Originally Posted by

**danny arrigo** I think there's a problem here. If you want the solution subject to the following IC's, $\displaystyle y(0)=1,\;\;y'(0)=1$, then the solution you give is right (with a slight change- should be $\displaystyle - \frac{1}{3}x^3$). However, if you want the solution with the IC's you give $\displaystyle y(0)=1,\;\;y'(0)=-1$, I believe the correct solution is

$\displaystyle

y(x) = -\frac{1}{2}e^{-x} + \frac{1}{54}e^{3x} - \frac{1}{3}x^3 + \frac{2}{3}x^2 - \frac{14}{9}x + \frac{40}{27}$