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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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    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.2013-01-17

2 Answers 2

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$yy'+\frac{y}{x}+k=0 \quad\quad (1)$ 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 $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{dy}{dt}}{-e^{-t}}=-e^t\dfrac{dy}{dt}$

$\therefore-e^ty\dfrac{dy}{dt}+e^ty+k=0$

$y\dfrac{dy}{dt}-y=ke^{-t}$

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

Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf