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Solve the non-linear first order differential equation $$\frac{dy}{dx}=\frac{x+2y-5}{2x+xy-4} $$

I tried substituting $x=X+h$ and $y=Y+k$ but the $xy$ term is creating problem.

How to solve it?

Any suggestion is appreciated.

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  • $\begingroup$ try $y=vx$ may be it'll work $\endgroup$ – Kushal Bhuyan Dec 11 '15 at 17:38
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    $\begingroup$ @Quintic) It is NOT homogeneous. So it does not work. $\endgroup$ – Empty Dec 11 '15 at 17:40
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    $\begingroup$ I bet that there is a typo. I bet it is not $xy$, but simply $y$. $\endgroup$ – JJacquelin Dec 11 '15 at 18:20
  • $\begingroup$ If @JJacquelin is right, then it is an exact differential equation $\endgroup$ – Jeybe Dec 11 '15 at 18:35
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    $\begingroup$ If it is a training exercise, of course it cannot be $xy$. But who can say if it is a training for Professor Emeritus ? $\endgroup$ – JJacquelin Dec 11 '15 at 19:03
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$\dfrac{dy}{dx}=\dfrac{x+2y-5}{2x+xy-4}$

$(x+2y-5)\dfrac{dx}{dy}=(y+2)x-4$

This belongs to an Abel equation of the second kind.

Let $u=x+2y-5$ ,

Then $x=u-2y+5$

$\dfrac{dx}{dy}=\dfrac{du}{dy}-2$

$\therefore u\left(\dfrac{du}{dy}-2\right)=(y+2)(u-2y+5)-4$

$u\dfrac{du}{dy}-2u=(y+2)u-2y^2+y+6$

$u\dfrac{du}{dy}=(y+4)u-2y^2+y+6$

Let $s=y+4$ ,

Then $\dfrac{du}{dy}=\dfrac{du}{ds}\dfrac{ds}{dy}=\dfrac{du}{ds}$

$\therefore u\dfrac{du}{ds}=su-2(s-4)^2+s-4+6$

$u\dfrac{du}{ds}=su-2s^2+17s-30$

Let $t=\dfrac{s^2}{2}$ ,

Then $s=\pm\sqrt{2t}$

$\dfrac{du}{ds}=\dfrac{du}{dt}\dfrac{dt}{ds}=s\dfrac{du}{dt}$

$\therefore su\dfrac{du}{dt}=su-2s^2+17s-30$

$u\dfrac{du}{dt}=u-2s+17-\dfrac{30}{s}$

$u\dfrac{du}{dt}-u=\mp2\sqrt{2t}+17\mp\dfrac{30}{\sqrt{2t}}$

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

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