# Linear Differential Equation $y'''−3y′+2y=\cos t+e^t$

I'm trying to find the solution to this non-homogenous third-order linear differential equation.

I know the solution is supposed to be:

$$c_1e^t+c_2te^t+c_3^{-2t}+\frac{e^tt^2}{6}-\frac{\sin(t)}{5}+\frac{\cos(t)}{10}$$

So far I've solved the left side of the equation to get the first half of the answer:

$$c_1e^t+c_2te^t+c_3^{-2t}$$

I don't know how to get the solutions from the right-hand side though. Thanks.

• Have you ever encountered the method of undetermined coefficients? – JohnK Oct 26 '13 at 20:22

You can try the method of undetermined coefficients by setting $y_p= Ae^t+Bte^t+Ct^2e^t+D\sin t+E\cos t$. You take the derivatives of $y_p$ and substitute into the equation. By equating the coefficients of same terms you can determine constants $A, B, C, D$ and $E$.
• Not quite - you're missing a $t^2$ term.. – nbubis Oct 26 '13 at 20:27
• Thank you nbubis I edited it. I supposed $r=1$ is the single root of the caharacteristic equation. – daulomb Oct 26 '13 at 20:32