## Calculus 8th Edition

$y=c_{1}e^{-4x}+c_{2}e^{2x}+\dfrac{1}{4}x^2+\dfrac{1}{8}x-\dfrac{1}{32}$
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, $y=c_{1}e^{-4x}+c_{2}e^{2x}+\dfrac{1}{4}x^2+\dfrac{1}{8}x-\dfrac{1}{32}$