Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE

Technique of variation of parameters main theorem for

This preview shows page 13 - 18 out of 27 pages.

Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Motivating Example 1 Motivating Example: Consider the nonhomogeneous problem y 00 + 4 y = 3 csc( t ) , which is inappropriate for the Method of Undetermined Coefficients The homogeneous solution is y c ( t ) = c 1 cos(2 t ) + c 2 sin(2 t ) Generalize this solution to the form y ( t ) = u 1 ( t ) cos(2 t ) + u 2 ( t ) sin(2 t ) , where the functions u 1 and u 2 are to be determined Differentiate y 0 ( t ) = - 2 u 1 ( t ) sin(2 t ) + 2 u 2 ( t ) cos(2 t ) + u 0 1 ( t ) cos(2 t ) + u 0 2 ( t ) sin(2 t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (13/27)
Image of page 13

Subscribe to view the full document.

Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Motivating Example 2 Motivating Example: The general solution has the form y ( t ) = u 1 ( t ) cos(2 t ) + u 2 ( t ) sin(2 t ) Since there is one general solution, there must be a condition relating u 1 and u 2 The computations are simplified by taking the relationship u 0 1 ( t ) cos(2 t ) + u 0 2 ( t ) sin(2 t ) = 0 This simplifies the derivative of the general solution to y 0 ( t ) = - 2 u 1 ( t ) sin(2 t ) + 2 u 2 ( t ) cos(2 t ) Differentiating again yields: y 00 ( t ) = - 4 u 1 ( t ) cos(2 t ) - 4 u 2 ( t ) sin(2 t ) - 2 u 0 1 ( t ) sin(2 t )+2 u 0 2 ( t ) cos(2 t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (14/27)
Image of page 14
Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Motivating Example 3 Motivating Example: The differential equation is y 00 + 4 y = 3 csc( t ) , so substituting the general solution gives - 4 u 1 ( t ) cos(2 t ) - 4 u 2 ( t ) sin(2 t ) - 2 u 0 1 ( t ) sin(2 t ) +2 u 0 2 ( t ) cos(2 t ) + 4( u 1 ( t ) cos(2 t ) + u 2 ( t ) sin(2 t )) = 3 csc( t ) , which simplifies to - 2 u 0 1 ( t ) sin(2 t ) + 2 u 0 2 ( t ) cos(2 t ) = 3 csc( t ) This equation is combined with our earlier simplifying condition u 0 1 ( t ) cos(2 t ) + u 0 2 ( t ) sin(2 t ) = 0 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (15/27)
Image of page 15

Subscribe to view the full document.

Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Motivating Example 4 Motivating Example: The previous equations give two linear algebraic equations in u 0 1 and u 0 2 u 0 1 ( t ) cos(2 t ) + u 0 2 ( t ) sin(2 t ) = 0 - 2 u 0 1 ( t ) sin(2 t ) + 2 u 0 2 ( t ) cos(2 t ) = 3 csc( t ) The first equation gives u 0 2 ( t ) = - u 0 1 ( t ) cot(2 t ) It follows that (with trig identities) u 0 1 ( t ) = - 3 2 csc( t ) sin(2 t ) = - 3 cos( t ) and (with trig identities) u 0 2 ( t ) = 3 cos( t ) cot(2 t ) = 3 2 csc( t ) - 3 sin( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (16/27)
Image of page 16
Cauchy-Euler Equation Review Variation of Parameters Motivating Example Technique of Variation of Parameters Main Theorem for Nonhomogeneous DE Motivating Example 5 Motivating Example: We solve the equations for u 0 1 and u 0 2 u 0 1 ( t ) = - 3 cos( t ) , so u 1 ( t ) = - 3 sin( t ) + c 1 Simlarly, u 0 2 ( t ) = 3 2 csc( t ) - 3 sin( t ) , so u 2 ( t ) = 3 2 ln | csc( t ) - cot( t ) | + 3 cos( t ) + c 2 It follows that the general solution is y ( t ) = u 1 ( t ) cos(2 t ) + u 2 ( t ) sin(2 t ) = - 3 sin( t ) cos(2 t ) + 3 2 sin(2 t ) ln | csc( t ) - cot( t ) | + 3 cos( t ) sin(2 t ) + c 1 cos(2 t ) + c 2 sin(2
Image of page 17

Subscribe to view the full document.

Image of page 18
  • Fall '08
  • staff

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Ask Expert Tutors You can ask You can ask ( soon) You can ask (will expire )
Answers in as fast as 15 minutes