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Math Help - [SOLVED] I need help for derive d2y/dx2 in integral form

  1. #1
    perfect
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    [SOLVED] I need help for derive d2y/dx2 in integral form

    Hello!
    I have problem:
    Derive the solution of the ordinary differential equation
    d2y/dx2 =f(x), x>0, y(x)=0, dy/dx (0)=0,
    in form y(x)= integral from 0 to x [(x-t)f(t)dt].
    tnxs
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by perfect View Post
    Hello!
    I have problem:
    Derive the solution of the ordinary differential equation
    d2y/dx2 =f(x), x>0, y(x)=0, dy/dx (0)=0,
    in form y(x)= integral from 0 to x [(x-t)f(t)dt].
    tnxs
    Some clarification is needed. It is not possible in general to write the solution of:

    <br />
\frac{d^2y}{dx^2}=f(x)<br />

    with y(0)=0,\ y'(0)=0, in the form

    y(x)=\int_0^x (x-t) f(t) dt.

    To show this just put f(x)=x.

    RonL
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  3. #3
    Super Member Rebesques's Avatar
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    perfect:
    Derive the solution of the ordinary differential equation
    d2y/dx2 =f(x), x>0, y(x)=0, dy/dx (0)=0,
    in form y(x)= integral from 0 to x [(x-t)f(t)dt]

    Are you asked to actually prove the formula, or just show this is the solution? Because the latter is easy: Differentiate twice (under the integral sign) to get y"=f, and since y also gives y(0)=0, y'(0)=0, from the uniqueness theorem this is the only solution!

    Yes, I know it's cheating, but it's a way. Now if you are asked to derive the formula, write
    y(x)=y(x)-y(0)=\int_{0}^{x}y'(t)dt

    and integrating by parts,

    y(x)=\int_{0}^{x}y'(t)(t)'dt=y'(t)t\bigg|_{0}^x-\int_0^xy''(t)tdt=xy'(x)-\int_0^xf(t)tdt, (1)

    since y''=f. Now one integration of the differential equation gives us y'(x)=\int_0^xf(t)dt, and substitute into (1) to get

    y(x)=x\int_0^xf(t)dt-\int_0^xf(t)tdt=\int_0^x(x-t)f(t)dt.



    Captainblack:
    To show this just put...
    The function must also satisfy the initial conditions.
    Last edited by Rebesques; August 17th 2007 at 01:33 AM. Reason: bad eyesight
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  4. #4
    Grand Panjandrum
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    Captainblack:
    To show this just put...
    The function must also satisfy the initial conditions.
    In fact I was misreading the question statement in another way (the function does not have to satisfy the initial condition but the solution does).

    RonL
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