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

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The discussion revolves around solving the partial differential equation (PDE) $$yu_x + 2xyu_y = y^2$$ with the initial condition $$u(0,y) = y^2$$. Participants suggest checking the PDE for accuracy, noting that a $y$ can be canceled, simplifying it to $$u_x + 2xu_y = y$$. The characteristic equations derived include $$dx/ds = y$$, $$dy/ds = 2xy$$, and $$du/ds = y^2$$. There is a correction regarding the calculation of $$dy/dx$$, which should be $$2x$$ instead of $$x/y$$. The conversation emphasizes the importance of verifying each step in the solution process.
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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.
 
Last edited:
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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 $$u(0,y)=y^2$$
 
Last edited:

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