Answer
$y(x)=c_1e^{4x}+c_2xe^{4x}+\frac{5}{2}x^2e^{4x}$
Work Step by Step
Given: $y''-8y'+16y=5e^{4x}$
Solve the auxiliary equation for the differential equation.
$r^2-8r+16=0$
Factor and solve for the roots.
$(r-4)^2=0$
Roots are: $r_1=4$, as a multiplicity of 2
The general solution is
$y(x)=c_1e^{4x}+c_2xe^{4x} $
We have $F(x)=8\cos 2x$
Obtain:
$(D-4)^2y_p(x)=5e^{4x}$
Therefore, the general solution for $(D-4)^2y_p(x)=0$ is:
$y(x)=c_1e^{4x}+c_2xe^{4x}+A_0x^2e^{4x}$
The trial solution can be computed as by plugging back into the given differential equation.
$y_p(x)=A_0x^2e^{4x}$
So, we have:
$(D-4)^2y_p(x)=5e^{4x}\\
(D-4)^2(A_0x^2e^{4x})=5e^{4x}$
On comparing coefficients, we get:
$A_0=\frac{5}{2}$
Therefore, the general solution for the given differential equation is:
$y(x)=c_1e^{4x}+c_2xe^{4x}+\frac{5}{2}x^2e^{4x}$
