Answer
See below
Work Step by Step
Given: $y'''+y''-10y'+8y=24e^{-3x}$
Substitute: $y'''(x)+2y''(x)-y'(x)-2y(x)\\
=-27A_0e^{-3x}+9A_0e^{-3x}+30A_0e^{-3x}+8A_0e^{-3x}\\=20A_0e^{-3x}\\
=24A_0e^{-3x}$
Then $A_0=\frac{24}{20}=\frac{6}{5}$
Hence, $y_p(x)=\frac{6}{5}e^{-3x}$
Find the solution.
Substitute: $r^3e^{rx}+2r^2e^{rx}-re^{rx}-2e^{rx}$
Put: $r^3e^{rx}+r^2e^{rx}-10re^{rx}+8e^{rx}=0\\
e^{rx}(r^3+r^2-10r+8)=0 \\$
Since $e^{rx} \ne0 \rightarrow r^3+r^2-10r+8=0\\
\rightarrow (r+4)(r-1)(r-2)=0\\
\rightarrow r=1,r=2,r=-4$
We found $y_1(x)=C_1e^{x}\\
y_2(x)=C_2e^{2x}\\
y_3(x)=C_3e^{-4x}$
The solutions to the given problem are: $y=C_1e^{x}+C_2e^{2x}+C_3e^{-4x}+\frac{6}{5}e^{-3x}$
