Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given Problem is to solve this separable differential equation:


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

share|cite|improve this question
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
up vote 6 down vote accepted

That's not the way to solve separable equations, this is the general procedure:



Now that's what you integrate:


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:

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

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

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer
+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.

share|cite|improve this answer

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.

share|cite|improve this answer

Do like this:

$\int \frac{y'}{y}dy=\int\frac{1}{4x-x^2}dx$,



where $C$ denotes an arbitrary constant.


$\frac{1}{4x-x^2}=\frac{1}{4}(\frac{1}{x}+\frac{1}{4-x})$ (check it!)





which implies


with $K=e^{C'}$ arbitrary.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.