partial differential equation

**vuze88** · April 5th 2010, 05:09 PM

A function $f$ of two variables is said to be homogeneous of degree n if $f(tx,ty)=t^nf(x,y)$ whenever $t>0$ .

Show that such a function $f$ satisfies the equation $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf$

**tonio** · April 5th 2010, 06:04 PM

Originally Posted by vuze88

A function $f$ of two variables is said to be homogeneous of degree n if $f(tx,ty)=t^nf(x,y)$ whenever $t>0$ .

Show that such a function $f$ satisfies the equation $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf$

Derive both sides of $f(tx,ty)=t^nf(x,y)$ wrt $t$ and then evaluate the result at $t=1$ .

Tonio

**vuze88** · April 8th 2010, 01:46 AM

Originally Posted by tonio

Derive both sides of $f(tx,ty)=t^nf(x,y)$ wrt $t$ and then evaluate the result at $t=1$ .

Tonio

Sorry i still dont fully understand. Can you explain this in more detail. Also im not sure whether the f on the RHS means f(x,y) or f(tx,ty)

**tonio** · April 8th 2010, 02:07 AM

Originally Posted by vuze88

Sorry i still dont fully understand. Can you explain this in more detail. Also im not sure whether the f on the RHS means f(x,y) or f(tx,ty)

It is exactly as I wrote it: derivate the left side wrt t, derivate the right side wrt t, they're equal, of course, and now put t=1 .

Tonio

**vuze88** · April 8th 2010, 02:37 AM

lol yeah but how to i differentiate with respect to t when the function is in terms of tx and ty

**HallsofIvy** · April 8th 2010, 03:30 AM

Use the chain rule, of course!

$\frac{df(ax)}{dx}= a\frac{df}{dx}$

**Pulock2009** · April 8th 2010, 03:37 AM

Originally Posted by HallsofIvy

Use the chain rule, of course!

$\frac{df(ax)}{dx}= a\frac{df}{dx}$

this is what Euler's theorem on partial derivatives says.

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