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ODE: Solving using Laplace Transform

  1. Oct 16, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Solve:
    [tex]y''+λ^2y = cos(λt), y(0) = 1, y'(π/λ) = 1 [/tex]
    where t > 0

    2. Relevant equations


    3. The attempt at a solution

    I start off by taking the Laplace transform of both sides. I get:

    [tex]L(y) = \frac{s}{(s^2+λ^2)^2}+\frac{sy(0)}{s^2+λ^2}+\frac{y'(0)} {s^2+λ^2}[/tex]

    Now take the inverse Laplace transform

    [tex] y = \frac{1}{2λ}tsin(λt) + y(0)cos(λt)+\frac{y'(0)}{λ}sin(λt)[/tex]

    since y(0) = 1, we get

    [tex] y = \frac{1}{2λ}tsin(λt) +cos(λt)+\frac{y'(0)}{λ}sin(λt)[/tex]


    Now this is where I get stuck. I have the boundary condition for y'(pi/λ)=1, but I'm not sure how to use it in this scenario.

    If I take the derivative of y, then I get an expression with a y'(0) term in it that I'm not sure how to use...

    Any thoughts?
     
  2. jcsd
  3. Oct 16, 2016 #2
    Well, you end up with the solution of ##y'(0)## when you do that, no? If you put that back into your expression for ##y(t)##, isn't that the solution to your DE?
     
  4. Oct 16, 2016 #3

    vela

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    Remember that ##y'(0)## is just a constant.
     
  5. Oct 16, 2016 #4

    RJLiberator

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    Ah, beautiful, this was the hint I needed.

    So I am capable of taking the derivative of y, since y'(0) is just some constant.

    [tex] y'(t) = \frac{1}{2λ}[sin(λt)+λtcos(λt)]-λsin(λt)+y'(0)cos(λt) [/tex]

    Solving this for y'(π/λ) = 1 shows that y'(0) = -1-π/(2λ)

    And thus, I have my determined y.
     
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