ = − ( ) . Substituting, ( ) ( ) ( ) Ce t Ce t g t t t t 2 2 2 2 2 4 4 + ′ − + = − ⇒ = − . 33. Multiplying both sides of the equation by the integrating factor, μ ( ) t t = , we have ty t Ct t C = + = + − ( ) 1 1 . Differentiating gives us ( ) ty ′ = 1 . Therefore, ( ) ( ( ) ) ( ) ( ) ( )( ) ( ) ty t y t g t t t g t t ′ = ′ = ⋅ = = ⇒ = − − μ μ 1 1 1 and μ ( ) ( ) ln ( ) ( ) t t e P t t p t t t P t = = ⇒ = ⇒ = = − 1 1 . 34. ( ) ( ) ( ) , e t e t t g t t y t t − − + − ′ + + − = ⇒ = = 1 1 0 0 . 35. y t e e t y y t t ( ) sin ( ) = − + + ⇒ = = − + + = − − 2 0 2 1 0 1 0 . If y t e e t t t ( ) sin = − + + − 2 , then ′ = + + − y e e t t t 2 cos . Substituting in ′ + = y y g t ( ) , ( cos ) ( sin ) cos sin ( ) 2 2 2 e e t e e t e t t g t t t t t t − − + + + − + + = + + = . 36. ′ + + = + = = + y t y t y e t t ( cos ) cos , ( ) , sin 1 1 0 3 μ . ( ) ( cos ) ( ) sin sin sin sin sin ( sin ) e y t e e e y e C y Ce t t t t t t t t t t t t + + + + + − + ′ = + = ′ ⇒ = + ⇒ = + 1 1 . y C C y e y t t t t ( ) lim ( ) ( sin ) 0 1 3 2 1 2 1 = + = ⇒ = ∴ = + = − + →∞ and .

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    12 • Chapter 2 First Order Differential Equations 37. Putting this D.E. in the conventional form, we have ′ + = − − y y e t 2 2 . For this D.E., p t ( ) = 2 . An integrating factor is, then, μ ( ) t e t = 2 . Multiplying the D.E. by μ ( ) t , we obtain e y e y e y e e t t t t t 2 2 2 2 2 2 ′ + = ′ = − ( ) . Integrating both sides yields e y e e C t t t 2 2 = − + . Solving for y gives us y e Ce t t = − + − − 1 2 , and with our initial condition, y C ( ) 0 2 1 1 = − = − + . Solving for C yields C = − 2 , and thus our final solution is y e e t t = − − − − 1 2 2 . Therefore, lim ( ) t y t →∞ = − 1 . 38. y ce te y c t t = + ⇒ = − − 0 . ′( ) = − + − − − − y t y e e te t t t 0 . ′( ) = ⇒ − + − ( ) = − y y e 1 0 1 1 0 0 1 ∴ = y 0 0. 39. The general solution of the D.E. is y Ce y t C t = + ≠ = + = − λ λ λ λ 1 0 , , 0; . Therefore, the relevant limits are: lim t y →∞ = < does not exist for and λ λ 0 0 ; lim t y C →∞ = ⋅ + > 0 1 0 λ λ λ = 1 for . 40. On [ , ] 1 2 : ′ + = = y t y t y 1 3 1 1 , ( ) . An integrating factor is μ ( ) t t = . Multiplying the D.E. by μ ( ) t , we obtain ( ) , ( ) ty t ty t C y t Ct y C C ′ = ⇒ = + ⇒ = + = + = ⇒ = − 3 1 1 1 0 2 3 2 1 . Therefore, the solution for 1 2 ≤ ≤ t is y t = 2 and y ( ) 2 4 = . On[ , ] 2 3 : ′ + = = y t y y 1 0 2 4 , ( ) . An integrating factor is μ ( ) t t = . Multiplying the D.E. by μ ( ) t , we obtain ( ) , ( ) ty ty C y Ct y C C ′ = ⇒ = ⇒ = = = ⇒ = − 0 2 2 4 8 1 . Therefore, the solution for 2 3 ≤ ≤ t is y t = 8 . 41. On [ , ] 0 π : ′ + = = y t y t y (sin ) sin , ( ) 0 3 . An integrating factor is μ ( ) cos t e t = − . Multiplying the D.E. by μ ( ) t , we obtain e y e t y e y t e t t t t − − − − ′ + = ′ = cos cos cos cos (sin ) ( ) (sin ) . Integrating both sides yields e y e C t t − − = + cos cos . Solving for y gives us y Ce t = + 1 cos , and with our initial condition, y Ce C e ( ) 0 3 1 2 1 = = + ⇒ = − . Therefore, the solution for 0 ≤ ≤ t π is y e t = + − 1 2 1 cos and y e ( ) π = + − 1 2 2 . On [ , ] π π 2 : ′ + = − = + − y t y t y e (sin ) sin , ( ) π 1 2 2 . Multiplying the D.E. by μ ( ) cos t e t = − , we obtain e y e t y e y t e t t t t − − − − ′ + = ′ = − cos cos cos cos (sin ) ( ) ( sin ) .

