Answer
See below
Work Step by Step
Given: $y'''+2y''-y'-2y=4e^{2x}$
Substitute: $y'''(x)+2y''(x)-y'(x)-2y(x)\\
=8A_0e^{2x}+8A_0e^{2x}-2A_0e^{2x}-2A_0e^{2x}\\=
12A_0e^{2x}\\
=4e^{2x}$
Then $A_0=\frac{4}{12}=\frac{1}{3}$
Hence, $y_p(x)=\frac{1}{3}e^{2x}$
Find the solution.
Substitute: $r^3e^{rx}+2r^2e^{rx}-re^{rx}-2e^{rx}$
Put: $r^3e^{rx}+2r^2e^{rx}-re^{rx}-2e^{rx}=0\\
e^{rx}(r^3+2r^2-r-2)=0 \\$
Since $e^{rx} \ne0 \rightarrow r^3+2r^2-r-2=0\\
\rightarrow (r+2)(r-1)(r+1)=0\\
\rightarrow r=-2,r=1,r=-1$
We found $y_1(x)=C_1e^{x}\\
y_2(x)=C_2e^{-x}\\
y_3(x)=C_3e^{-2x}$
The solutions to the given problem are: $y=C_1e^{x}+C_2e^{-x}+C_3e^{-2x}+\frac{1}{3}e^{2x}$
