How could this differential equation be solved (hint, method)?

Question

$$Ay''+B(y')^n+C=0$$ $n \in\mathbb N$, $A,B,C \in\mathbb R$

I tried to substitute $y'=f$, then

$Af'+Bf^n+C=0$, then for $A\neq 0$

$f'=-bf^n-c$, where $b=B/A$ and $c=C/A$.

I don't quite know how to manipulate with the $n$ power of the function.

Answer 1

The equation in $f$ is separable: $$ \frac{A\,f'}{B\,f^n+C}=-1,\quad \int\frac{A\,df}{B\,f^n+C}=-x+C. $$ Now you have to do:

1. Carry out the integration.
2. Solve for $f$ as a function of $x$.
3. Integrate $f$ to find $y$.

Except for special values of $n$, it will be almost impossible to get an explicit solution.

Answer 2

$$Ay''+B(y')^n+C=0$$ $$Az'+Bz^n+C=0$$ $$-\frac AB\int \frac {dz}{z^n+C/B}=\int dx$$ $$\int \frac {dz}{z^n+C/B}=-\frac BAx+K$$ Now integration depends on the value of n

For $n=2$ it's pretty easy..