How can i solve this separable differential equation? Given Problem is to solve this separable differential equation:
$$y^{\prime}=\frac{y}{4x-x^2}.$$
My approach: was to build the integral of y':
$$\int y^{\prime} = \int \frac{y}{4x-x^2}dy = \frac{y^2}{2(4x-x^2)}.$$
But now i am stuck in differential equations, what whould be the next step? And what would the solution looks like? Or is this already the solution? I doubt that.
P.S. edits were only made to improve language and latex
 A: That's not the way to solve separable equations, this is the general procedure:
$$\frac{dy}{dx}=\frac{y}{4x-x^2}$$
$$\frac{dy}{y}=\frac{dx}{4x-x^2}$$
Now that's what you integrate:
$$\int\frac{dy}{y}=\int\frac{dx}{4x-x^2}$$
The left one is immediate, the second one can be done by separating the fraction into two fractions as 1/x and 1/(4-x), which yields to two more logarithms:
$$4\log y + C = \log(x)-\log(x-4)$$
$$y = C\left(\frac{x}{x-4}\right)^\frac{1}{4}$$
A: It is separable in that you can separate everything that has $y$ in it from everything that has $x$ in it, i.e.,
$$
\frac{y'}{y} = \frac{1}{4x-x^2},
$$ 
and this is:
$$
(\ln y)' = \frac{1}{4x-x^2}.
$$
Integrating both sides with respect to $x$:
$$
\int (\ln y)'\,dx = \int \frac{1}{4x-x^2}\,dx,
$$
gives:
$$
\ln y = \int \frac{1}{4x-x^2}\,dx.
$$
Aside from all other answers, you can do it this way in case you don't like to separate $dy/dx$ as a fraction.
A: For $y \ne 0$ we have
$$
y'=\frac{y}{4x-x^2} \iff \frac{y'}{y}=\frac{1}{4x-x^2}=\frac{1}{4}\left(\frac{1}{x}+\frac{1}{4-x}\right).
$$
Integrating the two sides of the latter identity we get:
$$
\ln|y/a|=\frac{1}{4}\left(\ln|x|-\ln|4-x|\right)=\ln\sqrt[4]{\left|\frac{x}{4-x}\right|},
$$
where $a$ is a nonzero constant. Hence
$$
y(x)=b\sqrt[4]{\left|\frac{x}{4-x}\right|},
$$
with $b$ a real constant.
A: You seem to be slightly confused - where did the $dy$ come from? Why did you make that choice? You should apply separation of variables to solve this problem. 
That is, given 
$$ y' = \frac{y}{4x-x^{2}}$$
Write this as: 
$$ \frac{dy}{dx} = \frac{y}{4x-x^{2}}$$
Separating the variables, we have:
$$ \frac{dy}{y} = \frac{dx}{4x-x^{2}}$$
NOW we can integrate:
$$\int \frac{dy}{y} = \int \frac{dx}{4x-x^{2}}$$
$$\implies \ln y = \int \frac{dx}{4x-x^{2}}$$
From here, I recommend factoring the expression on the denominator of the right hand side and using partial fractions decomposition. I think you can take it from here, but feel free to post if you are still lost.
A: Separate variables:
$$
\frac{dy}{dx} = \frac{y}{4x-x^2}
$$
$$
\frac{dy}{y} = \frac{dx}{4x-x^2} = \frac{dx}{x(4-x)}
$$
Then integrate both sides, using partial fractions on the right side.
A: Do like this:
$\int \frac{y'}{y}dy=\int\frac{1}{4x-x^2}dx$, 
i.e.
$ln(y)=\int\frac{1}{4x-x^2}dx+C$,
where $C$ denotes an arbitrary constant.
As
$\frac{1}{4x-x^2}=\frac{1}{4}(\frac{1}{x}+\frac{1}{4-x})$ (check it!)
then
$ln(y)=\int(\frac{1}{4}(\frac{1}{x}+\frac{1}{4-x}))dx+C$,
or
$ln(y)=\frac{1}{4}(\ln(x)-\ln(4-x))+C'=\ln(\frac{x}{4-x})^{\frac{1}{4}}+C'$,
which implies
$y(x)=K(\frac{x}{4-x})^{\frac{1}{4}}$,
with $K=e^{C'}$ arbitrary.
