VariationOfParameters

# VariationOfParameters - u xu 1 2 = 2 2 2 1 2 2 u xu u = u...

This preview shows page 1. Sign up to view the full content.

VARIATION OF PARAMETERS Method for solving a non-homogeneous second order differential equation This method is more difficult than the method of undetermined coefficients but is useful in solving more types of equations such as this one with repeated roots. 49. p185 ′′ - ′ + = y y y e x 4 4 2 2 Find a particular solution. The characteristic equation of the associated homogeneous equation is: r r 2 4 4 0 - + = which factors to ( )( ) r r - - = 2 2 0 having repeated roots r = 2 . so that the general solution of the associated homogeneous equation is y c e c xe c x x = + 1 2 2 2 From this equation we determine our y 1 and y 2 values to be y e x 1 2 = and y xe x 2 2 = from which we get y e x 1 2 2 ' = and y xe e x x 2 2 2 2 ' = + From the textbook we have u y u y 1 1 2 2 0 ' ' + = and u y u y f x 1 1 2 2 ' ' ' ' ( ) + = into which we may substitute u e u xe x x 1 2 2 2 0 ' ' + = u e u xe e e x x x x 1 2 2 2 2 2 2 2 2 ' '( ) + + = reduce and solve simultaneously
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: u xu 1 2 ' ' + = 2 2 2 1 2 2 u xu u ' ' ' + + = u xu 1 2 ' ' = -2 2 2 2 2 2 ( ' ) ' '-+ + = xu xu u u x 1 2 ' = -u 2 2 ' = Integrate to find u 1 and u 2 u x dx x 1 2 2 = -= -∫ u dx x 2 2 2 = = ∫ Using the formula for a particular solution y x u y u y p ( ) = + 1 1 2 2 we have y x x e x xe p x x ( ) ( ) = -+ 2 2 2 2 which simplifies to the expression y x x e p x ( ) = 2 2 We then append this term to the general solution for the associated homogeneous equation to obtain the general solution y c e c xe x e c x x x = + + 1 2 2 2 2 2 If initial values are given, they can be applied to this solution and its derivative....
View Full Document

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern