Need help with a differential equation

[ramparts]

I've been wrestling with this one for a while....if you have a diff eq of the form y''(t) = c y^-2 (a la Newton's law of gravitation), how do you solve for y(t)?

[OfficeShredder]

You have d^2y/dt^2 = c/y^2 you can rewrite it as

y^2d^2y = cdt^2 and integrate twice.

[ramparts]

Doesn't work, for the same reason that solving for d^2 r /dt^2 = kr (like a spring) doesn't yield lnr - r = kt^2/2. I'm pretty sure this is because the integral of d^2 r is not dr. I believe the solution is periodic but I'm not sure how to get it...

[HallsofIvy]

Since t does not appear explicitely in the equation, you can define
v= dy/dt and then d^2y/dt^2= dv/dt. By the chain rule, dv/dt= dv/dy dy/dt= v dv/dy. So your equation becomes vdv/dy= c/y^2 which is separable: vdv= (c/y^2)dy. Integrating on both sides (1/2)v^2= -c/y+ C.
(In an actual physical situation, v= speed and this equation is just "conservation of energy" (with the ma ss factored out): (1/2)v^2+ c/y= C is "kinetic energy plus potential energy is a constant".)

Then v^2= (dy/dt)^2= -2c/y+ C= (Cy- 2c)/y
dy/dt= sqrt{(Cy- 2c)/y} . Multiply both numerator and denominator inside the square root by y: dy/dt= sqrt{(Cy^2- 2cy)/y^2} and then
ydy/sqrt{Cy^2- 2cy}= dt is relatively easy to integrate.

[ramparts]

Thanks! I don't know how I missed your response way back when; as it turns out, I figured the same solution out for a class a couple of months ago (I was so proud of myself), and was all ready to post it here, too. But yeah, that's right. Although I disagree that that last integral is particularly easy without a substitution of some sort ;)

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