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$$yy'+\frac yx+k=0$$

How to solve this differential equation for $y$ in terms of $x$ and $k$ where $k$ is a parameter of $x$?

$y(x)=y$ is a function and $x(k)=x$ is a gamma function

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@TomOldfield - I can't figure it out with my high school calculus knowledge – Victor Dec 30 '12 at 20:44
So is $y$ a function of $x$ and $k$, or is $y$ a function of $x$ with parameter $k$? – Daryl Dec 30 '12 at 22:25
@Victor If both $x(k)$ and $y(x)$ are unknown, then I suspect that this problem is ill-posed. – Daryl Dec 31 '12 at 0:07
The letter $k$ denotes a parameter, not a variable. - The ODE in question is equation $1\cdot237$ in Kamke's classical catalogue (here $a=0$, $b=1$, $c=k$). The prospects for a simple explicit solution are not excellent. – Christian Blatter Dec 31 '12 at 16:37
As wolfram alpha does not produce a closed form solution in its time limit I suspect high school calculus knowledge won't be enough! The statement ($x(k)=x$ is a gamma function) needs clarification. – Maesumi Jan 17 '13 at 20:32
up vote 2 down vote accepted

$$yy'+\frac{y}{x}+k=0 \quad\quad (1)$$ Change of function : $y(x)=\frac{1}{f(x)}$

$\frac{1}{f}\left(-\frac{f'}{f^2}\right)+\frac{1}{xf}+k=0$ $$f'=kf^3+\frac{1}{x}f^2$$ This is an Abel's differential equation of first kind which is knonw as ''non-sovable'' form, meaning that the solutions are not known on the form of a finite number of standard functions.

So, no closed form is known for the solutions of ODE $(1)$.

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This belongs to an Abel equation of the second kind.

Let $x=e^{-t}$ ,

Then $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{dy}{dt}}{-e^{-t}}=-e^t\dfrac{dy}{dt}$



This belongs to an Abel equation of the second kind in the canonical form.

Please follow the method in or in

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