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
The auxiliary equation is : $r^{2}+2r-8=0 \implies r=-4,2$
Thus, $y_c=c_{1}e^{-4x}+c_{2}e^{2x}$
Now , the particular solution is: $y_{p}=Ax^2+Bx+C$
$y_{p}'=2Ax+B; y_{p}''=2A$
Write the main equation.
$-8Ax^2+(4A-8B)X+2A+2B-8C=-2x^2+1$
$A=\dfrac{1}{4}$ and $4A-8B=0 \implies B=\dfrac{1}{8}$
Compare the constants and we get $B=\dfrac{-1}{32}$
$y=y_c+y_p$
or, $y=c_{1}e^{-4x}+c_{2}e^{2x}+\dfrac{1}{4}x^2+\dfrac{1}{8}x-\dfrac{1}{32}$
