variation of parameters
play

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

Jul 26, 2023 •263 likes •1.39k 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

Recommend

UMBC A B M A L T F O U M B C I M Y O R T 1 (10/13/04) I E S R C E O V

UMBC A B M A L T F O U M B C I M Y O R T 1 (10/13/04) I E S R C E O V

Advanced VLSI Design Process Variations CMPE 640 Process Variations The parameters of individual transistors vary from: Lot to lot (interprocess variation) Wafer to wafer (interprocess variation) Die to die (intraprocess variation)

321 views • 8 slides
Nonhomogeneous linear systems of DEs Diagonalization, Variation of Parameters ITI 11/04/2020

Nonhomogeneous linear systems of DEs Diagonalization, Variation of Parameters ITI 11/04/2020

Diagonalization Variation of parameters Nonhomogeneous linear systems of DEs Diagonalization, Variation of Parameters ITI 11/04/2020 ITI Nonhomogeneous systems Diagonalization Variation of parameters Diagonalization 1 Variation of

432 views • 25 slides
Lecture 4.9: Variation of parameters for systems Matthew Macauley Department of Mathematical

Lecture 4.9: Variation of parameters for systems Matthew Macauley Department of Mathematical

Lecture 4.9: Variation of parameters for systems Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 2080, Differential Equations M. Macauley (Clemson) Lecture 4.9: Variation of

394 views • 6 slides
abatement cost curves to variation of G4M parameters Mykola Gusti, Nikolay Khabarov and Nicklas

abatement cost curves to variation of G4M parameters Mykola Gusti, Nikolay Khabarov and Nicklas

Sensitivity of marginal abatement cost curves to variation of G4M parameters Mykola Gusti, Nikolay Khabarov and Nicklas Forsell 4th International Workshop on Uncertainty in Atmospheric Emissions, Krakw , 7th-9th October, 2015 Acknowledgements.

658 views • 46 slides
12 th November 2018 Terminal Business Mix [ Amt. in ` ` Crore] Y-o-Y Variation Sl. No.

12 th November 2018 Terminal Business Mix [ Amt. in ` ` Crore] Y-o-Y Variation Sl. No.

12 th November 2018 Terminal Business Mix [ Amt. in ` ` Crore] Y-o-Y Variation Sl. No. Parameters Sep-18 Jun-18 Sep-17 Amount % 1 Business 298823 294065 312175 -13352 -4.28 Domestic 282257 274424 280393 1864 0.66 Overseas

601 views • 16 slides
Me & the UO Language Variation & Computation Lab Sociophonetician and

Me & the UO Language Variation & Computation Lab Sociophonetician and

T. Kendall (U Oregon) Social and Cognitive Aspects of Language Variation and Change Me & the UO Language Variation & Computation Lab Sociophonetician and sociolinguist In terms of speech technology, researching variation and

202 views • 5 slides
Assessing the Statistical Power/Precision Of Multisite Trials for Estimating Parameters Of

Assessing the Statistical Power/Precision Of Multisite Trials for Estimating Parameters Of

Assessing the Statistical Power/Precision Of Multisite Trials for Estimating Parameters Of Crosssite ITT Impact Distributions Howard Bloom Jessaca Spybrook Presented to the Workshop on Learning about and from Variation in Program

664 views • 23 slides
Prof. Dr. Atilla Ansal MICROZONATION MAIN PURPOSE: To estimate variation of the selected

Prof. Dr. Atilla Ansal MICROZONATION MAIN PURPOSE: To estimate variation of the selected

MICROZONATION WITH RESPECT GEOTECHNICAL HAZARDS Lecture Notes by Prof. Dr. Atilla Ansal MICROZONATION MAIN PURPOSE: To estimate variation of the selected parameters for land use and urban planning for the mitigation of earthquake risk to

1.1k views • 84 slides
Camera Parameters INEL 6088 Computer Vision Camera Parameters Extrinsic parameters: define

Camera Parameters INEL 6088 Computer Vision Camera Parameters Extrinsic parameters: define

Camera Parameters INEL 6088 Computer Vision Camera Parameters Extrinsic parameters: define the location and orientation of the camera with respect to the world reference frame Intrinsic parameters: link the pixel coordinates of an image

498 views • 15 slides
Outline Introduction Variation Among Batches Variation Within Batches Experimenting

Outline Introduction Variation Among Batches Variation Within Batches Experimenting

Dennis R. Buckmaster Purdue University Agricultural & Biological Engineering Outline Introduction Variation Among Batches Variation Within Batches Experimenting on the farm How Example analysis Summary 1 Goals of

583 views • 22 slides
Statistics in medicine Sources of variation Lecture 1- part 1: Describing variation, and

Statistics in medicine Sources of variation Lecture 1- part 1: Describing variation, and

11/4/2016 Outline Statistics in medicine Sources of variation Lecture 1- part 1: Describing variation, and Types of variables graphical presentation Fatma Shebl, MD, MS, MPH, PhD Assistant Professor Chronic Disease Epidemiology

791 views • 23 slides
Two-Port Networks Definitions Impedance Parameters Admittance Parameters Hybrid

