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I am having trouble and I am unsure how to solve this PDE. $x\frac{\partial v}{\partial x} +y\frac{\partial v}{\partial y} = 2xy(x^2-y^2)$ I know you will use method of characteristics, but I am being thrown off by the function on the right side.

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  • $\begingroup$ What are the boundary conditions? $\endgroup$ – User8128 May 7 '16 at 21:00
  • $\begingroup$ there are none it just say find the general solution of v=v(x,y). sorry i should have been more specific. $\endgroup$ – daultongray8 May 7 '16 at 21:02
  • $\begingroup$ See this question (from earlier today!): math.stackexchange.com/questions/1775228/… $\endgroup$ – Hans Lundmark May 8 '16 at 2:21
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I think I already answered to a question like this, but where ?.

Change $\begin{cases} t=\frac{y}{x} \\u(x,y)=U(x,t)\end{cases}\quad\to\quad \begin{cases} u_x=U_x-U_t\frac{y}{x^2} \\u_y=U_y\frac{1}{x} \end{cases}$ $$xu_x+y_y=xU_x=2xy(x^2-y^2)=2x^2t(x^2-x^2t^2)=2x^4t(1-t^2)$$ $$U_x=2x^3t(1-t^2)$$ $$U=\frac{x^4}{2}t(1-t^2)+f(t)$$ $$u(x,y)=\frac{x^4}{2}\frac{y}{x}\left(1-\frac{y^2}{x^2}\right)+f\left(\frac{y}{x}\right)=\frac{xy}{2}\left(x^2-y^2\right)+f\left(\frac{y}{x}\right)$$ where $f$ is any differentiable function.

Note :

The idea to the change of variable $t=\frac{y}{x}$ comes from the preliminary solving of the related homoheneous PDE $xu_x+yu_y=0$ which general solution is $f\left(\frac{y}{x}\right)$. So, we replace the "one variable" function $f$ by a "two variables" function $U\left(x,\frac{y}{x}\right)=U(x,t)$ : For inhomogeneous linear PDEs, this is similar to the "variation of parameter" for inhomogeneous linear ODEs.

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