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variation of parameters

Variation of Parameters Bernd Schr oder logo1 Bernd Schr oder - PowerPoint PPT Presentation

Jul 26, 2023 •263 likes •1.4k views

Overview An Example Double Check Further Discussion Variation of Parameters Bernd Schr oder logo1 Bernd Schr oder Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double

  1. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Solution of the homogeneous equation. y ′′ + 4 y ′ + 4 y = 0 λ 2 e λ x + 4 λ e λ x + 4 e λ x = 0 λ 2 + 4 λ + 4 = 0 √ − 4 ± 16 − 16 = = − 2 λ 1 , 2 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  2. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Solution of the homogeneous equation. y ′′ + 4 y ′ + 4 y = 0 λ 2 e λ x + 4 λ e λ x + 4 e λ x = 0 λ 2 + 4 λ + 4 = 0 √ − 4 ± 16 − 16 = = − 2 λ 1 , 2 2 c 1 e − 2 x + c 2 xe − 2 x = y h logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  3. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  4. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the Wronskian. e − 2 t � e − 2 t − 2 te − 2 t � − te − 2 t � − 2 e − 2 t � W ( y 1 , y 2 )( t ) = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  5. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the Wronskian. e − 2 t � e − 2 t − 2 te − 2 t � − te − 2 t � − 2 e − 2 t � W ( y 1 , y 2 )( t ) = e − 4 t − 2 te − 4 t + 2 te − 4 t = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  6. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the Wronskian. e − 2 t � e − 2 t − 2 te − 2 t � − te − 2 t � − 2 e − 2 t � W ( y 1 , y 2 )( t ) = e − 4 t − 2 te − 4 t + 2 te − 4 t = e − 4 t = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  7. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  8. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 1 ( t ) dt W ( y 1 , y 2 )( t ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  9. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) sin ( t ) 1 1 � � e − 2 t dt a 2 ( t ) y 1 ( t ) dt = W ( y 1 , y 2 )( t ) e − 4 t 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  10. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) sin ( t ) 1 1 � � e − 2 t dt a 2 ( t ) y 1 ( t ) dt = W ( y 1 , y 2 )( t ) e − 4 t 1 � e 2 t sin ( t ) dt = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  11. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  12. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � e 2 t sin ( t ) dt logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  13. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � 1 1 � e 2 t sin ( t ) dt 2 e 2 t sin ( t ) − 2 e 2 t cos ( t ) dt = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  14. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � 1 1 � e 2 t sin ( t ) dt 2 e 2 t sin ( t ) − 2 e 2 t cos ( t ) dt = � 1 � 1 � 1 2 e 2 t sin ( t ) − 1 2 e 2 t cos ( t ) − 2 e 2 t � � = − sin ( t ) dt 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  15. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � 1 1 � e 2 t sin ( t ) dt 2 e 2 t sin ( t ) − 2 e 2 t cos ( t ) dt = � 1 � 1 � 1 2 e 2 t sin ( t ) − 1 2 e 2 t cos ( t ) − 2 e 2 t � � = − sin ( t ) dt 2 1 2 e 2 t sin ( t ) − 1 4 e 2 t cos ( t ) − 1 � e 2 t sin ( t ) dt = 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  16. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � 1 1 � e 2 t sin ( t ) dt 2 e 2 t sin ( t ) − 2 e 2 t cos ( t ) dt = � 1 � 1 � 1 2 e 2 t sin ( t ) − 1 2 e 2 t cos ( t ) − 2 e 2 t � � = − sin ( t ) dt 2 1 2 e 2 t sin ( t ) − 1 4 e 2 t cos ( t ) − 1 � e 2 t sin ( t ) dt = 4 5 1 2 e 2 t sin ( t ) − 1 � e 2 t sin ( t ) dt 4 e 2 t cos ( t ) = 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  17. