Knowing that any solution of $y \frac{\partial ^2 u}{\partial y^2} + 2 \frac{\partial u}{\partial y} = \frac{2}{x}$ is of the form $u(x,y) = \frac{\varphi(x) + y^2}{xy} + \psi(x)$,
where $\varphi, \ \psi $ are $C^2$,
find the solution of
$(1) \ \ y \frac{\partial ^2 u}{\partial y^2} + 2 \frac{\partial u}{\partial y} = \frac{2}{x}, \ u(x,x)=g(x)$
and of
$(2) \ \ y \frac{\partial ^2 u}{\partial y^2} + 2 \frac{\partial u}{\partial y} = \frac{2}{x}, \ u(y^3, y)=g(y)$.
Here $g: \mathbb{R} \rightarrow \mathbb{R}$ is a given function.
I'm having problems solving this.
I've solved a similar problem $\frac{\partial u}{\partial x} = 2x \frac{\partial u}{\partial y}, \ u(4,y)=g(y)$, ($g$ is given) where the solution was $u(x,y)=\varphi(y+x^2)$ and I just plugged $\varphi$ to the equation: $\varphi(y+16)=u(4,y)=g(y), \ \ \varphi(y)=g(y-16), \ \ u(x,y)=g(x^2+y-16)$.
But here it doesn't seem to work. Could you explain to me how to solve it?
Thank you!
This is a second order equation. The general solution depends on two arbitrary functions ($\phi$ and $\psi$.) If you impose only one condition, you will get infinite answers depending on one arbitrary function. Consider (1). You want $$ u(x,x)=\frac{\phi(x)+x^2}{x^2}+\psi(x)=g(x). $$ You can choose $\psi$ arbitrarily and let $$ \phi(x)=x^2(g(x)-\psi(x)-1). $$
On the other hand, if you impose both conditions (1) and (2), you will get a system of two equations with the two unknowns $\phi$ and $\psi$, and you will find (in most cases) a unique solution $u$.