280hwfornotesANS

Homework 0.1 : A) Practice with the following integrals. 1. 1 2 ln | 2 x + 1 | + C 2. √ x 2 + 1 + C 3. 1 4 ln( x 4 + 1) + C 4. 1 2 arctan( x 2 ) + C 5. x + 2 − 2 ln | x + 2 | + C 6. 1 2 ln | x − 1 | − 1 2 ln | x + 1 | + C 7. ln | x − 2 | − 1 2 ln( x 2 + 1) + arctan( x ) + C 8. − 1 x + ln | x | − ln | x + 1 | + C. 9. 1 3 x 2 e 3 x − 2 9 xe 3 x + 2 27 e 3 x + C. 10. 1 3 e x 3 + C. 11. arcsin( e x ) + C 12. sin(ln( x )) + C 13. 1 3 ln | sec(3 x ) + tan(3 x ) | + C 14. sec( x ) + C 15. x tan( x ) − ln | sec( x ) | + C. 16. x ln( x ) − x + C. 1

Homework 1.1 : A) For each ODE, find the order and classify as linear or nonlinear. 1. 1 -st order, linear. 2. 2 nd order, nonlinear. 3. 5 -th order, nonlinear. 4. 3 -rd order, linear. 5. 2 -nd order, linear. 6. 1 -st order, nonlinear. 7. 3 -rd order, linear. 8. 1 -st order, linear. 9. 2 -nd order, nonlinear. 10. 2 -nd order, linear. 11. 1 -st order, nonlinear. 12. 3 -rd order, linear. B) Show that each function is a solution to the IVP. On what interval is the solution defined? 13. I = ( −∞ , ∞ ) . 14. I = (0 , 2) . 15. I = (2 , ∞ ) . 16. I = ( − π 2 , π 2 ) . 17. I = ( −∞ , ∞ ) . 18. I = (0 , ∞ ) 19. I = ( −∞ , 1) . 20. I = ( π 2 , 3 π 2 ) . 21. I = ( − π 4 , π 4 ) . 22. I = ( − 3 π 4 , π 4 ) . 23. I = ( −∞ , 3) . 24. I = (2 , ∞ ) . 25. I = ( −∞ , ∞ ) . 26. I = ( −∞ , 0) . 27. I = ( − 5 , 5) . 28. I = ( − 2 , ∞ ) . 29. I = ( − 1 , 1) . 30. I = (1 , ∞ ) 2

Homework 1.2 : A) Solve for an explicit solution y ( x ) . 1. y = ( 1 4 arctan ( x 2 ) + C ) 2 2. y = tan( − . 5 x − 2 + C ) 3. y = (tan(3 x + C )) 1 3 4. y = − 1 arctan( x )+ C . 5. y = ( 1 2 e − x + C ) 2 6. y = √ ( x 2 + C ) 2 + 1 7. y = sin(2 √ x + C ) 8. y = 2 / (ln 2 ( x ) + C ) 9. y = 1 2 ln ( e x 2 + C ) . B) Find the explicit solutions y ( x ) to the initial value problems. 10. y = 4 x 3 . 11. y = 3 e 5 x − 2 . 12. y = ( x 2 + 1) 2 − 1 = x 4 + 2 x 2 . 13. y = 2 1+ln( x 2 +1) . 14. y = arctan(ln( x ) + 1) . 15. y = sin ( x 2 + π 6 ) . 16. y = tan ( 1 − 1 x ) . 17. y = 1 − e x 1+ e x . C) Find the explicit solution y ( x ) by separation of variables. Simplify so that there are no logarithms in your final answer. 18. y = Cx 5 e − x 19. y = Cx 3 ( x > 0) 20. y = C/x 21. y = 1 2 + C x 2 22. y = 2 + Cx 3 23. y = 3 + C (1 + . 2 x ) 5 24. y = C √ x 2 + 1 25. y = ( 1 + Ce 3 x ) 1 3 26. y = C x e 2 x 2 27. y = Cx 3 e − x 28. y = 3 + Cx 2 e x 29. y = Ce x x +1 30. y = Ce x ( x − 2) 3 31. y = C sec 3 ( x ) 32. y = arcsin( Ce − x 2 ) 33. y = 3 2 e 3 x +1 . 34. y = 6 e 2 x 3 e 2 x − 1 . 35. y = 1 − e 2 x e 2 x +1 . 3

