Question

I hope this is an appropriate place to post this question, however, I am stuck on solving an apparently simple ODE.

I have checked numerous texts, references, software packages and colleagues before posting this...

$y(t)^n+a(t)\frac{dy(t)}{dt}=ba(t)$

If the RHS had a $y$ term it would simply be Bernoulli's equation. Does the $n$ term prevent a solution?

Answer 1

First, rewrite your equation as $$ \frac{dy(t)}{dt}=b+f(t)(y(t))^n, \qquad \qquad \qquad \qquad \qquad (*) $$ where $f(t)=-1/a(t)$.

This is a special case of the so-called Chini equation (Equation 1.55 in the Kamke's book mentioned below) $$ \frac{dy(t)}{dt}=f(t)(y(t))^n+g(t)y(t)+h(t) $$ which generalizes the Riccati and the Abel equations and is in general not solvable by quadratures but some of its special cases are, see e.g. the book (in German)

E. KAMKE, Differentialgleichungen: Lösungen und Lösungsmethoden, Band I: Gewöhnliche Differentialgleichungen, Leipzig, 1951,

and this list and references therein.

It is known that if the Chini invariant $$ C=f(t)^{-n-1}h(t)^{-2n+1}(f(t) dh(t)/dt-h(t)df(t)/dt+n f(t)g(t)h(t))^n n^{-n} $$ is independent of $t$, there is a straightforward recipe (described in the Kamke's book) for solving the equation. However, in the case under study (both for general $a(t)$ and for $a(t)$ linear in $t$ as suggested by the original poster in the comment to this reply) this invariant for (*) does depend on $t$ (unless I messed up the computations :)), so the recipe in question does not apply.

The only case when $C$ is independent of $t$ occurs (again modulo possible errors in computations :)) if $((1/f(t))df(t)/dt)^n=\alpha f(t)$, i.e., when $f=(-\alpha(t+\beta)/n)^n$, where $\alpha$ and $\beta$ are arbitrary constants.

Now let us turn to the particular cases with small values of $n$.

For $n=1$ you have a linear inhomogeneous ODE which is easily solved.

For $n=2$ you get a special case of the so-called general Riccati equation $$ \frac{dy(t)}{dt}=b+f(t)(y(t))^2, $$ solving which is equivalent to solving a second-order linear ODE. Indeed, upon introducing a new independent variable $\tau(t)=\int f(t) dt$ you end up with the "standard" Riccati equation $$ \frac{dy(\tau)}{d\tau}=h(\tau)+(y(\tau))^2, $$ where the function $h$ is defined so that $h(\tau(t))=b/ f(t)$, and putting $y(\tau)=-(1/z(\tau))dz(\tau)/d\tau$ yields $$ d^2z(\tau)/d\tau^2+h(\tau)z(\tau)=0. $$ However, this linear equation in general is not necessarily solvable by quadratures.

For $n=3$ (*) is a special case of the Abel differential equation of the first kind, see e.g. here and references therein for details.