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How can we solve the equation:

$$\frac{dy}{dx} = \frac{y^3}{2(xy^2-x^2)}$$

I get the idea of dividing by $y^2$, But it doesn't become any more solvable (not homogenous). $$\frac{dy}{dx} = \frac{y}{2(x-\frac{x^2}{y^2})}$$

Substituting $\frac{x}{y} = t$ causes even more complications.

I get an idea of the question to convert into homogenous, but cant form the equation. Please give me a hint!

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  • $\begingroup$ Yes, But the same problem occurred as with $\frac{dy}{dx}$, I got a mix of terms. $\endgroup$ – samjoe Nov 5 '16 at 3:31
  • $\begingroup$ Try putting $y^2 = t$ $\endgroup$ – Max Payne Nov 5 '16 at 3:39
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Nice question!

Given: $$\frac{dx}{dy}=\frac{2x}{y}-\frac{2x^2}{y^3}$$ $$=>\frac{dx}{dy}-\frac{2x}{y}=-\frac{2x^2}{y^3}$$ On dividing by $-x^2$, $$=>\frac{dx}{dy}\frac{(-1)}{x^2}+\frac{2}{xy}=\frac{2}{y^3}$$ Substituting $v=\frac{1}{x}$; $\frac{dv}{dy}=\frac{dx}{dy}\frac{(-1)}{x^2}$gives: $$\frac{dv}{dy}+\frac{2v}{y}=\frac{2}{y^3}$$

Now simply use the method for solving linear differential equations.

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  • $\begingroup$ Nice solution ! $\endgroup$ – Max Payne Nov 5 '16 at 4:04
  • $\begingroup$ One question, in this linear equation, is $P = \frac{1}{y}$ and $Q = \frac{2}{y^3}$ ? $\endgroup$ – samjoe Nov 5 '16 at 4:06
  • $\begingroup$ That's right! Make sure the function involves constants as well (2/y) @samjoe $\endgroup$ – Kugelblitz Nov 5 '16 at 4:07
  • $\begingroup$ @samjoe $P = \frac{2}{y}$ $\endgroup$ – Max Payne Nov 5 '16 at 4:08
  • $\begingroup$ @MaxPayne Thanks; your solution is, however, more intuitive imo. $\endgroup$ – Kugelblitz Nov 5 '16 at 4:09
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By putting $y^2 = t$, it will reduce to homogenous:

$$\frac{t'}{2y} = \frac{ty}{xt-x^2}$$ $$\therefore \ t' = \frac{2t^2}{xt-x^2}$$

Now you can divide by $x^2$, and put $t=ux.$

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