(Wrong site) Can't find mistake in resolution of second order non-homogenous linear differential equation I'm trying to solve: $y'' + 10y' + 25y = (3x+2)e^{-5x}$
I start by rewriting as $(D^2 + 10D + 25)y = (3x + 2)e^{-5x}$
We have a non-distinct real root of multiplicity 2 which is -5.
The homogeneous solution is then $y_{h} = (C_{1} + C_{2})e^{-5x}$
We now find the right candidate for the undermined coefficients method.
$F(x) = (3x+2)e^{-5x}$
$(3x+2)$ is in the form of a first-degree polynomial and $e^{-5x}$ is an exponential function.
So our particular solution is $y_{p} = (Ax + B)e^{-5x}$
My problem is that when I replace $y_{p}$ into the differential equation of line 1, I get 0, so I can't continue the problem...
Thank you.
 A: You could try finding the solution of this ODE with Mathematica:
eqn = y''[x] + 10 y'[x] + 25 y[x] == (3 x + 2) Exp[-5 x]
y[x] /. First@DSolve[eqn, y, x]
% // Factor
(* 1/2 \[ExponentialE]^(-5 x) (2 x^2+x^3+2 Subscript[\[ConstantC], \
1]+2 x Subscript[\[ConstantC], 2]) *)

A: Homogeneous part:
If we propose $y=e^{\lambda x}$ as a solution, then the characteristic equation is given by $\left(\lambda + 5 \right)^{2}=0$, whose solution is $\lambda=-5$ with multiplicity two. So, the solution to the homogeneous part has the following form
$$ y_{h}\left(x\right)= \left(c_{1} + c_{2} x\right)e^{-5x}. $$
Inhomogeneous part:
The inhomogeneous term should be considered to be a polynomial of degree one by $e^{-5x}$, also $e^{-5x}$ and $xe^{-5x}$ are solutions of the homogeneous part. We propose as a particular solution the next function
$$y_{p}=x^{2}\left( A x + B\right)e^{-5x}, $$
with $A$ and $B$ to be determined. After some calculations, we obtain the following linear system for $A$ and $B$
$$2 B + 6 A x = 2 + 3 x, $$
whose solution is: $A=\displaystyle \frac{1}{2}$ and $B=1$. Therefore, the particular solution is
$$ y_{p}\left(x\right)=\frac{1}{2} e^{-5 x} x^2 (x+2),$$
and the general solution is given by
$$y_{g}\left( x\right)=\frac{1}{2} e^{-5 x} \left(2 (c_{1}+c_{2}x)+x^2(x+2)\right).$$
A: Misprint:
The characteristic equation is given by $\left(\lambda^{2}+5\right)$, whose solution is $\lambda=-5$ with multiplicity two.
Sorry it's my post (Eduardo), but I didn't read that this question was in another community.
