$\displaystyle y'' + 2y' = x$ sub to y(0) = 1 and y'(0) = 0
answer is $\displaystyle y = \frac{9}{8} -\frac{x}{4} + \frac{x^2}{4} - \frac{exp(-2x)}{8}$ but cant see how they did it,,,some version of inverse laplace?
$\displaystyle y'' + 2y' = x,~~y(0) = 1,~y'(0) = 0.$
First, find the image of the equation by Laplace:
$\displaystyle \mathcal{L}\left\{y(x)\right\} = y(s),$
$\displaystyle \mathcal{L}\left\{y'(x)\right\} = sy(s) - y(0) = sy(s) - 1,$
$\displaystyle \mathcal{L}\left\{y''(x)\right\} = s^2y(s) - y(0)s - y'(0) = s^2y(s) - s.$
$\displaystyle \mathcal{L}\left\{x\right\} = \frac{1}{s^2}.$
So, you have
$\displaystyle s^2y(s) - s + 2\left(sy(s) - 1\right) = \frac{1}{s^2};$
$\displaystyle \left( s^2 + 2s\right)y(s) = \frac{1}{s^2} + s + 2;$
$\displaystyle y(s) = \frac{s^3 + 2s^2 + 1}{s^3 (s + 2)} = \frac{9}{8s} + \frac{1}{2s^3} - \frac{1}{4s^2} - \frac{1}{8(s + 2)}.$
Finally you have
$\displaystyle y(x) = {\mathcal{L}^{-1}}\left\{y(s)\right\} = \mathcal{L}^{-1}\left\{\frac{9}{8s} + \frac{1}{2s^3} - \frac{1}{4s^2} - \frac{1}{8(s + 2)} \right\} = \frac{9}{8} - \frac{x}{4} + \frac{x^2}{4} - \frac{e^{-2x}}{8}.$