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!

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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 $t=ux.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.