Solve $u_{xx} + y u_{xy} = 0$ for $u(0,y)=y^3$ Solve the IVP
$$u_{xx} + y u_{xy} = 0 \text{ for } u(0,y)=y^3$$
Is the solution unique?
My attempt:
Let $u_x = v$. Substituting on equation:
$$v_x + y v_y = 0$$
We solve this using the characteristics method: $x=x(s), \; y=y(s),\; w(s)=v(x,y)$
$$\Rightarrow \frac{dx}{ds}=1, \; \frac{dy}{ds}=y, \; \frac{dw}{ds}=0$$
Doing all the calculation,
$$v(x,y) = f(y e^{-x})$$
where $f: R \rightarrow R$ is an arbitrary function. Now,
$$u_x = v \Rightarrow u(x,y) = \int_?^x f(y e^{-\eta}) d \eta + C(y) $$
My doubt is:


*

*What are the integration limits? Usually, I use 0 to $x$ but I don't know why.

*Using the initial conditions: $$u(0,y) = \int_?^0 f(y e^{-\eta}) d \eta + C(y) = y^3 $$. If I use the limits that I am used to, we must have $C(y)=y^3$ and $f$ still arbitrary (and integrable). But I don't know if it is rigorous. In this case, the solution won't be unique (there are many for any $f$ integrable)


Thanks in advance!
 A: Hint:
let $$u=XY$$
so
$$u_{xx}=X''Y$$
$$u_{xy}=X'Y'$$
then you get
$$X''Y+yX'Y'=0$$
divide by  $$X'Y$$
hence 
$$\frac{X''}{X'}+y\frac{Y'}{Y}=0$$
now you can let $$\frac{X''}{X'}=\lambda $$
and $$y\frac{Y'}{Y}=-\lambda $$
you can solve this equation by euler method as follow
$$Y=c_1y^m$$
$$Y'=c_1my^{m-1}$$
substitute to get
$$y(c_1my^{m-1})=-\lambda (c_1y^m)$$
$$m=-\lambda $$
$$Y=c_1y^{-\lambda }$$
now we will solve the $\frac{X''}{X'}=\lambda$
The general solution is 
$$X=c_2e^{\lambda x}-\frac{c_3}{\lambda}$$ 
hence
$$u=(c_1y^{-\lambda})(c_2e^{\lambda x}-\frac{c_3}{\lambda})$$
when the $u(0,y)=y^3$
$$(c_1y^{-\lambda})(c_2-\frac{c_3}{\lambda})=y^3$$
$$\lambda=-3$$
and
$$c_1(c_2-\frac{c_3}{\lambda})=1$$
or
$$(k_1-\frac{k_2}{\lambda})=1$$
then
$$u=(c_1y^{3})(c_2e^{-3 x}+\frac{c_3}{3})$$
or you can write it as
$$u=(y^{3})(k_1e^{-3 x}+\frac{k_2}{3})$$
A: Since you know $u(0,y)$ and $u_x$, you can get $u(x,y)$ by
$$ u(x,y) = u(0,y) + \int\limits_{0}^{x}{u_x(\eta,y)\text{ d}\eta} = y^3 + \int\limits_{0}^{x}{f(ye^{-\eta})\text{ d}\eta}. $$
Also, you should double check exactly how "arbitrary" $f$ can be (in particular, it probably should be at least differentiable on $(0,\infty)$).
A: *

*Use the fundamental theorem of calculus: $$u(x,y) - u(0,y) = \int_0^xu_x(\eta,y)\,d\eta \Longrightarrow u(x,y) = \int_0^xf(ye^{-\eta})\,d\eta + y^3.$$

*You are correct, you don't get uniqueness, indeed for any integrable and differentiable function $f$ you have that $u(x,y)$ defined as above solves the system. This follows by construction but you can easily verify it by noticing that the initial condition is trivially satisfied and that \begin{align}
u_{xx}(x,y) = &\  -f'(ye^{-x})ye^{-x} \\
u_{xy}(x,y) = &\  f'(ye^{-x})e^{-x}
\end{align}