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    Chapter 2 First Order Differential Equations • 13 Integrating both sides yields e y e C t t − − = − + cos cos . Solving for y gives us y Ce t = − + 1 cos , and with our initial condition, y e Ce C e e ( ) π = + = − + ⇒ = + − − − 1 2 1 2 2 2 1 1 1 . Therefore, the solution for π π ≤ ≤ t 2 is y e e t t = − + + + − 1 2 2 1 1 cos cos . 42. On [ , ] 0 1 : ′ = = y y 2 0 1 , ( ) . y t C y C C = + = = ⇒ = 2 0 1 1 , ( ) . Therefore, the solution for 0 1 ≤ ≤ t is y t = + 2 1 and y ( ) 1 3 = . On [ , ] 1 2 : ′ + = = y t y y 1 2 1 3 , ( ) . An integrating factor is μ ( ) t t = . Multiplying the D.E. by μ ( ) t , we obtain ( ) , ( ) ty t ty t C y t Ct y C C ′ = ⇒ = + ⇒ = + = + = ⇒ = − 2 1 1 3 2 2 1 . Therefore, the solution for 1 2 ≤ ≤ t is y t t = + 2 . 43. On [ , ] 0 1 : ′ + − = = y t y y ( ) , ( ) 2 1 0 0 3 . An integrating factor is μ ( ) t e t t = − 2 . Multiplying the D.E. by μ ( ) t , we obtain e y e t y e y t t t t t t 2 2 2 2 1 0 − − − ′ + − = ′ = ( ) ( ) . Integrating both sides yields e y C t t 2 − = . Solving for y gives us y Ce t t = − 2 , and with our initial condition, y C ( ) 0 3 = = . Therefore, the solution for 0 1 ≤ ≤ t is y e t t = − 3 2 and y ( ) 1 3 = . On [ , ] 1 3 : ′ + = ′ = = y y y y ( ) , ( ) 0 0 1 3 . Integrating gives us y C = = 3. Therefore, the solution for 1 3 ≤ ≤ t is y = 3 and y ( ) 3 3 = . On [ , ] 3 4 : ′ + − ( ) = = y y y t 1 0 3 3 , ( ) . An integrating factor is μ ( ) ln t e t t = = − 1 . Multiplying the D.E. by μ ( ) t , we obtain 1 1 1 2 0 t t t y y y ′ − = ′ = ( ) . Integrating both sides yields 1 t y C = . Solving for y gives us y Ct = , and with our initial condition, y C C ( ) ( ) 3 3 3 1 = = ⇒ = . Therefore, the solution for 3 4 ≤ ≤ t is y t = . 44. y t t Si t Si ( ) ( ) ( ) = − + { } 1 3 Section 2.3 1 (a). To begin, Q ( ) 0 0 = and ′ = − Q Q ( . )( ) ( ) 0 2 3 100 3 . Putting the second equation in the conventional form, we have ′ + = Q Q 0 03 0 6 . . . Multiplying both sides of this equation by the integrating factor μ ( ) . t e t = 0 03 gives us ( ) . . . e Q e t t 0 03 0 03 0 6 ′ = . Integrating both sides

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