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

[blackthunder]

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.

 

[Mathwebster]

First, check your PDE. If it's written down correct, you can cancel out a $y$. Also, check this step

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

 

[blackthunder]

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 $$u(0,y)=y^2$$