# Finding a Particular Solution to $u''+2u'+u=9x^2e^{-x}$

I am trying to find a particular solution of the ODE $$u''+2u'+u=9x^2e^{-x}.$$

I know the homogeneous solutions is of the form $$u_H(x)=Ae^{-x}+Bxe^{-x} \ \ \ A,B\in\mathbb{R}.$$ For a particular solution, I could use the method of undetermined coefficients and guess that $$u_P(x)=x^2(Cx^2+Dx+E)e^{-x} \ \ \ C,D,E\in\mathbb{R}.$$ I believe this would yeild an answer, but the computation required is very tedious.

I was wondering if anyone had a suggestion for another way to find a particular solution when the forcing term (RHS) is a product of an exponential and a polynomial.

## 2 Answers

For th particular solution, make $$u(x)=e^{-x}\,y(x)$$ to get $$e^{-x} \left(y''(x)-9 x^2\right)=0 \implies y''(x)=9 x^2$$ which looks easy.

• I'm not sure how this is easier. – user557493 Oct 29 '18 at 9:42

Starting from your homogeneous solution, consider what happens if $$A$$ is not a constant but instead some function $$A(x)$$. This gives the function $$u(x) = A(x)e^{-x} + Bxe^{-x}$$ Then, computing you get (if I'm not mistaken) $$u'' + 2u' + u = e^{-x} A''(x)$$ We therefore want to choose $$A(x) = 3x^4/4$$ (for instance). A direct calculation confirms that this works (note that this is a not a unique solution, as you should expect).