# Change of variable

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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