# How to show that a given function is a solution for a partial differential equation.

Let $g : \mathbb R \to \mathbb R$ be a differentiable function and let $f(x,y)=x^ng(\frac {y}{x})$, where $n\in\Bbb Z^+$, show that $f$ is a solution to the PDE

$$x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf$$

i know what to use, but i have a problem in the simplification, i am stopped , the far i do was:

$x\frac {\partial f}{\partial x}=nx^ng(\frac {y}{x})-x^{n-1}g´(\frac {y}{x})$

$y\frac {\partial f}{\partial y}=\frac {y}{x}g´(\frac {y}{x})$

and thats what i got, besides the sum and a vague simplification but i can´t really approach the answer

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What is the problem here? Use product and chain rule to compute $\partial_xf$, and chain rule to compute $\partial_y$. –  Davide Giraudo Sep 7 '12 at 10:33
You computed both partial derivatives wrongly. –  Harald Hanche-Olsen Sep 7 '12 at 11:02
then, how it is? –  Sebastian Griotberg Sep 7 '12 at 11:15
Note: $$\frac{\partial}{\partial x} g(y/x) = - g'(y/x) (y/x^2)$$ so you missed a factor of $y$; and that in your $y$ derivative you dropped a factor of $x^n$ that is in the definition of $f$. –  Willie Wong Sep 7 '12 at 11:32

I think the comments have pointed out where you went wrong, but I will just give a bit more detail (but as you've listed it as a homework question, I'm not going to do it all).

Basically, you have just calculated the partial derivatives incorrectly. By the product rule, you obtain:

$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(x^ng\left(\frac{y}{x}\right)\right) = \frac{\partial}{\partial x}(x^n)g\left(\frac{y}{x}\right) + x^n\frac{\partial}{\partial x}\left(g\left(\frac{y}{x}\right)\right).$$

Now, you need to use the chain rule to calculate $\frac{\partial}{\partial x}\left(g\left(\frac{y}{x}\right)\right)$. That is:

$$\frac{\partial}{\partial x}\left(g\left(\frac{y}{x}\right)\right) = g'\left(\frac{y}{x}\right)\frac{\partial}{\partial x}\left(\frac{y}{x}\right).$$

To calculate $\frac{\partial f}{\partial y}$ is slightly simpler as you don't need to use the product rule ($x^n$ is not a function of $y$), but you still need to use the chain rule to calculate $\frac{\partial}{\partial y}\left(g\left(\frac{y}{x}\right)\right)$.

Substituting your expressions for $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ into the left hand side of your equation, you should get the right hand side, which shows that $f$ is indeed a solution to the given differential equation.

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Let $h (x, y) = y / x$, so that $f (x, y) = x^n g (y/x)$ can be written in the form $f (x, y) = x^n g ( h (x,y))$. The partial derivatives can be computed using the chain rule, as follows

$\partial_x f (x,y) = n x^{n-1} g (h(x,y)) + x^n g'(h(x,y)) \partial_x h (x,y)$

$\partial_y f (x,y) = x^n g'(h(x,y)) \partial_y h (x,y)$

and, since $\partial_x h (x,y) = -y / x^2$ and $\partial_y h (x,y) = 1 / x$, we obtain

$\partial_x f (x,y) = n x^{n-1} g(y/x) - x^{n-2} y \, g'(y/x)$

$\partial_y f (x,y) = x^{n-1} g'(y/x)$

and, therefore, we have

$x\, \partial_x f (x,y) = n x^n g(y/x) - x^{n-1} y\, g'(y/x)$

$y\, \partial_y f (x,y) = x^{n-1} y\, g'(y/x)$

and, finally, we obtain

$x\, \partial_x f (x,y) + y\, \partial_y f (x,y) = n x^n g(y/x) = n f (x,y)$.

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