Originally Posted by

**dankelly07** Just wondering if anyone could have a look through nd see if I've done this right? When it comes to the general solution & conditon I'm not sure if I've done this right.

$\displaystyle

\begin{gathered}

y' - 2xy = x \hfill \\

y' = 2xy + x \hfill \\

\smallint \frac{{dy}}

{y} = \smallint 3x \hfill \\

\log (y) = \frac{3}

{2}x^2 \hfill \\

y(x) = A\exp \{ \frac{3}

{2}x^2 \} \hfill \\

\end{gathered}

$

satisfies

$\displaystyle

y(0) = 0

$

is the answer y(x)=A??

I think I might have gone wrong somewhere