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � 1 1 � e 2 t sin ( t ) dt 2 e 2 t sin ( t ) − 2 e 2 t cos ( t ) dt = � 1 � 1 � 2 e 2 t sin ( t ) − 1 1 2 e 2 t cos ( t ) − 2 e 2 t � � = − sin ( t ) dt 2 1 2 e 2 t sin ( t ) − 1 4 e 2 t cos ( t ) − 1 � e 2 t sin ( t ) dt = 4 5 1 2 e 2 t sin ( t ) − 1 � e 2 t sin ( t ) dt 4 e 2 t cos ( t ) = 4 2 5 e 2 t sin ( t ) − 1 � e 2 t sin ( t ) dt 5 e 2 t cos ( t ) = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  18. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  19. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 2 ( t ) dt W ( y 1 , y 2 )( t ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  20. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 2 ( t ) dt W ( y 1 , y 2 )( t ) sin ( t ) 1 � te − 2 t dt = e − 4 t 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  21. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 2 ( t ) dt W ( y 1 , y 2 )( t ) sin ( t ) 1 � te − 2 t dt = e − 4 t 1 � te 2 t sin ( t ) dt = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  22. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 2 ( t ) dt W ( y 1 , y 2 )( t ) sin ( t ) 1 � te − 2 t dt = e − 4 t 1 � te 2 t sin ( t ) dt = � 2 � 2 � 5 e 2 t sin ( t ) − 1 5 e 2 t sin ( t ) − 1 5 e 2 t cos ( t ) 5 e 2 t cos ( t ) dt = t − logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  23. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 2 ( t ) dt W ( y 1 , y 2 )( t ) sin ( t ) 1 � te − 2 t dt = e − 4 t 1 � te 2 t sin ( t ) dt = � 2 � 2 � 5 e 2 t sin ( t ) − 1 5 e 2 t sin ( t ) − 1 5 e 2 t cos ( t ) 5 e 2 t cos ( t ) dt = t − 2 5 te 2 t sin ( t ) − 1 5 te 2 t cos ( t ) − 2 e 2 t sin ( t ) dt + 1 � � e 2 t cos ( t ) dt = 5 5 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  24. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  25. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � e 2 t cos ( t ) dt logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  26. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � 1 1 � e 2 t cos ( t ) dt 2 e 2 t cos ( t )+ 2 e 2 t sin ( t ) dt = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  27. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � 1 1 � e 2 t cos ( t ) dt 2 e 2 t cos ( t )+ 2 e 2 t sin ( t ) dt = � 1 � 1 � 1 2 e 2 t cos ( t )+ 1 2 e 2 t sin ( t ) − 2 e 2 t cos ( t ) dt = 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  28. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � 1 1 � e 2 t cos ( t ) dt 2 e 2 t cos ( t )+ 2 e 2 t sin ( t ) dt = � 1 � 1 � 2 e 2 t cos ( t )+ 1 1 2 e 2 t sin ( t ) − 2 e 2 t cos ( t ) dt = 2 1 2 e 2 t cos ( t )+ 1 4 e 2 t sin ( t ) − 1 � e 2 t cos ( t ) dt = 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  29. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � 1 1 � e 2 t cos ( t ) dt 2 e 2 t cos ( t )+ 2 e 2 t sin ( t ) dt = � 1 � 1 � 1 2 e 2 t cos ( t )+ 1 2 e 2 t sin ( t ) − 2 e 2 t cos ( t ) dt = 2 1 2 e 2 t cos ( t )+ 1 4 e 2 t sin ( t ) − 1 � e 2 t cos ( t ) dt = 4 5 1 2 e 2 t cos ( t )+ 1 � e 2 t cos ( t ) dt 4 e 2 t sin ( t ) = 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  30. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. � 1 1 � e 2 t cos ( t ) dt 2 e 2 t cos ( t )+ 2 e 2 t sin ( t ) dt = � 1 � 1 � 1 2 e 2 t cos ( t )+ 1 2 e 2 t sin ( t ) − 2 e 2 t cos ( t ) dt = 2 1 2 e 2 t cos ( t )+ 1 4 e 2 t sin ( t ) − 1 � e 2 t cos ( t ) dt = 4 5 1 2 e 2 t cos ( t )+ 1 � e 2 t cos ( t ) dt 4 e 2 t sin ( t ) = 4 2 5 e 2 t cos ( t )+ 1 � e 2 t cos ( t ) dt 5 e 2 t sin ( t ) = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  31. