# Change of variable

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• Apr 21st 2006, 08:56 PM
askmemath
Change of variable
• Apr 22nd 2006, 06:00 AM
topsquark
There's probably a more elegant way, but this does work:

$\displaystyle x^2\frac{dy}{dx}=y^2+4xy$

This is essentially a Bernoulli equation, so if we go ahead and let $\displaystyle v = y^{1-2}=\frac{1}{y}$ we obtain a linear equation:

$\displaystyle y=\frac{1}{v}$
$\displaystyle \frac{dy}{dx}=-\frac{1}{v^2}\frac{dv}{dx}$

So
$\displaystyle -x^2\frac{1}{v^2}\frac{dv}{dx}=\frac{1}{v^2}+4x \frac{1}{v}$

$\displaystyle x^2\frac{dv}{dx}+4xv=-1$

$\displaystyle (4xv+1)dx+x^2dv=0$

Now, $\displaystyle \frac{1}{x^2}(4x-2x)$ depends only on x, so the integrating factor will be:
$\displaystyle exp \left ( \int dx \frac{1}{x^2}(4x-2x) \right )=e^{2ln|x|}=x^2$

So $\displaystyle (4x^3v+x^2)dx+x^4dv=0$ is an exact differential equation.

Thus $\displaystyle x^4v+\frac{1}{3}x^3=C$

or
$\displaystyle v=\frac{C-\frac{1}{3}x^3}{x^4}$

Thus $\displaystyle y=\frac{3x^4}{3C-x^3}$

-Dan
• Apr 22nd 2006, 07:50 AM
Rebesques
Note that the right hand side of

$\displaystyle \frac{dy}{dx}=\frac{y^2+4xy}{x^2}$

is homogeneous in x and y, and change variables to z=y/x.