OK, so I'm working on this revision question which looked straight forward, but now it's confusing me.
I've used the integrating factor method with respect to $x$ and arrived at $u=\frac{5y}{2}+f(y)$
I'm unsure how to proceed at finding a particular solution. I tried solving it like a normal algebra problem, but I have two solutions for $y$. I've integrated with respect to $x$ because of the $\frac{\partial u}{\partial x}$ term, but it seems to profit me nothing. Am I approaching this the wrong way? I have nothing to plug the $x$ value into for part b).
Edit: My workings:
IF=$e^{\int 4y dx}=e^{4yx}$
$\frac{\partial u}{\partial x} e^{4yx}u=10y^2e^{4yx}$
$\int \frac{\partial u}{\partial x} e^{4yx}u dx=\int 10y^2e^{4yx} dx$
$e^{4yx}u = \frac{10y^2e^{4yx} }{4y}+f(y)$
$u = \frac{10y^2e^{4yx} }{4ye^{4yx}}+f(y)$
$u=\frac{5y}{2}+f(y)$
You make an algebra mistake in solving for $u$, you forget to divide $f(y)$ by $e^{4yx}$ as well.
Note that $$ u_x+4yu=10y^2\implies u_xe^{4yx}+4yue^{4yx}=10y^2e^{4yx} $$ or rewriting $$ \partial_x(ue^{4yx})=10y^2e^{4yx} $$ and integrating both sides in $x$ yields $$ ue^{4yx}=\frac52ye^{4yx}+f(y) $$ for some unknown function of $y$, $f(y)$, as you found. So, $$ u(x,y)=\frac52 y+e^{-4yx}f(y) $$ is your general solution.
For the particular, plug in and enforce, $$ u(1/4,y)=\frac52y+e^{-y}f(y)=5y \implies f(y)=\frac{5y}{2}e^y $$