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  32. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 2 ( t ) dt W ( y 1 , y 2 )( t ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  33. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 2 ( t ) dt W ( y 1 , y 2 )( t ) 2 5 te 2 t sin ( t ) − 1 5 te 2 t cos ( t ) − 2 e 2 t sin ( t ) dt + 1 � � e 2 t cos ( t ) dt = 5 5 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  34. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 2 ( t ) dt W ( y 1 , y 2 )( t ) 2 5 te 2 t sin ( t ) − 1 5 te 2 t cos ( t ) − 2 e 2 t sin ( t ) dt + 1 � � e 2 t cos ( t ) dt = 5 5 2 5 te 2 t sin ( t ) − 1 5 te 2 t cos ( t ) = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  35. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 2 ( t ) dt W ( y 1 , y 2 )( t ) 5 te 2 t sin ( t ) − 1 2 5 te 2 t cos ( t ) − 2 e 2 t sin ( t ) dt + 1 � � e 2 t cos ( t ) dt = 5 5 � 2 � 2 5 te 2 t sin ( t ) − 1 5 te 2 t cos ( t ) − 2 5 e 2 t sin ( t ) − 1 5 e 2 t cos ( t ) = 5 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  36. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 2 ( t ) dt W ( y 1 , y 2 )( t ) 2 5 te 2 t sin ( t ) − 1 5 te 2 t cos ( t ) − 2 e 2 t sin ( t ) dt + 1 � � e 2 t cos ( t ) dt = 5 5 � 2 � 2 5 te 2 t sin ( t ) − 1 5 te 2 t cos ( t ) − 2 5 e 2 t sin ( t ) − 1 5 e 2 t cos ( t ) = 5 � 2 � + 1 5 e 2 t cos ( t )+ 1 5 e 2 t sin ( t ) 5 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  37. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Computing the integrals. f ( t ) 1 � a 2 ( t ) y 2 ( t ) dt W ( y 1 , y 2 )( t ) 2 5 te 2 t sin ( t ) − 1 5 te 2 t cos ( t ) − 2 e 2 t sin ( t ) dt + 1 � � e 2 t cos ( t ) dt = 5 5 � 2 � 2 5 te 2 t sin ( t ) − 1 5 te 2 t cos ( t ) − 2 5 e 2 t sin ( t ) − 1 5 e 2 t cos ( t ) = 5 � 2 � + 1 5 e 2 t cos ( t )+ 1 5 e 2 t sin ( t ) 5 2 5 te 2 t sin ( t ) − 1 5 te 2 t cos ( t ) − 3 25 e 2 t sin ( t )+ 4 25 e 2 t cos ( t ) = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  38. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  39. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Stating the general solution. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  40. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Stating the general solution. � x � x f ( t ) f ( t ) 1 1 y p = − y 1 ( x ) a 2 ( t ) y 2 ( t ) dt + y 2 ( x ) a 2 ( t ) y 1 ( t ) dt W ( y 1 , y 2 )( t ) W ( y 1 , y 2 )( t ) x 0 x 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  41. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Stating the general solution. � x � x f ( t ) f ( t ) 1 1 y p = − y 1 ( x ) a 2 ( t ) y 2 ( t ) dt + y 2 ( x ) a 2 ( t ) y 1 ( t ) dt W ( y 1 , y 2 )( t ) W ( y 1 , y 2 )( t ) x 0 x 0 � 2 � 5 xe 2 x sin ( x ) − 1 5 xe 2 x cos ( x ) − 3 25 e 2 x sin ( x )+ 4 25 e 2 x cos ( x ) = − e − 2 x logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  42. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Stating the general solution. � x � x f ( t ) f ( t ) 1 1 y p = − y 1 ( x ) a 2 ( t ) y 2 ( t ) dt + y 2 ( x ) a 2 ( t ) y 1 ( t ) dt W ( y 1 , y 2 )( t ) W ( y 1 , y 2 )( t ) x 0 x 0 � 2 � 5 xe 2 x sin ( x ) − 1 5 xe 2 x cos ( x ) − 3 25 e 2 x sin ( x )+ 4 25 e 2 x cos ( x ) = − e − 2 x � 2 � 5 e 2 x sin ( x ) − 1 5 e 2 x cos ( x ) + xe − 2 x logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  43. