Answer
See below
Work Step by Step
Given: $y''-y'-2y=15e^{2x}$
Solve the auxiliary equation for the differential equation.
$r^2-r-2=0$
Factor and solve for the roots.
$(r-2)(r+1)=0$
Roots are: $r_1=-1$, as a multiplicity of 1 and $r_2=2$ as a multiplicity of 1
The general solution is
$y(x)=c_1e^{-x}+c_2e^{2x} $
We have $F(x)=15e^{2x}$
Obtain:
$(D-2)^2(D+1)y_p(x)=15e^{2x}$
Therefore, the general solution for $(D-2)^2(D+1)y_p(x)=0$ is:
$y(x)=c_1e^{-x}+c_2e^{2x}+A_0xe^{2x}$
The trial solution can be computed as by plugging back into the given differential equation.
$y_p(x)=A_0xe^{2x}$
So, we have:
$(D-2)^2(D+1)y_p(x)=15e^{2x}\\
(D-2)^2(D+1)(A_0xe^{2x})=15e^{2x}$
On comparing coefficients, we get:
$A_0=5$
Therefore, the general solution for the given differential equation is:
$y(x)=c_1e^{-x}+c_2e^{2x}+5xe^{2x}$
Since $y(0)=0\\
y'(0)=8$
we obtain:
$c_1+c_2=0\\
-c_1+2c_2=3$
Solve:
$c_1=-1\\
c_2=1$
The solution to the initial problem is
$y(x)=-e^{-x}+e^{2x}+5xe^{2x}$
