Var_Parameter - 23 Variation of Parameters(A Better...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
23 Variation of Parameters (A Better Reduction of Order Method for Nonhomogeneous Equations) “Variation of parameters” is another way to solve nonhomogeneous linear differential equations, be they second order, ay ′′ + by + cy = g , or even higher order, a 0 y ( N ) + a 1 y ( N 1 ) + · · · + a N 1 y + a N y = g . One advantage of this method over the method of undetermined coefficients from chapter 21 is that the differential equation does not have to be simple enough that we can ‘guess’ the form for a particular solution. In theory, the method of variation of parameters will work whenever g and the coefficients are reasonably continuous functions. As you may expect, though, it is not quite as simple a method as the method of guess. So, for ‘sufficiently simple’ differential equations, you may still prefer using the guess method instead of what we’ll develop here. We will first develop the variation of parameters method for second-order equations. Then we will see how to extend it to deal with differential equations of even higher order. 1 As you will see, the method can be viewed as a very clever improvement on the reduction of order method for solving nonhomogeneous equations. What might not be so obvious is why the method is called “variation of parameters”. 23.1 Second-Order Variation of Parameters Derivation of the Method Suppose we want to solve a second-order nonhomogeneous differential equation ay ′′ + by + cy = g 1 It is possible to use a “variation of parameters” method to solve first-order nonhomogeneous linear equations, but that’s just plain silly. 453
Image of page 1

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

View Full Document Right Arrow Icon
454 Variation of Parameters over some interval of interest, say, x 2 y ′′ 2 xy + 2 y = 3 x 2 for x > 0 . Let us also assume that the corresponding homogeneous equation, ay ′′ + by + cy = 0 , has already been solved. That is, we already have an independent pair of functions y 1 = y 1 ( x ) and y 2 = y 2 ( x ) for which y h ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) is a general solution to the homogeneous equation. For our example, x 2 y ′′ 2 xy + 2 y = 3 x 2 , the corresponding homogeneous equation is the Euler equation x 2 y ′′ 2 xy + 2 y = 0 . You can easily verify that this homogeneous equation is satisfied if y is either y 1 = x or y 2 = x 2 . Clearly, the set { x , x 2 } is linearly independent, and, so, the general solution to the corresponding homogeneous homogeneous equation is y h = c 1 x + c 2 x 2 . Now, in using reduction of order to solve our nonhomogeneous equation ay ′′ + by + cy = g , we would first assume a solution of the form y = y 0 u where u = u ( x ) is an unknown function ‘to be determined’, and y 0 = y 0 ( x ) is any single solution to the corresponding homogeneous equation. However, we do not just have a single solution to the corresponding homogeneous equation — we have two: y 1 and y 2 (along with all linear combinations of these two). So why don’t we use both of these solutions and assume, instead, a solution of the form y = y 1 u + y 2 v where y 1 and y 2 are the two solutions to the corresponding homogeneous equation already found, and u = u ( x )
Image of page 2
Image of page 3
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