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Stating the general solution. � x � x f ( t ) f ( t ) 1 1 y p = − y 1 ( x ) a 2 ( t ) y 2 ( t ) dt + y 2 ( x ) a 2 ( t ) y 1 ( t ) dt W ( y 1 , y 2 )( t ) W ( y 1 , y 2 )( t ) x 0 x 0 � 2 � 5 xe 2 x sin ( x ) − 1 5 xe 2 x cos ( x ) − 3 25 e 2 x sin ( x )+ 4 25 e 2 x cos ( x ) = − e − 2 x � 2 � 5 e 2 x sin ( x ) − 1 5 e 2 x cos ( x ) + xe − 2 x 25 sin ( x ) − 4 3 = 25 cos ( x ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  44. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Stating the general solution. � x � x f ( t ) f ( t ) 1 1 y p = − y 1 ( x ) a 2 ( t ) y 2 ( t ) dt + y 2 ( x ) a 2 ( t ) y 1 ( t ) dt W ( y 1 , y 2 )( t ) W ( y 1 , y 2 )( t ) x 0 x 0 � 2 � 5 xe 2 x sin ( x ) − 1 5 xe 2 x cos ( x ) − 3 25 e 2 x sin ( x )+ 4 25 e 2 x cos ( x ) = − e − 2 x � 2 � 5 e 2 x sin ( x ) − 1 5 e 2 x cos ( x ) + xe − 2 x 25 sin ( x ) − 4 3 = 25 cos ( x ) y = − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  45. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  46. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  47. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x = y logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  48. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x = y 25 sin ( x )+ 3 4 25 cos ( x ) − 2 c 1 e − 2 x + c 2 � e − 2 x − 2 xe − 2 x � y ′ = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  49. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x = y 25 sin ( x )+ 3 4 25 cos ( x ) − 2 c 1 e − 2 x + c 2 � e − 2 x − 2 xe − 2 x � y ′ = 1 = y ( 0 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  50. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x = y 25 sin ( x )+ 3 4 25 cos ( x ) − 2 c 1 e − 2 x + c 2 � e − 2 x − 2 xe − 2 x � y ′ = y ( 0 ) = − 4 1 = 25 + c 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  51. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x = y 25 sin ( x )+ 3 4 25 cos ( x ) − 2 c 1 e − 2 x + c 2 � e − 2 x − 2 xe − 2 x � y ′ = y ( 0 ) = − 4 c 1 = 29 1 = 25 + c 1 , 25 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  52. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x = y 25 sin ( x )+ 3 4 25 cos ( x ) − 2 c 1 e − 2 x + c 2 � e − 2 x − 2 xe − 2 x � y ′ = y ( 0 ) = − 4 c 1 = 29 1 = 25 + c 1 , 25 y ′ ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  53. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x = y 25 sin ( x )+ 3 4 25 cos ( x ) − 2 c 1 e − 2 x + c 2 � e − 2 x − 2 xe − 2 x � y ′ = y ( 0 ) = − 4 c 1 = 29 1 = 25 + c 1 , 25 y ′ ( 0 ) = 3 = 25 − 2 c 1 + c 2 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  54. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x = y 25 sin ( x )+ 3 4 25 cos ( x ) − 2 c 1 e − 2 x + c 2 � e − 2 x − 2 xe − 2 x � y ′ = y ( 0 ) = − 4 c 1 = 29 1 = 25 + c 1 , 25 y ′ ( 0 ) = 3 c 2 = 2 c 1 − 3 = 25 − 2 c 1 + c 2 , 0 25 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  55. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x = y 25 sin ( x )+ 3 4 25 cos ( x ) − 2 c 1 e − 2 x + c 2 � e − 2 x − 2 xe − 2 x � y ′ = y ( 0 ) = − 4 c 1 = 29 1 = 25 + c 1 , 25 y ′ ( 0 ) = 3 c 2 = 2 c 1 − 3 25 = 55 = 25 − 2 c 1 + c 2 , 0 25 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  56. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x = y 25 sin ( x )+ 3 4 25 cos ( x ) − 2 c 1 e − 2 x + c 2 � e − 2 x − 2 xe − 2 x � y ′ = y ( 0 ) = − 4 c 1 = 29 1 = 25 + c 1 , 25 y ′ ( 0 ) = 3 c 2 = 2 c 1 − 3 25 = 55 25 = 11 = 25 − 2 c 1 + c 2 , 0 5 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  57. