2nd order ODE with constant coefficients and tricky RHS

[Batmaniac]

Homework Statement

$$\frac {d^2x} {dt^2} -x = te^{-t}$$

Find the general solution (I only need help on finding the particular solution).

Homework Equations

Well, I can easily find the complimentary solution via the characteristic equation, my problem lies in the particular solution.

If the RHS (right hand side) is a simple exponential function, I know to guess $$Ae^{-t}$$ then take the derivatives and sub them back into the equation then solve for A. I also know if the RHS is a simple polynomial of t, I guess that the particular solution is of the form $$At^{n} + Bt^{n-1} + ... +$$.

The Attempt at a Solution

Given my above understanding, I tried guessing that the solution is $$Ate^{-t} + Bt$$ and $$Ate^{-t} + Be^{-t}$$ and both of them gave me no solution for B when trying to equate coefficients. So I have no idea what my guess should be.

Any help? Thanks!

 

[mighty2000]

I can understand your frustration with finding the particular solution for this differential equation. It can be tricky at times, but I am here to help you through it.

First, let's rewrite the equation in a more familiar form:
x'' - x = te^(-t)

Now, we can use the method of undetermined coefficients to find the particular solution. Since the RHS is a product of t and e^(-t), we will try a particular solution of the form At^2e^(-t). This is because the highest power of t in the RHS is 1, so we need to have a term with a power of t that is at least 1 higher in our particular solution.

Now, let's plug this into the equation and solve for A:
x'' - x = te^(-t)
(2Ae^(-t) - At^2e^(-t)) - (At^2e^(-t) - At^2e^(-t)) = te^(-t)
2Ae^(-t) = te^(-t)
A = 1/2

Therefore, our particular solution is x_p = (1/2)t^2e^(-t).

I hope this helps you in finding the particular solution for this differential equation. Keep practicing and you will become more confident in solving these types of problems. Good luck with your studies!