MHB Help with PDE: $$yu_x+2xyu_y=y^2$$

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Hi, need some help here so thanks to any replies.

PDE: $$yu_x+2xyu_y=y^2$$

edit: Forgot to mention the condition $$u(0,y)=y^2$$

a) characteristic equations:
$$dx/ds=y$$ $$dy/ds=2xy$$ $$du/ds=y^2$$

b) find dy/dx and solve

$$dy/dx=dy/ds * ds/dx = x/y$$
$$ydy=xdx$$
$$y^2/2=x^2/2 +c$$
$$y=\pm \sqrt{x^2+2c}$$

c) general solution

d) solve PDE

e) find u(x,y)
here I should be finding du/ds and solve after letting x(s)=? and y(s)=?

I need some help, so thanks guys.
 
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First, check your PDE. If it's written down correct, you can cancel out a $y$. Also, check this step

$\dfrac{dy}{dx} = \dfrac{x}{y}$
 
Jester said:
First, check your PDE. If it's written down correct, you can cancel out a $y$. Also, check this step

$\dfrac{dy}{dx} = \dfrac{x}{y}$

so the pde can be restated as: $$u_x+2xu_y=y$$

and yes, i stuffed up dy/dx.
dy/dx=2x

edit:
I forgot to mention the condition $$u(0,y)=y^2$$
 
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There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
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