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Question:

By changing the variables from $x$, $y$ to $s$, $t$, where $s = xy$ and $t = x/y$, solve the equation

$x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y} = 2x^2$

The only part of solving this type of equation I am having trouble seeing how to do is how to get the values of $u_x$ and $u_y$ but I can carry on from there to solve the equation.

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Is that first derivative supposed to be with respect to t, or is that a typo? It seems from the wording that the only variables that should be present in the given problem are x and y. –  Barry Smith Apr 25 '12 at 20:26
    
Woops! It was a typo, fixed it now. –  Nicky Apr 25 '12 at 20:28

1 Answer 1

up vote 1 down vote accepted

We can just use the chain rule, so

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial x}=yu_s+\frac{1}{y}u_t$$

and

$$\frac{\partial u}{\partial y}=\frac{\partial u}{\partial s}\frac{\partial s}{\partial y}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial y}=xu_s-\frac{x}{y^2}u_t.$$

We obtain

$$x\left(yu_s+\frac{1}{y}u_t\right)-y\left(xu_s-\frac{x}{y^2}u_t\right)=2x^2$$

$$\iff 2\frac{x}{y}u_t=2x^2 \iff u_t=s.$$

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