How should I solve $\left\{x\frac{\partial }{\partial x}+\left(y+x^2\right)\frac{\partial }{\partial y}\right\}u\left(x,y\right)=u$? $\left\{x\frac{\partial }{\partial x}+\left(y+x^2\right)\frac{\partial }{\partial y}\right\}u\left(x,y\right)=u$
$x\frac{\partial u}{\partial x}+\left(y+x^2\right)\frac{\partial u}{\partial y}=u$
Let's solve for
$x\frac{\partial u}{\partial x}+\left(y+x^2\right)\frac{\partial u}{\partial y}=0$
$\frac{dx}{x}=\frac{dy}{\left(y+x^2\right)}$
$y=x^2+Cx$
$\frac{y-x^2}{x}=C$
$u=\Phi \left(\frac{y-x^2}{x}\right)$
But that's just if it's = 0, so I don't know what to do if that's = u or something else...
 A: A general solution using the method of characteristics is
$$ u = x\;  \Phi \left( \frac{y-x^2}{x}\right) $$
EDIT: 
The characteristic curves are $(y-x^2)/x = c$, as you essentially found.  Along one of these curves, parametrized as $x = t$, $y = t^2 + c t$, we have 
$$ \eqalign{\dfrac{d}{dt} u(t, t^2+ct) &=  \dfrac{\partial u}{\partial x}(t,t^2+ct) +  (2 t + c) \dfrac{\partial u}{\partial y}(t,t^2+ct)\cr &=  \left.\dfrac{\partial u}{\partial x} + \dfrac{y+x^2}{x} \dfrac{\partial u}{\partial y} \right|_{x=t,y=t^2+ct}\cr &= \left. \frac{1}{x} \left( x \dfrac{\partial u}{\partial x} + (y+x^2) \dfrac{\partial u}{\partial y} \right)\right|_{x=t,y=t^2+ct}\cr &= \frac{u(t,t^2+ct)}{t}}$$
The general solution of $\dfrac{dv}{dt} = \dfrac{v}{t}$ is
$ v = C t $.
Since the "constant" $C$ depends on which characteristic curve it is, i.e. on $c = (y-x^2)/x$, and $t = x$ on the curve, we get the solution
$$ u(x,y) = x\; \Phi\left( \frac{y-x^2}{x}\right) $$
A: Giving 
$$
u_h(x,y) = \Phi\left(\frac{y-x^2}{x}\right)
$$
now making a change of variables
$$
\cases{a = x\\
b=\frac{y-x^2}{x}}
$$
on
$$
x\frac{\partial u}{\partial x}+\left(y+x^2\right)\frac{\partial u}{\partial y}=u
$$
we get at
$$
u = a \frac{\partial u}{\partial a}
$$
with solution
$$
u(a,b) = a\Phi(b)
$$
