please help with this differential eq. problem

hi, i am trying to solve:

m(d^2/dt^2)=F(x) (1)

where F(x) = m/x^3. v=0, x=1, t=0

the book says... multiply (1) by v=dx/dt and integrate w.r.t. t to get

mv(dv/dt) = [F(x)](dx/dt) or mv dv = F(x)dx

and finally m(v^2)/2 = integr {F(x)dx} + const.

i'm real stuck with this type of questions please can you help me see the solution method??!!

Reply 1

Babithas

(d^2x/dt^2) = dv/dt
Dx/dt= V

This means you can rewrite the
m(d^2x/dt^2)=F(x) as M (Dv/dt) = F(x)
So F(x) (dx/dt) = Mv (dv/dt)

Int mV(dv/dt) dt = m V^2/2 + c

what exam board is this?

Reply 2

pepsigirl

okay, thanks for your responses... i have worked through the problem up to a point but an integration is still giving me some trouble... here is what i have done and what troubles me...

(m/2)[(dx/dt)^2]=integr[m/(x^3)]dx =>

(1/2)[(dx/dt)^2]=-(1/2x^2)+c/2 =>

(dx/dt)^2 = -1/x^2 +c =>

...first? do I put +/- and know work two cases for the positive and negative root?

v = dx/dt = {[-1+(x^2)]^(1/2)/x}

here i use the initial conditions that v=0 and x=1

0 = (-1+c) => c=1

so... dx/dt = (-1+x^2)/x^2 => {x/[-1+(x^2)]^(1/2)}dx=dt

and make the substitution x^2 = sinu which works out however, in the next exercise (please allow me to insert this question here) i end up with

{(x^2)/[1-(x^4)]^(1/2)}dx=dt and if i similarly put x^2 = sinu need to integrate the sqrt.root of sinu and don't know how. please help.

p.s. i am not doing it for an examination, i haven't studied maths before but am entering an MSc in maths course in a couple of weeks and doing some revision (kidding.. not actually)

Reply 3

pepsigirl

much appreciated, what would you suggerst for this?
{(x^2)/[1-(x^4)]^(1/2)}dx=dt

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