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Second Order ODE, Complex Roots, Change of Variables

  1. Oct 20, 2014 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations


    3. 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: Oct 20, 2014
  2. jcsd
  3. Oct 20, 2014 #2
    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
     
    Last edited: Oct 20, 2014
  4. Oct 20, 2014 #3
    I made a mistake copying my solution, I am editing now, there should not be a fraction in my separation
     
  5. Oct 20, 2014 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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.
     
    Last edited: Oct 20, 2014
  6. Oct 20, 2014 #5
    Thanks for the help!
     
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