Second order differential equation via substitution

[bitrex]

Homework Statement

Substitute p = \frac{dx}{dt} to solve x\prime\prime + \omega^2x = 0

Homework Equations

\frac{dp}{dx} = v + x\frac{dv}{dx}

v = \frac{p}{x}

The Attempt at a Solution

p = \frac{dx}{dt}, \frac{dp}{dt} = \frac{d^2x}{dt^2}

\frac{dp}{dt} = \frac{dp}{dx}\frac{dx}{dt} = \frac{dp}{dx}p

\frac{dp}{dx} + \frac{\omega^2x}{p} = 0

v + x\frac{dv}{dx} = \frac{-\omega^2}{v}

\frac{-v}{\omega^2 + v^2}dv = \frac{1}{x} dx

\frac{-1}{2}ln(\omega^2 + v^2) = ln|x| + C

\frac{1}{\sqrt{\omega^2 + v^2}} = x + C

I get tripped up here and I'm not sure how to go forward, with regards to all the various substitutions I've made! I see the beginnings of an integral involving trigonometric substitution, so I hope I may be on the right track. A hint would be much appreciated.

 

[gabbagabbahey]

I don't understand why you are making a second substitution. You have a separable 1st order ODE for p(x):

\frac{dp}{dx} + \frac{\omega^2x}{p} = 0

Just solve it.

 

[bitrex]

Wow, so I do. That's what I get for doing this when I'm tired!

 

[bitrex]

To finish up:

p dp = -\omega^2x dx

p^2 = -\omega^2 x^2

p = i \omega x + C_1

t = \int \frac{1}{i\omega x + C_1} dx

i \omega t = ln|i\omega x + C_1| + C_2

e^{i\omega t} = C_2(i\omega x + C_1) = C_2x + C_1C_2

\frac{1}{C_2}cos(\omega t) + \frac{1}{C_2}isin(\omega t) - C_1

x = Acos(\omega t) + Bsin(\omega t) if we set \frac{1}{C2} = A, \frac{1}{C1} = B, and let C_1 = \frac{1}{C_2}isin(\omega t) - \frac{1}{C_1}sin(\omega t).

 

[gabbagabbahey]

To finish up:

p dp = -\omega^2x dx

p^2 = -\omega^2 x^2

p = i \omega x + C_1

Why didn't the constant make an appearance in your second line?

p dp = -\omega^2x dx \implies p^2=-\omega^2x^2+C_1 \implies p=\sqrt{C_1-\omega^2 x^2}

 

[bitrex]

Carelessness. I'll try it again.

 

[bitrex]

I think I'm a little closer now. Continuing from the dropped constant and substituting into p = dx/dt we have:

\frac{1}{\sqrt{C_1 - \omega^2 x^2}} dx = dt

Make the substitution x = \frac{\sqrt{C_1}}{\omega} sin\theta

\int \frac{\sqrt{C_1} cos\theta}{\omega C_1 cos\theta} d\theta = \int \frac{\sqrt{C_1}}{\omega C_1} d\theta = \frac{\sqrt{C_1}}{\omega C_1}\theta + C_2

So t = \frac{\sqrt{C_1}}{\omega C_1} sin^-1(\frac{x \omega}{\sqrt{C_1}} + C_2)

And solving for X in terms of t I get:

\frac{\sqrt{C_1}}{\omega} sin (\frac{\omega t C_1 - \omega C1 C2}{\sqrt{C_1}})<br />

Not sure this is entirely correct, it's closer though! The first term we can call B because the constant swallows up the omega and square root, but as for the terms inside the sin() I'm not so sure. I know I can find the second solution by substituting y2 = y1*v into the original equation, where y1 = B sin (wt).

 

[gabbagabbahey]

I think I'm a little closer now. Continuing from the dropped constant and substituting into p = dx/dt we have:

\frac{1}{\sqrt{C_1 - \omega^2 x^2}} dx = dt

Make the substitution x = \frac{\sqrt{C_1}}{\omega} sin\theta

\int \frac{\sqrt{C_1} cos\theta}{\omega C_1 cos\theta} d\theta = \int \frac{\sqrt{C_1}}{\omega C_1} d\theta = \frac{\sqrt{C_1}}{\omega C_1}\theta + C_2

The C_1 in the denominator should be \sqrt{C_1}, so you get

\int\frac{dx}{\sqrt{C_1 - \omega^2 x^2}}=\frac{\theta}{\omega}+C_2

So t = \frac{\sqrt{C_1}}{\omega C_1} sin^-1(\frac{x \omega}{\sqrt{C_1}} + C_2)

Why is C_2 inside the arcsine?

\frac{\theta}{\omega}+C_2=\frac{1}{\omega}\sin^{-1}\left(\frac{\omega x}{\sqrt{C_1}}\right)+C_2

 

[bitrex]

I see it now. It's been months since I've worked on this stuff, and as you can tell I'm pretty out of practice. Ugly careless mistakes.