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  y'=(2x-y+3) / (x+3y-2)

if the coefficients are same this equation can be solved by replacement by a variable function (V) but here in this case coefficient are not same. how would it be solved ? (just curious and cannot think of any solution)

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    this is an equation of Abel type – 2017-01-24
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    Also after shift $u = x + a$, $w = y + b$ you can make similar equations homogeneous. But if it's exact, it is much easier to solve it as an exact equation :) – 2017-01-24

1 Answers 1

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Hint

Change variable defining $x+3y-2=z$. Replace and simplify. The equation will become $$z'=\frac{7 (x+1)}{z}$$ which is separable and looks quite simple.

Edit

Suppose the equation to be $$y'=\frac{a x+b y+c}{x+d y+e}$$ $$x+d y+e=z \implies y=\frac{z-x-e}{d}\implies y'=\frac{z'-1}{d}$$ Replace an simplify. You should arrive to $$-d (a x+c)+b (e+x)+z \left(z'-b-1\right)=0$$ which becomes separable if $b=-1$ (which is your case). In such a case,

$$z \,z'=x (a d+1)+(c d+e)$$ which is simple too.