# Solving for $v$ in $v'(t) + R(t)\cdot v^{2/3}=J(t)$

I have a differential equation that I can't figure out how to solve. It is a first order non-linear ordinary differential equation.

Here it is: $$v'(t) + R(t)\cdot v^{2/3}=J(t)$$

I want to solve for v.

It looks like a linear one, but then you see the 2/3. Is there a way to solve this? I know how to integrate R and J if that helps.

$$R(t) = -3^{2/3}$$ and $$J(t) = Qt+C_1 \textrm{, where }Q\textrm{ is a constant and }C_1\textrm{ is a constant of integration.}$$

This equation can also be written as $$u'(t) - 1 = J(t)\cdot u^{-2}$$ or $$w'(t)-2w^{1/2}=2J(t)\cdot w^{-1/2}$$

using change of variables where $v = \frac{u^3}{3}$, and $w=u^2$.

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"I know how to integrate R and J if that helps." - okay. So, what do they look like? – J. M. Oct 8 '11 at 3:24
If an ode cannot be solved by analytic methods, then it can always be approximated by a numerical method. – pedja Oct 8 '11 at 3:58
@LeoIzen: $R(t)$ is a constant? – Jayesh Badwaik Aug 2 '12 at 15:15

Let $u=v^{\frac{-1}{3}}$,

Then $v=u^{-3}$

$v'=-3u^{-4}u'$

$\therefore-3u^{-4}u'+R(t)u^{-2}=J(t)$

$-3u^{-4}u'=J(t)-R(t)u^{-2}$

$u'=-\dfrac{J(t)u^4}{3}+\dfrac{R(t)u^2}{3}$

This belongs to a "Chini-like" equation as mentioned here and which is more complicated than Abel equation of the first kind.

I still don't know whether existing a substitution so that $u^2$ term can be eliminated and keeping $u^3$ term still vanished, leading it exactly belongs to a Chini equation.

It still don't know whether existing method can solve Chini equation generally, while Abel equation of the first kind eventually can be solved generally starting in 2011 August, see this for details.

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