# Help with finding particular solution of a 2nd order linear ode

1. Jul 6, 2011

### iqjump123

1. The problem statement, all variables and given/known data
obtain the general solution y(x) of
y''-2y'+y=e^(2x)/(e^x+1)^2

2. Relevant equations
variation of parameters

3. The attempt at a solution
I have obtained the continuous equation.
I tried two methods of variation of parameters, but both of them got me stuck.

1. using the wronskian method. However, this method gave me an answer of 0.
2. Using the method involving setting a'(x) and b'(x) to 0- however, this method doesn't work if the complimentary solution involves a multiple of x (b'(x)xe^x).

This one seems to be more complicated than a simple constant of coefficients- wolfram got me a straightforward answer, however. Any help will be appreciated. TIA!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 7, 2011

### Ray Vickson

My personal preference in such cases is to always use the Green's Function method. In your case, let Ly(x) be the operator on the left of your DE. To solve Ly(x) = f(x), try solving LG(x) = delta(x-t) for G = G(x;t), where delta(x-t) = Dirac delta and t is just a parameter. Then Y(x) = integral_{t=-infinity..infinity} G(x;t) f(t) dt is a particular solution of LY(x) = f(x).

For more on Green's functions and the step-by-step solution of a problem almost like yours, see the pedagogical article http://www.mathphysics.com/pde/green/g15.html . This article also relates the Green's function to the Wronskian.

RGV

3. Aug 22, 2011

### iqjump123

Hey Ray Vickson,

I appreciate your help. It is just that I haven't tried using greens function to solve an ODE, and therefore would like to see if there is an easier and/or more straightforward way?

4. Aug 22, 2011

### Screwdriver

The Wronskian method didn't give me zero. You should find that:

$$W(e^{x}, x e^{x}) = e^{2x}$$

Then the integrals are:

$$Y_{p}(x) = -e^{x} \int \frac{x e^{x}}{(e^{x}+1)^2}dx + xe^{x} \int \frac{e^{x}}{(e^{x}+1)^2}dx$$