Answer
See below
Work Step by Step
Given: $y''+y'-2y=4x^2+5$
Substitute: $y''(x)+y'(x)-2y(x)\\
=2A_2+2A_2x+A_1-2(A_0+A_1x+A_2x^2)\\
=x^2(-2A_2)+x(2A_2-2A_1)+(2A_2+A_1-2A_0)\\
=4x^2+5$
Then $-2A_2=4\\
2A_2-2A_1=0\\
2A_2+A_1-2A_0=5$
Then $A_0=-\frac{9}{2} \\
A_1=-2\\
A_2=-2$
Hence, $y_p(x)=-2x^2-2x-\frac{9}{2}$
Find the solution.
Substitute: $r^2e^{rx}+re^{rx}-2e^{rx}$
Put: $r^2e^{rx}+re^{rx}-2e^{rx}=0\\
e^{rx}(r^2+r-2)=0 \\$
Since $e^{rx} \ne0 \rightarrow r^2+r-2=0\\
\rightarrow (r+2)(r-1)=0\\
\rightarrow r=-2,r=1$
We found $y_1(x)=C_1e^{-2x}\\
y_2(x)=C_2e^{x}\\
$
The solutions to the given problem are: $y=C_1e^{-2x}+C_2e^{x}+2x^2-2x-\frac{9}{2}$
