Answer
$y(x)=c_1e^{-x}+c_2e^{2x}+\frac{5}{3}xe^{2x}$
Work Step by Step
Given: $y''-y'-2y=5e^{2x}$
Solve the auxiliary equation for the differential equation.
$r^2-r-2r=0$
Factor and solve for the roots.
$(r+1)(r-2)=0$
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 $F(x)=5e^{2x}$
Obtain:
$(D+1)(D-2)^2y_p(x)=5e^{2x}$
Therefore, the general solution for $(D+1)(D-2)^2y_p(x)=0$ 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:
$(D+1)(D-2)^2y_p(x)=5e^{2x}\\
(D+1)(D-2)^2(A_0xe^{2x})=5e^{2x}$
On comparing coefficients, we get:
$A_0=\frac{5}{3}$
Therefore, the general solution for the given differential equation is:
$y(x)=c_1e^{-x}+c_2e^{2x}+\frac{5}{3}xe^{2x}$
