Answer
See below
Work Step by Step
Given: $y'''+5y''+6y'=6e^{-x}$
Substitute: $y'''(x)+5y''(x)+6y'(x)\\
=-A_0e^{-x}+5A_0e^{-x}-6A_0e^{-x}\\=-2A_0e^{-x}\\
=6A_0e^{-x}$
Then $A_0=\frac{-2}{6}=\frac{-1}{3}$
Hence, $y_p(x)=-\frac{1}{3}e^{-x}$
Find the solution.
Substitute: $r^3e^{rx}+5r^2e^{rx}+6re^{rx}$
Put: $r^3e^{rx}+5r^2e^{rx}+6re^{rx}=0\\
e^{rx}(r^3+5r^2+6r)=0 \\$
Since $e^{rx} \ne0 \rightarrow r^3+5r^2+6r=0\\
\rightarrow r(r+2)(r+3)=0\\
\rightarrow r=0,r=-2,r=-3$
We found $y_1(x)=C_1e^{0x}=C_1\\
y_2(x)=C_2e^-{2x}\\
y_3(x)=C_3e^{-3x}$
The solutions to the given problem are: $y=C_1+C_2e^{-2x}+C_3e^{-3x}-\frac{1}{3}e^{-x}$