D) Find the solutions to the IVP’s in implicit form (Do not solve for y !). 36. y sin( y ) + cos( y ) = 1 4 x 2 − 5 . 37. 1 3 y 3 + y = x 2 − x + 10 . 38. y sin( y ) + cos( y ) = x + 1 . 39. ye − y + e − y = 1 x + 2 . 40. y ln( y ) − y = − 3 cos( x ) + 2 . 41. 2 ln( y ) − 7 y = − 1 3 x 3 + 2 . 42. 2 ln( y ) + 3 y − 1 = x 2 + 2 . 43. y − ln( y + 1) = 1 2 x 2 − 2 . 44. ye y − e y = 2 e 2 x − 3 . 45. 2 ln( y − 2) − ln( y − 1) = x − ln(2) . 46. ln( y ) + arctan( y ) = x + π 4 . E) Write the explicit solution to the IVP using a definite integral. 47. y = ( 1 2 ∫ x 3 cos ( 1 t ) dt + 8 ) 2 3 . 48. y = ∫ x 1 e t t 2 dt + 5 . 49. y = ∫ x π sin( t ) t dt + 1 . 50. y = ( 3 ∫ x 0 cos( t 2 ) dt + 8 ) 1 3 . 51. y = exp ( ∫ x 0 1 t 3 +1 dt + 1 ) . 52. y = ( 4 − ∫ x e 1 ln( t ) dt ) − 1 . 53. y = ( 1 2 ∫ x 8 sin( t 2 ) dt + 3 ) 2 . 54. y = √ ∫ x 1 √ 1 + t 3 dt + 9 . 55. y = ln ( 1 + ∫ x 1 e t 2 dt ) . 4

Homework 1.3: A) Use the method of integrating factors to find the general solution. 1. y = − 1 2 x 2 + Ce x 2 x 2 . 2. y = 1 2 e − 3 x + Ce − 5 x 3. y = − x − 1 + Ce x 4. y = xe 3 x + Ce 3 x 5. y = x + Cx − 2 6. y = 1 2 x ln 2 ( x ) + Cx 7. y = 2 x 3 + Cx 5 8. y = − x 2 + Cx 3 . 9. y = e x 3 3 x 3 + C x 3 . 10. y = 1 3 x 2 + Cx 2 e − 3 x . 11. y = 1 x − 1 x 2 + Ce − x x 2 . 12. y = sin( x ) x + cos( x ) x 2 + C x 2 . 13. y = 1 + C ( x + 2) − 3 . 14. y = 1 x − 1 + Ce − x x − 1 . 15. y = 1 2 x 3 + Ce − x 2 x 3 . 16. y = 1 + C √ x 2 +1 . 17. y = 1 1 − x + C ( x +1 1 − x ) . 18. y = sin( x ) + C cos( x ) . 19. y = x − cos( x ) sec( x )+tan( x ) + C sec ( x )+tan( x ) . B) Use the method of integrating factors to solve the IVP’s. 20. y = − 1 2 x + 1 4 x 3 . 21. y = e 3 x + 2 e x . 22. y = 2 x − 1 + 2 e − 2 x . 23. y = 1 x − cos( x 2 ) x . 24. y = 2 − 4 √ x . 25. y = 1 + 2 e − x 2 . 26. y = 1 8 x 3 − 1 8 x − 5 . 27. y = x 3 − x 2 . 28. y = x 2 − 1 2 x 2 cos(2 x ) . 29. y = sin( x ) x 2 + cos( x ) x 3 + 1 x 3 . 30. y = 1 + 8 ( x +1) 3 . 31. y = 2 x − 3 − ( x − 1) ln( x − 1) . 32. y = x − 1 + 2 2 x − 1 . 33. y = 1 2 e x − 2 x 2 e x . 34. y = 3 x − 3 xe − 2 x +2 . 35. y = 1 3 x 3 +1 x 2 − 1 . 36. y = 2 √ x 2 + 9 − 1 . 37. y = 1 − 2 e cos( x ) 38. y = x cos( x ) − π 4 cos( x ) C) Find the solution y ( x ) to the initial value problem in terms of a definite integral . 39. y = x 3 ∫ x 2 t − 4 sin( t ) dt + 3 x 3 . 40. y = e x 2 ∫ x 1 e − t 2 dt + 2 e x 2 41. y = e − sin( x ) ∫ x π t 2 e sin( t ) dt + 3 e − sin( x ) 42. y = x ∫ x 3 e 2 t t 2 dt + 2 x 43. y = 1 x 2 ∫ x 3 t sec( t ) dt − 9 x 2 44. y = 1 √ x ∫ x 4 1 √ t e t 2 dt + 2 √ x 45. y = 1 √ x 4 +8 ∫ x − 1 1 √ t 4 +8 dt + 3 √ x 4 +8 5