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 Finding c 1 , c 2 . − 4 25 cos ( x )+ 3 25 sin ( x )+ c 1 e − 2 x + c 2 xe − 2 x = y 25 sin ( x )+ 3 4 25 cos ( x ) − 2 c 1 e − 2 x + c 2 � e − 2 x − 2 xe − 2 x � y ′ = y ( 0 ) = − 4 c 1 = 29 1 = 25 + c 1 , 25 y ′ ( 0 ) = 3 c 2 = 2 c 1 − 3 25 = 55 25 = 11 = 25 − 2 c 1 + c 2 , 0 5 − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x = y logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  58. Overview An Example Double Check Further Discussion Solve the Initial Value Problem y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0 y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  59. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  60. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? � � − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  61. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? � � − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x 4 � 4 e − 2 x − 2 xe − 2 x �� 25 sin ( x )+ 3 25 cos ( x ) − 58 25 e − 2 x + 11 � + 4 5 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  62. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? � � − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x 4 � 4 e − 2 x − 2 xe − 2 x �� 25 sin ( x )+ 3 25 cos ( x ) − 58 25 e − 2 x + 11 � + 4 5 + 4 25 cos ( x ) − 3 25 sin ( x )+ 116 25 e − 2 x + 11 − 4 e − 2 x + 4 xe − 2 x � ? � = sin ( x ) 5 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  63. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? � � − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x 4 � 4 e − 2 x − 2 xe − 2 x �� 25 sin ( x )+ 3 25 cos ( x ) − 58 25 e − 2 x + 11 � + 4 5 + 4 25 cos ( x ) − 3 25 sin ( x )+ 116 25 e − 2 x + 11 − 4 e − 2 x + 4 xe − 2 x � ? � = sin ( x ) 5 � � − 16 25 + 12 25 + 4 cos ( x ) 25 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  64. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? � � − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x 4 � 4 e − 2 x − 2 xe − 2 x �� 25 sin ( x )+ 3 25 cos ( x ) − 58 25 e − 2 x + 11 � + 4 5 + 4 25 cos ( x ) − 3 25 sin ( x )+ 116 25 e − 2 x + 11 − 4 e − 2 x + 4 xe − 2 x � ? � = sin ( x ) 5 � � � 12 � − 16 25 + 12 25 + 4 25 + 16 25 − 3 cos ( x )+ sin ( x ) 25 25 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  65. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? � � − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x 4 � 4 e − 2 x − 2 xe − 2 x �� 25 sin ( x )+ 3 25 cos ( x ) − 58 25 e − 2 x + 11 � + 4 5 + 4 25 cos ( x ) − 3 25 sin ( x )+ 116 25 e − 2 x + 11 − 4 e − 2 x + 4 xe − 2 x � ? � = sin ( x ) 5 � � � 12 � − 16 25 + 12 25 + 4 25 + 16 25 − 3 cos ( x )+ sin ( x ) 25 25 � 116 � 25 − 232 25 + 44 5 + 116 25 − 44 e − 2 x + 5 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  66. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? � � − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x 4 � 4 e − 2 x − 2 xe − 2 x �� 25 sin ( x )+ 3 25 cos ( x ) − 58 25 e − 2 x + 11 � + 4 5 + 4 25 cos ( x ) − 3 25 sin ( x )+ 116 25 e − 2 x + 11 − 4 e − 2 x + 4 xe − 2 x � ? � = sin ( x ) 5 � � � 12 � − 16 25 + 12 25 + 4 25 + 16 25 − 3 cos ( x )+ sin ( x ) 25 25 � 116 � � 44 � 25 − 232 25 + 44 5 + 116 25 − 44 5 − 88 5 + 44 e − 2 x + xe − 2 x + 5 5 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  67. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? � � − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x 4 � 4 e − 2 x − 2 xe − 2 x �� 25 sin ( x )+ 3 25 cos ( x ) − 58 25 e − 2 x + 11 � + 4 5 + 4 25 cos ( x ) − 3 25 sin ( x )+ 116 25 e − 2 x + 11 − 4 e − 2 x + 4 xe − 2 x � ? � = sin ( x ) 5 � � � 12 � − 16 25 + 12 25 + 4 25 + 16 25 − 3 cos ( x )+ sin ( x ) 25 25 � 116 � � 44 � 25 − 232 25 + 44 5 + 116 25 − 44 5 − 88 5 + 44 ? e − 2 x + xe − 2 x + = sin ( x ) 5 5 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  68. