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I've been investigating a lot of interesting physically inspired second order differential equations that don't have t in the equation at all and thus are separable (one example being $y''=-\frac{k}{y^2}$).

I know that if $y' = f(y)$ I can solve this by $\int {\frac {dy}{f(y)}} = \int dt$

For the second order equation $y'' = f(y)$ I have been removing a variable since t is not involved by the following steps, but it can get messy.

$$y' = v$$ $$v' = f(y)$$ $$\frac{v'}{y'} = \frac{\frac{dv}{dt}}{\frac{dy}{dt}} = \frac{dv}{dy} = \frac{f(y)}{v}$$

$$\int{v dv} = \int{f(y) dy}$$

And after solving for v(y), then trying to solve for y(t), by substitution and a further 1st order separable equation.

I was wondering if there is some easier approach for a 2nd order of the form $y''=f(y)$?

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I'm more used to calling this sort of DE's autonomous, but anyway.., –  J. M. Aug 30 '11 at 13:25

1 Answer 1

There is no general approach for all ODE's of the form $ y'' = f(y) $. In fact, even when the functions $f(y)$ are restricted to relatively simple functions, the solutions can be very exotic. For example, if $f(y) = 2k^3 y^3-(1+k^2)y$ where $k$ is some fixed constant, the resulting solution (with initial conditions $y(0)=y'(0)=0$ ) is something called the Jacobi elliptic function $\operatorname{sn}(z,k)$. That's just one case for what is a relatively simple $f(y)$ (it was a 3rd degree polynomial). People devote a lot of effort into studying methods to solve DE's like yours, for specific special families of functions $f(y)$, with each special family having special techniques to solve them.

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Thanks, I thought that might be the case, but I was curious to find out for sure. Even y''=-k/y^2 results in an implicit equation for y that can't be solved. –  Tony Peterson Aug 30 '11 at 13:44

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