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