- #1

- 351

- 3

## Homework Statement

Solve:

[itex] \frac{d^{2}y}{dx^{2}} + \omega^{2}y = 0 [/itex]

Show that the general solution can be written in the form:

[itex] y(x) = A\sin(\omega x + \alpha) [/itex]

Where A and alpha are arbitrary constants

## Homework Equations

## The Attempt at a Solution

I know that I will need to change variables for this problem, so I let

[itex] v = \frac{dy}{dx} [/itex]

Then, using the chain rule I found:

[itex] \frac{d^{2}y}{dx^{2}} = \frac{dv}{dx} = \frac{dv}{dy}\frac{dy}{dx} = v\frac{dv}{dy} [/itex]

Substituting this into the equation I found:

[itex] v\frac{dv}{dy} + \omega^{2} y = 0 [/itex]

This looks to me like a separable equation, so I made the following separation:

[itex] v\frac{dv}{dy} = -\omega^{2} y [/itex]

[itex] vdv = -\omega^{2} ydy[/itex]

I integrated both sides and obtained:

[itex] \frac{v^2}{2} = -\omega^{2} \frac{y^2}{2}[/itex]

After some manipulation I obtained:

[itex] v = +- sqrt(2(-\frac{\omega^{2}y}{2} + C)) [/itex]

Since:

[itex] v = \frac{dy}{dx} [/itex]

I made that substitution and found that:

[itex] \frac{dy}{dx} = +- sqrt(2-\frac{\omega^{2}y}{2} + C)) [/itex]

However, I am a little unsure of myself when it comes to problems like this, so I just wanted to see if I was on the right track with this problem or completely off base. My next step would be to integrate that square root, but I don't see how that would give me anything close to what the book wants for an answer.

Last edited: