Answer
$y(x)=c_1e^x+c_2e^{-x}-x^3-5x$
Work Step by Step
The auxiliary equation is:
$r^2-1=0$
We factor:
$(r+1)(r-1)=0$
Thus $r=\pm 1$
Therefore, the complementary equation is:
$y_c(x)=c_1*e^x+c_2*e^{-x}$
We attempt:
$y_p(x)=Ax^3+Bx^2+Cx+D$
Thus:
$y_p'(x)=3Ax^2+2Bx$
$y_p''(x)=6Ax+2B$
We plug into the differential equation:
$(6Ax+2B)-(Ax^3+Bx^2+Cx+D)=x^3-x$
If we examine the coefficients, we get:
$A=-1, B=0, C=-5, D=0$
Therefore, the general solution is:
$y(x)=c_1e^x+c_2e^{-x}-x^3-5x$
