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Math Help - Second Order ODE - Change of Variable

  1. #1
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    Second Order ODE - Change of Variable

    I am having problems with this ODE:

    Derive the solution of the ordinary differential equation
     \frac{d^2y}{dx^2} = f(x)
    x > 0 ,  y(0) = 0
    \frac {dy}{dx} (0) = 0

    In the form:

    y(x) = \int_{0}^{x} (x-t)f(t)dt

    Can't use particular integral/complementry function... I have tried to use a laplace transform but get stuck on integgrating an unknown function f(x)... also dont understand the change of variable to 't' - the  (x-t) suggests to me a laplacian shift. I am stuck... Thanks for looking!
    Last edited by padawan; August 10th 2010 at 01:10 PM.
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  2. #2
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    My first instinct was to substitute y=\int_0^x (x-t)f(t)dt into the ODE and show it satisfies the ODE. But integrating by parts and differentiating with respect to x shows that \frac{dy}{dx} = f(x) Unless I have made an error somewhere, that will leave only the trivial solution. Is there perhaps a typo in the question?
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  3. #3
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    Well, it is a fact that if you differentiate the integral form given there twice, you end up back at the DE you started with. You could just reverse the steps of differentiating the integral form, though I'd say such a thing wouldn't be all that intellectually satisfying. I mean, if I were to integrate the ODE, I'd probably end up with

    y(x)=\int_{0}^{x}\int_{0}^{t}f(s)\,ds\,dt,

    which would work just as well, except that it's two integrations. Now, if the function f were known, and if its first two antiderivatives were elementary, then probably most anyone would solve it the way I just did. The advantage of the form given is that you've reduced a second-order ODE to a first-order integral equation. It's clever.
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  4. #4
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    Thanks guys... I guess Imissed the obvious there. There is no typo, the trivial solution will be fine for me! and I will work through Ackbeet's solution! thanks again
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  5. #5
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    Reply to badgerigar at post # 2:

    I suppose you could integrate by parts, but it'd be tricky. You'd have integrals of f in there. Why not split the integral into two pieces? The x will come out of the first integral. Then you're differentiating. You can use the fundamental theorem of the calculus to avoid having actually to compute the integral.

    I really don't see where the \frac{dy}{dx}=f(x) came from. Could you please show your steps to get that?
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    Well, it is a fact that if you differentiate the integral form given there twice, you end up back at the DE you started with. You could just reverse the steps of differentiating the integral form, though I'd say such a thing wouldn't be all that intellectually satisfying. I mean, if I were to integrate the ODE, I'd probably end up with

    y(x)=\int_{0}^{x}\int_{0}^{t}f(s)\,ds\,dt,

    which would work just as well, except that it's two integrations. Now, if the function f were known, and if its first two antiderivatives were elementary, then probably most anyone would solve it the way I just did. The advantage of the form given is that you've reduced a second-order ODE to a first-order integral equation. It's clever.
    I'll pick it up here with a switch of variables

    y(x)=\int_{0}^{x}\int_{0}^{s}f(t)\,dt\,ds.

    Now switch the order of integration

    y(x)=\int_{0}^{x}\int_{t}^{x}f(t)\,ds\,dt, and integrate once.
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  7. #7
    MHF Contributor chisigma's Avatar
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    In terms of Laplace Tranform the DE is written as...

    s^{2}\ Y(s) - F(s)= 0 (1)

    ... and the 'other identity' ...

     Y(s) = \frac{F(s)}{s^{2}} (2)

    That means that for any  f(x) that is L-transformable the solution of the DE...

    y^{''} = f(x) , y(0)=0 , y^{'}(0)=0 (3)

    ... is...

    \displaystyle y(x)=\int_{0}^{x} (x-t)\ f(t)\ dt (4)

    Kind regards

    \chi \sigma
    Last edited by chisigma; August 11th 2010 at 06:53 AM. Reason: more correct procedure...
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  8. #8
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    Thumbs up

    A big thanks to everyone that has contributed... I'll work through the different methods- can never learn enough and can certainly learn from the guys in here
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  9. #9
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    Yeah, I found my mistake. I somehow lost one of the integrals in there somewhere. The method I used, with the mistake corrected, looks like this
    y= \int_0^x(x-t)f(t)dt
    =x\int_0^xf(t)dt-\int_0^xtf(t)dt
    integrating by parts
    u=t
    dv=f(t)dt
    du=dt
    v=\int f(t)dt
    =x\int_0^xf(t)dt-t|^x_0\int_0^xf(t)dt+\int_0^x\int f(t)dtdt
    =\int_0^x\int f(t)dtdt

    which is a derivation of what Ackbeet was able to see with no calculation whatsoever.
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  10. #10
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    \chi \sigma:I can see these proofs starting with the answer, but I still can't see your method:

    I can follow to:

    Y(s) = \frac{F(s)}{s^{2}}

    but fail to see how this equates to the integral form with limits of 0 and x... I have tried breaking the Laplace transforms into the \int_{o}^{\infty}f(x)e^{-st} but this does not seem to simplify for me... is this the approach you used \chi\sigma?
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  11. #11
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by padawan View Post
    \chi \sigma:I can see these proofs starting with the answer, but I still can't see your method:

    I can follow to:

    Y(s) = \frac{F(s)}{s^{2}}

    but fail to see how this equates to the integral form with limits of 0 and x... I have tried breaking the Laplace transforms into the \int_{o}^{\infty}f(x)e^{-st} but this does not seem to simplify for me... is this the approach you used \chi\sigma?
    A defect of this forum is that often the answers is written quickly for not to be beaten form somebody else! ...

    We agree that is...

    Y(s) = \frac{F(s)}{s^{2}} (1)

    ... and now You have to remember that if F(s)= \mathcal {L} \{f(t)\} and G(s)= \mathcal {L} \{g(t)\} is...

    \mathcal {L}^{-1} \{F(s) \cdot G(s)\} = \int_{0}^{t} g(t-\tau) \cdot f(\tau)\ d \tau (2)

    Because is {L}^{-1} \{F(s)\} = f(t) and {L}^{-1} \{ \frac{1}{s^{2}}\} = t , from (1) and (2) if follows that...

    \mathcal {L}^{-1} \{\frac{F(s)}{s^{2}}\} = \int_{0}^{t} (t-\tau) \cdot f(\tau)\ d \tau (3)

    Kind regards

    \chi \sigma
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  12. #12
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    Thats great \chi \sigma... I follow it now. The one thing I did not look at was the Laplace convolution whish I guess this is the inverse form.

    Thanks for that.. I really appreciate the help.
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