Answer
$y(x)=c_1+c_2e^{-4x}+e^{-4x}$
Work Step by Step
Given: $y''+4y'=4x^{2}$
Solve the auxiliary equation for the differential equation.
$r^2+4r=0$
Factor and solve for the roots.
$r(r+4)=0$
Roots are: $r_1=0$, as a multiplicity of 1 and $r_2=-4$, as a multiplicity of 1
The general solution is
$y(x)=c_1+c_2e^{-4x}$
We have $F(x)=4x^2$
Obtain:
$D^3(D^2+4D)y_p(x)=4x^2$
Therefore, the general solution for $D^4(D+4)y_p(x)=0$ is:
$y(x)=c_1+c_2e^{-4x}+A_0+B_0e^{-4x}$
The trial solution can be computed as by plugging back into the given differential equation.
$y_p(x)=A_0+B_0e^{-4x}$
So, we have:
$D^4(D+4)y_p(x)=4x^{2}\\
D^4(D+4)(A_0+B_0e^{-4x})=4x^{2}$
On comparing coefficients, we get:
$A_0=0\\
B_0=1$
Therefore, the general solution for the given differential equation is:
$y(x)=c_1+c_2e^{-4x}+e^{-4x}$
