Answer
$y(x)=c_1e^{-2x}+c_2xe^{-2x}+\frac{5}{6}x^3e^{-2x}$
Work Step by Step
Given: $y''+4y'+4y=5xe^{-2x}$
Solve the auxiliary equation for the differential equation.
$r^2+4r+4=0$
Factor and solve for the roots.
$(r+2)^2=0$
Roots are: $r_1=-2$, as a multiplicity of 2
The general solution is
$y(x)=c_1e^{-2x}+c_2xe^{-2x}$
We have $F(x)=5xe^{-2x}$
Obtain:
$(D+2)^4y_p(x)=5xe^{-2x}$
Therefore, the general solution for $(D+2)^4y_p(x)=0$ is:
$y(x)=c_1e^{-2x}+c_2xe^{-2x}+A_0x^2e^{-2x}+B_0x^3e^{-2x}$
The trial solution can be computed as by plugging back into the given differential equation.
$y_p(x)=A_0x^2e^{-2x}+B_0x^3e^{-2x}$
So, we have:
$(D+2)^4y_p(x)=5xe^{-2x}\\
(D+2)^4(A_0x^2e^{-2x}+B_0x^3e^{-2x})=5xe^{-2x}$
On comparing coefficients, we get:
$A_0=0\\
B_0=\frac{5}{6}$
Therefore, the general solution for the given differential equation is:
$y(x)=c_1e^{-2x}+c_2xe^{-2x}+\frac{5}{6}x^3e^{-2x}$
