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
$\endgroup$ 26 Answers
$\begingroup$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:
$$4\log y + C = \log(x)-\log(x-4)$$
$$y = C\left(\frac{x}{x-4}\right)^\frac{1}{4}$$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ 0 $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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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