Consider this Abel differential equation:$$4\left[ y(z)+ z^2\right] y'(z)-(1+6z)y(z)=0$$Is there a way to separate variables?
I just tried some change of variables, but with no good results.
$\endgroup$1 Answer
$\begingroup$Hint:
Approach $1$:
$4(y(z)+z^2)y'(z)-(1+6z)y(z)=0$
$4(y+z^2)\dfrac{dy}{dz}=(6z+1)y$
Let $x=\dfrac{z}{4}$ ,
Then $\dfrac{dy}{dz}=\dfrac{dy}{dx}\dfrac{dx}{dz}=\dfrac{1}{4}\dfrac{dy}{dx}$
$\therefore(y+16x^2)\dfrac{dy}{dx}=(24x+1)y$
Let $u=y+16x^2$ ,
Then $y=u-16x^2$
$\dfrac{dy}{dx}=\dfrac{du}{dx}-32x$
$\therefore u\left(\dfrac{du}{dx}-32x\right)=(24x+1)(u-16x^2)$
$u\dfrac{du}{dx}-32xu=(24x+1)u-384x^3-16x^2$
$u\dfrac{du}{dx}=(56x+1)u-384x^3-16x^2$
Let $s=56x+1$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{ds}\dfrac{ds}{dx}=56\dfrac{du}{ds}$
$\therefore56u\dfrac{du}{ds}=su-384\left(\dfrac{s-1}{56}\right)^3-16\left(\dfrac{s-1}{56}\right)^2$
$56u\dfrac{du}{ds}=su-\dfrac{3s^3-9s^2+9s-3}{1372}-\dfrac{s^2-2s+1}{196}$
$56u\dfrac{du}{ds}=su-\dfrac{3s^3-2s^2-5s+4}{1372}$
Let $t=\dfrac{s^2}{112}$ ,
Then $\dfrac{du}{ds}=\dfrac{du}{dt}\dfrac{dt}{ds}=\dfrac{s}{56}\dfrac{du}{dt}$
$\therefore su\dfrac{du}{dt}=su-\dfrac{3s^3-2s^2-5s+4}{1372}$
$u\dfrac{du}{dt}=u-\dfrac{3s^2}{1372}+\dfrac{s}{686}+\dfrac{5}{1372}-\dfrac{1}{343s}$
$u\dfrac{du}{dt}=u\mp\dfrac{12t}{49}\pm\dfrac{2\sqrt{7t}}{343}\pm\dfrac{5}{1372}\mp\dfrac{1}{1372\sqrt{7t}}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in or in
Approach $2$:
$4(y(z)+z^2)y'(z)-(1+6z)y(z)=0$
$4(y+z^2)\dfrac{dy}{dz}=(6z+1)y$
$\left(z+\dfrac{1}{6}\right)\dfrac{dz}{dy}=\dfrac{2z^2}{3y}+\dfrac{2}{3}$
Let $u=z+\dfrac{1}{6}$ ,
Then $u\dfrac{du}{dy}=\dfrac{2}{3y}\left(u-\dfrac{1}{6}\right)^2+\dfrac{2}{3}$
$u\dfrac{du}{dy}=\dfrac{2u^2}{3y}-\dfrac{2u}{9y}+\dfrac{1}{54y}+\dfrac{2}{3}$
Let $u=y^\frac{2}{3}v$ ,
Then $\dfrac{du}{dy}=y^\frac{2}{3}\dfrac{dv}{dy}+\dfrac{2v}{3y^\frac{1}{3}}$
$\therefore y^\frac{2}{3}v\left(y^\frac{2}{3}\dfrac{dv}{dy}+\dfrac{2v}{3y^\frac{1}{3}}\right)=\dfrac{2y^\frac{4}{3}v^2}{3y}-\dfrac{2y^\frac{2}{3}v}{9y}+\dfrac{1}{54y}+\dfrac{2}{3}$
$y^\frac{4}{3}v\dfrac{dv}{dy}+\dfrac{2y^\frac{1}{3}v^2}{3}=\dfrac{2y^\frac{1}{3}v^2}{3}-\dfrac{2v}{9y^\frac{1}{3}}+\dfrac{1}{54y}+\dfrac{2}{3}$
$y^\frac{4}{3}v\dfrac{dv}{dy}=-\dfrac{2v}{9y^\frac{1}{3}}+\dfrac{1}{54y}+\dfrac{2}{3}$
$v\dfrac{dv}{dy}=-\dfrac{2v}{9y^\frac{5}{3}}+\dfrac{1}{54y^\frac{7}{3}}+\dfrac{2}{3y^\frac{4}{3}}$
Let $t=\dfrac{1}{3y^\frac{2}{3}}$ ,
Then $\dfrac{dv}{dy}=\dfrac{dv}{dt}\dfrac{dt}{dy}=-\dfrac{2}{9y^\frac{5}{3}}\dfrac{dv}{dt}$
$\therefore-\dfrac{2v}{9y^\frac{5}{3}}\dfrac{dv}{dt}=-\dfrac{2v}{9y^\frac{5}{3}}+\dfrac{1}{54y^\frac{7}{3}}+\dfrac{2}{3y^\frac{4}{3}}$
$v\dfrac{dv}{dt}=v-\dfrac{1}{12y^\frac{2}{3}}-3y^\frac{1}{3}$
$v\dfrac{dv}{dt}=v-\dfrac{t}{4}\pm\dfrac{\sqrt3}{\sqrt{t}}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in or in
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