Answer
$y=c_{1}e^{-4x}+c_{2}e^{2x}+\dfrac{1}{4}x^2+\dfrac{1}{8}x-\dfrac{1}{32}$
Work Step by Step
Write the auxiliary equation.
$r^{2}+2r-8=0$
$r=-4,2$
Here, we have $y_c=c_{1}e^{-4x}+c_{2}e^{2x}$
$y_{p}=Ax^2+Bx+C$
$y_{p}'=2Ax+B$
$y_{p}''=2A$
The main equation becomes:
$-8Ax^2+(4A-8B)X+2A+2B-8C=-2x^2+1$ ...(1)
$A=\dfrac{1}{4}$ and $4A-8B=0$
This gives: $B=\dfrac{1}{8}$
Now compare the constants in equation (1):
$C=\dfrac{-1}{32}$
$y=y_c+y_p=c_{1}e^{-4x}+c_{2}e^{2x}+\dfrac{1}{4}x^2+\dfrac{1}{8}x-\dfrac{1}{32}$
