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Math Help - \theta''+\lambda\theta=0

  1. #1
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    \theta''+\lambda\theta=0

    Let \theta(x,\lambda) be the solution of

    \theta''+\lambda\theta=0
    \theta(0)=1
    \theta'(0)=0

    Use Green's formula to evaluate

    \displaystyle\int_0^L\theta^2(x,\lambda) \ dx

    and by specializing the result evaluate

    \displaystyle\int_0^L\cos^2\left(\frac{n\pi x}{L}\right) \ dx, \ \ \int_0^L\cos^2\left[\left(n+\frac{1}{2}\right)\frac{\pi x}{L}\right] \ dx

    where n is an integer.

    Help!

    Using Green's Formula, I obtain.

    \displaystyle\int_0^L\theta^2(x,\lambda) \ dx=\int_0^L[(A\theta)\bar{\theta}-\theta(A\bar{\theta}] \ dx=\theta'\bar{\theta}-\theta\bar{\theta}']_0^L=\theta'(L)\bar{\theta}(L)-\theta(L)\bar{\theta}'(L)

    where \displaystyle A=-\frac{d^2\theta}{dx^2}

    Now what?
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  2. #2
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    Sorry, but this question doesn't make a lot of sense as it's written. \displaystyle \theta is a function of two variables, so this should be a PDE, but you've written it as an ODE. Which variable has the function been differentiated with respect to?
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Sorry, but this question doesn't make a lot of sense as it's written. \displaystyle \theta is a function of two variables, so this should be a PDE, but you've written it as an ODE. Which variable has the function been differentiated with respect to?
    Lambda is the parameter the derivatives are wrt to x.
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  4. #4
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    Are you given any restriction on \displaystyle \lambda?
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Are you given any restriction on \displaystyle \lambda?
    Nope. The full question with all information is posted.
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