Answer
See below
Work Step by Step
Given: $y''+y'-6y=18e^{5x}$
Substitute: $y''(x)+y'(x)-6y(x)\\
=25A_0e^{5x}+5A_0e^{5x}-6A_0e^{5x}\\
=24A_0e^{5x}=18A_0e^{5x}$
Then $A_0=\frac{18}{24}=\frac{3}{4}$
Hence, $y_p(x)=\frac{3}{4}e^{5x}$
Find the solution.
Substitute: $r^2e^{rx}+re^{rx}-6e^{rx}$
Put: $r^2e^{rx}+re^{rx}-6e^{rx}=0\\
e^{rx}(r^2+r-6)=0 \\$
Since $e^{rx} \ne0 \rightarrow r^2+r-6=0\\
\rightarrow (r+3)(r-2)=0\\
\rightarrow r=3,r=2$
We found $y_1(x)=C_1e^{-3x}\\
y_2(x)=C_2e^{2x}\\
$
The solutions to the given problem are: $y=C_1e^{-3x}+C_2e^{2x}+\frac{3}{4}e^{5x}$
