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How do we solve the following first-order ordinary differential equation without a method for exact differential equation?

$\displaystyle\frac{dy}{dx}=\frac{a^2rx^2-absxy+ar}{a^2sy^2+abrxy-sa}$

where a,b,r and s are constants.

I want a solve which is a form of explicit function such as $y=f(x)$.

Is there a solution?

Thank you for your help.

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  • $\begingroup$ It usually helps a bit if you also mention what you have tried already and how it failed. Otherwise, your post looks like a homework exercise and could be closed or ignored. By the way two points: 1) Assuming a is not zero, it can be simplified. 2) Mathematica doesn't seem to be able to solve it, so probably (but not certainly!) the ODE doesn't have a closed form solution. $\endgroup$ – Rebel-Scum Nov 6 '17 at 14:54
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Hint:

$\dfrac{dy}{dx}=\dfrac{a^2rx^2-absxy+ar}{a^2sy^2+abrxy-sa}$

$\dfrac{dy}{dx}=\dfrac{arx^2-bsxy+r}{asy^2+brxy-s}$

Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=164:

Let $u=\dfrac{y}{x}$ ,

Then $y=xu$

$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$

$\therefore x\dfrac{du}{dx}+u=\dfrac{arx^2-bsx^2u+r}{asx^2u^2+brx^2u-s}$

$x\dfrac{du}{dx}=\dfrac{(ar-bsu)x^2+r}{(asu+br)x^2u-s}-u$

$x\dfrac{du}{dx}=\dfrac{(ar-bsu-bru^2-asu^3)x^2+su+r}{(asu^2+bru)x^2-s}$

Let $v=\dfrac{1}{x^2}$ ,

Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=-\dfrac{2}{x^3}\dfrac{du}{dv}$

$\therefore-\dfrac{2}{x^2}\dfrac{du}{dv}=\dfrac{(ar-bsu-bru^2-asu^3)x^2+su+r}{(asu^2+bru)x^2-s}$

$\left(asu^3+bru^2+bsu-ar-\dfrac{su+r}{x^2}\right)\dfrac{dv}{du}=\dfrac{2(asu^2+bru)}{x^2}-\dfrac{2s}{x^4}$

$(asu^3+bru^2+bsu-ar-(su+r)v)\dfrac{dv}{du}=2(asu^2+bru)v-2sv^2$

This belongs to an Abel equation of the second kind.

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