The differential equation is the following:
$ydx+(\frac{y^2}{4}-2x)dy=0$
It can be written as:
$\frac{y}{2x-\frac{y^2}{4}}=\frac{dy}{dx}$
However, this is not the linear shape. So we rewrite the equation as:
$\frac{2x-\frac{y^2}{4}}{y}=\frac{dx}{dy}\to\frac{dx}{dy}-\frac{2x}{y}=-\frac{y}{4}$
The integrating factor is:
$e^{\int{\frac{-2dy}{y}}}=\frac{k}{y^2}$ and k can't be zero (1).
If you apply the integrating factor to the original equation you get:
$\frac{k}{y}dx+(\frac{k}{4}-\frac{-2xk}{y^2})dy=0$
According to the definition of an exact differential equation, it must be true that the partial derivative in Y of $\frac{k}{y}$ is equal to the partial derivative in X of $(\frac{k}{4}-\frac{-2xk}{y^2})$. This leaves us with the equality:
$\frac{-k}{y^2}=\frac{-2k}{y^2}$
So the only possible solution is $k=0$, but this a contradiction (1).
How is it possible? Am I making a mistake? How can I get an integrating factor other than zero?
Best Answer
You can not just apply the integrating factor to the original equation. You need to apply it to the equation that you computed it for. $$ \frac1{y^2}\frac{xy}{xy}-\frac{2x}{y^3}=-\frac1{4y} $$ is nicely integrable.