Second Order ODE, Complex Roots, Change of Variables

[_N3WTON_]

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.

 

[_N3WTON_]

I just realized that perhaps I could solve this using an auxiliary equation, so I'm going to try that out real quick..
Edit: not working out

 

[_N3WTON_]

I made a mistake copying my solution, I am editing now, there should not be a fraction in my separation

 

[HallsofIvy]

To prove that this is the general solution, you need to do two things: show that it is a solution- that it satisfies the differential equation- and that every solution is of that form. The first part is easy: just take the second derivative of the function and put it into the equation.

For the second, note that [tex]A sin(\omega x+ \alpha)= A(cos(\alpha) sin(\omega x)+ sin(\alpha) cos(\omega x))= C sin(\omega x)+ D cos(\omega x)[/tex] where [itex]C= Acos(\alpha)[/itex] and [itex]D= Asin(\alpha)[/itex]. Since A and [itex]\alpha[/itex] can be any constants, so can C and D.

Then use the fact that, since this is a second order linear homogeneous differential equation, the set of all solutions forms a two dimensional vector space and that [itex]sin(\omega x)[/itex] and [itex]cos(\omega x)[/itex] are independent solutions and so form a basis for that vector space.

 

[_N3WTON_]

To prove that this is the general solution, you need to do two things: show that it is a solution- that it satisfies the differential equation- and that every solution is of that form. The first part is easy: just take the second derivative of the function and put it into the equation.

For the second, note that [tex]A sin(\omega x+ \alpha)= A(cos(\alpha) sin(\omega x)+ sin(\alpha) cos(\omega x))= C sin(\omega x)+ D cos(\omega x)[/tex] where [itex]C= Acos(\alpha)[/itex] and [itex]Asin(\alpha)[/itex]. Since A and [itex]\alpha[/itex] can be any constants, so can C and D.

Then use the fact that, since this is a second order linear homogeneous differential equation, the set of all solutions forms a two dimensional vector space and that [itex]sin(\omega x)[/itex] and [itex]cos(\omega x)[/itex] are independent solutions and so form a basis for that vector space.

Thanks for the help!