Answer
$y(x)=c_1e^x+c_2e^{-3x}-\frac{1}{8}xe^{-3x}-\frac{1}{4}x^2e^{-3x}$
Work Step by Step
Given: $y''+2y'-3y=2xe^{-3x}$
Solve the auxiliary equation for the differential equation.
$r^2+2r-3=0$
Factor and solve for the roots.
$(r-1)(r+3)=0$
Roots are: $r_1=1$, as a multiplicity of 1 and $r_2=-3$, as a multiplicity of 1
The general solution is
$y(x)=c_1e^x+c_2e^{-3x}$
We have $F(x)=2xe^{-3x}$
Obtain:
$(D-1)(D+3)^3y_p(x)=2xe^{-3x}$
Therefore, the general solution for $(D-1)(D^2+4)^3y_p(x)=0$ is:
$y(x)=c_1e^x+c_2e^{-3x}+A_0xe^{-3x}+B_0x^2e^{-3x}$
The trial solution can be computed as by plugging back into the given differential equation.
$y_p(x)=A_0xe^{-3x}+B_0x^2e^{-3x}$
So, we have:
$(D-1)(D+3)^3y_p(x)=2xe^{-3x}\\
(D-1)(D+3)^3(A_0xe^{-3x}+B_0x^2e^{-3x})=2xe^{-3x}$
On comparing coefficients, we get:
$A_0=-\frac{1}{8}\\
B_0=-\frac{1}{4}$
Therefore, the general solution for the given differential equation is:
$y(x)=c_1e^x+c_2e^{-3x}-\frac{1}{8}xe^{-3x}-\frac{1}{4}x^2e^{-3x}$