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? � � − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x 4 � 4 e − 2 x − 2 xe − 2 x �� 25 sin ( x )+ 3 25 cos ( x ) − 58 25 e − 2 x + 11 � + 4 5 + 4 25 cos ( x ) − 3 25 sin ( x )+ 116 25 e − 2 x + 11 − 4 e − 2 x + 4 xe − 2 x � ? � = sin ( x ) 5 � � � 12 � − 16 25 + 12 25 + 4 25 + 16 25 − 3 cos ( x )+ sin ( x ) 25 25 √ � 116 � � 44 � 25 − 232 25 + 44 5 + 116 25 − 44 5 − 88 5 + 44 e − 2 x + xe − 2 x + = sin ( x ) 5 5 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  69. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  70. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? − 4 25 + 29 y ( 0 ) = 25 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  71. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? − 4 25 + 29 y ( 0 ) = 25 = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  72. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? √ − 4 25 + 29 y ( 0 ) = 25 = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  73. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? √ − 4 25 + 29 y ( 0 ) = 25 = 1 25 sin ( x )+ 3 4 25 cos ( x ) − 58 25 e − 2 x + 55 � e − 2 x − 2 xe − 2 x � y ′ ( x ) = 25 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  74. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? √ − 4 25 + 29 y ( 0 ) = 25 = 1 25 sin ( x )+ 3 4 25 cos ( x ) − 58 25 e − 2 x + 55 � e − 2 x − 2 xe − 2 x � y ′ ( x ) = 25 25 − 58 3 25 + 55 y ′ ( 0 ) = 25 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  75. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? √ − 4 25 + 29 y ( 0 ) = 25 = 1 25 sin ( x )+ 3 4 25 cos ( x ) − 58 25 e − 2 x + 55 � e − 2 x − 2 xe − 2 x � y ′ ( x ) = 25 25 − 58 3 25 + 55 y ′ ( 0 ) = 25 = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  76. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? √ − 4 25 + 29 y ( 0 ) = 25 = 1 25 sin ( x )+ 3 4 25 cos ( x ) − 58 25 e − 2 x + 55 � e − 2 x − 2 xe − 2 x � y ′ ( x ) = 25 25 − 58 3 25 + 55 √ y ′ ( 0 ) = 25 = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  77. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? √ − 4 25 + 29 y ( 0 ) = 25 = 1 25 sin ( x )+ 3 4 25 cos ( x ) − 58 25 e − 2 x + 55 � e − 2 x − 2 xe − 2 x � y ′ ( x ) = 25 25 − 58 3 25 + 55 √ y ′ ( 0 ) = 25 = 0 Alternatively, use a computer to double check the result. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  78. Overview An Example Double Check Further Discussion Does y = − 4 25 cos ( x )+ 3 25 sin ( x )+ 29 25 e − 2 x + 11 5 xe − 2 x Solve y ′′ + 4 y ′ + 4 y = sin ( x ) , y ( 0 ) = 1, y ′ ( 0 ) = 0? √ − 4 25 + 29 y ( 0 ) = 25 = 1 25 sin ( x )+ 3 4 25 cos ( x ) − 58 25 e − 2 x + 55 � e − 2 x − 2 xe − 2 x � y ′ ( x ) = 25 25 − 58 3 25 + 55 √ y ′ ( 0 ) = 25 = 0 Alternatively, use a computer to double check the result. Beyond a certain level of complexity, that really is the way to go. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  79. Overview An Example Double Check Further Discussion Double Checking with a Computer Algebra System logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

  80. Overview An Example Double Check Further Discussion Double Checking with a Computer Algebra System logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Variation of Parameters

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