# 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

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So, note that $$\int \frac{y}{4x-x^2}dy = \frac{y^2}{2(4x-x^2)}$$ is not correct, because $x$ depends on $y$ (and $y$ on $x$). – GEdgar Jun 9 '13 at 18:51
The fact that it was called "separable" should tell you there's something wrong with your approach. "Separable" means that you can separate the two variables, but that's not what you did. – Michael Hardy Jun 9 '13 at 19:13

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:

$$4\log y + C = \log(x)-\log(x-4)$$

$$y = C\left(\frac{x}{x-4}\right)^\frac{1}{4}$$

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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.

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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.

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+1 because you introduced the absolute values. – Avitus Jun 9 '13 at 18:47

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.

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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.

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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.

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Your first integration $$\int \frac{y'}{y} dy = \int \frac{1}{y}\frac{dy}{dx} dy$$ doesn't make sense. – Shuhao Cao Jun 9 '13 at 18:43