Two-Port Networks Definitions Impedance Parameters Admittance Parameters Hybrid

Two-Port Networks Definitions Impedance Parameters Admittance Parameters Hybrid Parameters Transmission Parameters Cascaded Two-Port Networks Examples Applications J. McNames Portland State University ECE 222

373 views • 33 slides
Power variation and p -variation of sample functions of stochastic processes Rimas Norvai sa

Power variation and p -variation of sample functions of stochastic processes Rimas Norvai sa

Power variation and p -variation of sample functions of stochastic processes Rimas Norvai sa Department of Econometric Analysis Vilnius University 15 November 2014, Guanajuato, Mexico 5th Workshop on Game-Theoretic Probability and Related

739 views • 20 slides
Tracking Predictable Drifting Parameters Paulo Serra of a Time Series The Model Joint work with

Tracking Predictable Drifting Parameters Paulo Serra of a Time Series The Model Joint work with

Tracking Predictable Drifting Parameters of a Time Series Tracking Predictable Drifting Parameters Paulo Serra of a Time Series The Model Joint work with Eduard Belitser Results Examples of Parameter Variation Paulo Serra

348 views • 22 slides
Variation in sampling The death of toxicology Variation in sampling The death of

Variation in sampling The death of toxicology Variation in sampling The death of

Variation in sampling The death of toxicology Variation in sampling The death of toxicology Jennifer Button Product Specialist Phytochemistry, Forensic & Clinical TIAFT 2013 Funchal Madeira Garbage in, Garbage out (GIGO)

326 views • 18 slides
Molecular g gas a as a a d dynamical IMF v variation probe o of I IMF v variation Timo

Molecular g gas a as a a d dynamical IMF v variation probe o of I IMF v variation Timo

Molecular g gas a as a a d dynamical p probe o of Molecular g gas a as a a d dynamical IMF v variation probe o of I IMF v variation Timo mothy A y A. Da . Davis Ernest Rutherford Fellow & Lecturer, Cardiff University With

1.04k views • 37 slides
2.13 Point Kinematics Having introduced in the previous Sections the different representations of

2.13 Point Kinematics Having introduced in the previous Sections the different representations of

71 Basilio Bona - Dynamic Modelling If we use the quaternions, the Euler parameters, the Cayley-Klein parameters or the Rodrigues vectors, it in unclear what are the parameters that take the place of the i in (2.65). Using these alternative

222 views • 11 slides
One-Port v Admittance Parameters Network - Hybrid Parameters i' 1 Transmission

One-Port v Admittance Parameters Network - Hybrid Parameters i' 1 Transmission

Two-Port Networks One-Port Networks Definitions i 1 + Impedance Parameters One-Port v Admittance Parameters Network - Hybrid Parameters i' 1 Transmission Parameters A pair of terminals at which a signal (voltage or

238 views • 9 slides
Why all of this talk of populations, parameters, samples, and statistics? For simplicity, lets

Why all of this talk of populations, parameters, samples, and statistics? For simplicity, lets

Today - A little on standard error of the mean and variation in estimates of central tendency. - A rough and ready primer on linear mixed effects models, a commonly- used tool for statistical analysis of experimentally collected data in

503 views • 38 slides
Lake Pleasant Limnology and Down-Canal Water Quality Implications Spatial Variation in

Lake Pleasant Limnology and Down-Canal Water Quality Implications Spatial Variation in

Lake Pleasant Limnology and Down-Canal Water Quality Implications Spatial Variation in Reservoirs Thermal stratification (latitudinal variation) Longitudinal variation from incoming river to the dam A priori Information Taste and

989 views • 40 slides
all your variation points for free? Variation points Design for change/easily accomodated to

all your variation points for free? Variation points Design for change/easily accomodated to

Coordinate-free numerics: all your variation points for free? Variation points Design for change/easily accomodated to change Different PDEs Discretisations Numerical methods Time stepping technique Underlying grid

539 views • 9 slides
Modern human variation (Ch 12) There's variation in ALL humans Polytypic species: composed of

Modern human variation (Ch 12) There's variation in ALL humans Polytypic species: composed of

Modern human variation (Ch 12) There's variation in ALL humans Polytypic species: composed of populations that differ in the expression of one or more traits -Geographically variable pattern among human populations 1 How anthropologists

401 views • 23 slides
Block 9 Section 12 Hackett Territory Plan variation and Crown lease variation to regularise the

Block 9 Section 12 Hackett Territory Plan variation and Crown lease variation to regularise the

Block 9 Section 12 Hackett Territory Plan variation and Crown lease variation to regularise the existing use October 2016 Introductions 1. Key people 2. The site and its context 3. The Christian Community 4. Our current use of the site 5.

469 views • 18 slides
Documenting Interaction and Variation in Ampenan Sasak Khairunnisa Bradley McDonnell InLaLi

Documenting Interaction and Variation in Ampenan Sasak Khairunnisa Bradley McDonnell InLaLi

Documenting Interaction and Variation in Ampenan Sasak Khairunnisa Bradley McDonnell InLaLi February 18, 2020 Documenting variation Studying language variation has been an important part of linguistics, but it is often restricted to

430 views • 22 slides

More recommend