# Finding an implicit solution of the ODE $y' (x) = \frac{y^2-xy}{2xy^3+xy+x^2}$

I am currently trying to find an implicit solution of the nonlinear ODE:$$y' (x) = \frac{y^2-xy}{2xy^3+xy+x^2}$$ But to be quite honest, I am a little bit lost. I am pretty new to this field and have not learned many methods of solving differential equations so far. What I tried was, to reformulate the problem as $$xy-y^2+y'(2xy^3+xy+x^2)=0$$ and check if this is an exact equation. However, the integrability criterion does not hold: $$\frac d{dx}(2xy^3+xy+x^2) \neq \frac{d}{dy}(xy-y^2)$$ Therefore, I tried to find an integrating factor to make this equation exact. The approach that my integrating factor $m(x,y)$ only depends on $x$ or only depends on $y$ seems to fail. Another approach I have seen is to set $m(x,y) = x^\alpha y^\beta$ as the integrating factor and then find out what $\alpha$ and $\beta$ have to be. Unfortunately, when using this approach, I end up with utterly long equations, which do not seem easily solveable ( I managed to actually find values for $\alpha$ and $\beta$ using Maple and then verify that the ODE multiplied by that factor is exact numerically, but I do not think this is the way to go here.)

Any help would be greatly appreciated!

• it's a non-linear equation that may not have a closed form. are you sure this is solvable? Mar 27 '18 at 18:58
• @Vasya This is from an exercise sheet and it states there is an implicit solution. There might also be a typo in there, however. Mar 27 '18 at 19:02
• What are the values of $\alpha$ and $\beta$ you found with Maple? Mar 30 '18 at 12:09
• From what I see $x^a y^b$ doesn't work as an integrating factor Mar 30 '18 at 14:14

$\dfrac{dy}{dx}=\dfrac{y^2-xy}{2xy^3+xy+x^2}$

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

$(y-x)\dfrac{dx}{dy}=\dfrac{x^2}{y}+(2y^2+1)x$

This belongs to an Abel equation of the second kind.

Let $u=y-x$ ,

Then $x=y-u$

$\dfrac{dx}{dy}=1-\dfrac{du}{dy}$

$\therefore u\left(1-\dfrac{du}{dy}\right)=\dfrac{(y-u)^2}{y}+(2y^2+1)(y-u)$

$u-u\dfrac{du}{dy}=\dfrac{u^2}{y}-(2y^2+3)u+2y(y^2+1)$

$u\dfrac{du}{dy}=-\dfrac{u^2}{y}+2(y^2+2)u-2y(y^2+1)$

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

Then $\dfrac{du}{dy}=\dfrac{1}{y}\dfrac{dv}{dy}-\dfrac{v}{y^2}$

$\therefore\dfrac{v}{y}\left(\dfrac{1}{y}\dfrac{dv}{dy}-\dfrac{v}{y^2}\right)=-\dfrac{v^2}{y^3}+\dfrac{2(y^2+2)v}{y}-2y(y^2+1)$

$\dfrac{v}{y^2}\dfrac{dv}{dy}-\dfrac{v^2}{y^3}=-\dfrac{v^2}{y^3}+\dfrac{2(y^2+2)v}{y}-2y(y^2+1)$

$\dfrac{v}{y^2}\dfrac{dv}{dy}=\dfrac{2(y^2+2)v}{y}-2y(y^2+1)$

$v\dfrac{dv}{dy}=2y(y^2+2)v-2y^3(y^2+1)$

Let $s=y^2+2$ ,

Then $\dfrac{dv}{dy}=\dfrac{dv}{ds}\dfrac{ds}{dy}=2y\dfrac{dv}{ds}$

$\therefore2yv\dfrac{dv}{ds}=2y(y^2+2)v-2y^3(y^2+1)$

$v\dfrac{dv}{ds}=(y^2+2)v-y^2(y^2+1)$

$v\dfrac{dv}{ds}=sv-(s-2)(s-1)$

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

Then $\dfrac{dv}{ds}=\dfrac{dv}{dt}\dfrac{dt}{ds}=s\dfrac{dv}{dt}$

$\therefore sv\dfrac{dv}{dt}=sv-(s-2)(s-1)$

$v\dfrac{dv}{dt}=v-s+3-\dfrac{2}{s}$

$v\dfrac{dv}{dt}=v\pm\sqrt{2t}+3\pm\sqrt{\dfrac{2}{t}}$

This belongs to an Abel equation of the second kind in the canonical form.