Method of Variation of Params SLP 9_06_11

Method of Variation of Params SLP 9_06_11 - Prof S.L...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Prof. S.L. Phoenix 9/06/11 Math 3100 Method of Variation of Parameters for Higher Order Linear ODE’s Suppose we have the equation     1 1 1 1 n n n n L y y p x y p x y p x y g x  (1) which is eqn. (2) in E&P, except we use g x instead of f x . From the homogeneous equation 0 L y we can get n linearly independent complementary solutions 1 2 , ,..., n y x y x y x . To get a particular solution, ( ) y x Y x , we develop a general useful formula based on the method of variation of parameters. This method begins with the form 1 1 2 2 n n Y x u x y x u x y x u x y x   (2) where 1 2 , , , n u x u x u x  arbitrary, but so far unknown functions. At this stage, this form is general enough to cover all possibilities for the particular solution. In fact it looks like overkill , but as we shall see, we will eventually apply 1 n constraints. To be a particular solution we must get the form (2) to satisfy the differential equation (1). This means we will need to take n derivatives of Y x , which through repeated applications of the chain rule will cause a big mess unless we make some clever simplifications that we can get to work out in the end. For the first derivative we use the chain rule of differentiation to give us 1 1 2 2 1 1 2 2 n n n n Y x u x y x u x y x u x y x u x y x u x y x u x y x     (3) but we simplify this by setting the last half of the sum equal to zero, i.e., we apply the constraint 1 1 2 2 0 n n u x y x u x y x u x y x   (4) Thus we now only have 1 1 2 2 n n Y x u x y x u x y x u x y x   (5) Taking a derivative of (5) we get
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

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