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How to find all solutions of the PDE

$x^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}=(x+y)u$?

My way was:

The Lagrange-Charpit equations are: $\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{du}{(x+y)u}$.

There are two linear independent solutions. For the first one I got:

$\frac{du}{(x+y)u}=\frac{dx-dy}{x^2-y^2}=\frac{dx-dy}{(x+y)(x-y)} \Leftrightarrow \frac{1}{u}du-\frac{1}{x-y}d(x-y)=0 \Leftrightarrow d\mathrm{ln}(|\frac{u}{x-y}|)=0$.

So the first solution is $f(x,y,u)=\frac{u}{x-y}$.

Now I don't know how to find the second solution. I tried to manipulate the terms $\frac{dx}{x^2}, \frac{dy}{y^2}$ and $\frac{du}{u(x+y)}$ in a similar way as above but I don't get any simplification .

Can it be done in the same way or how can the second solution be found?

Tartulop
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2 Answers2

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$\frac{\frac{1}{x}dx}{x}=\frac{\frac{1}{y}dy}{y}=\frac{\frac{1}{u}du}{(x+y)}$

$\implies\frac{\frac{1}{x}dx+\frac{1}{y}dy}{x+y}=\frac{\frac{1}{u}du}{x+y} $

$\implies \frac{1}{x}dx+\frac{1}{y}dy=\frac{1}{u}du$

Solve and get second integral curve to form the solution surface.

Sourav Ghosh
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$$x^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}=(x+y)u$$

$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{du}{(x+y)u}$$

You correctly found a first characteristic equation in solving $\frac{du}{(x+y)u}=\frac{dx-dy}{x^2-y^2}$ leading to :

$$\frac{u}{x-y}=c_1$$.

A second characteristic equation comes from solving $\frac{dx}{x^2}=\frac{dy}{y^2}$ $$\frac{1}{y}-\frac{1}{x}=c_2$$ The solution of the PDE is : $$\frac{u}{x-y}=F\left(\frac{1}{y}-\frac{1}{x}\right)$$ $F$ is an arbitrary function until no boundary condition is specified. $$\boxed{u(x,y)=(x-y)F\left(\frac{1}{y}-\frac{1}{x}\right)}$$

JJacquelin
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