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Math Help - Leibniz rule

  1. #1
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    Leibniz rule

    Hi guys I am currently stuck at a question.

    i) Show that if x(t) satisfies the integral equation

    x(t)=a+bt+ \int_0^t (t-s)f(x(s)) ds

    then x(t) is a solution to the initial value problem

    x''(t)=f(x(t)) for t>0, with x(0)=a , \ x'(0)=b.

    ii) Prove the converse of the result in part (i).

    (Hint: You will need to do a change of order of integration in a double integral.)

    Part (i) I differentiated x(t) twice and got the result by applying Leibniz rule and the fundamental theorem of calculus.

    For part (ii) I don't even know where to start. Where would I be using a double integral? I tried proving it just by integrating x''(t) but I wasn't sure if i could do that.

    Any help would be appreciated.

    Thanks.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by sakodo View Post
    For part (ii) I don't even know where to start. Where would I be using a double integral? I tried proving it just by integrating x''(t) but I wasn't sure if i could do that.

    Necessarily

    x'(t)=\displaystyle\int_{0}^{t}f(x(s))\;ds +b


    Now, you'll have to integrate again.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by sakodo View Post
    Hi guys I am currently stuck at a question.

    i) Show that if x(t) satisfies the integral equation

    x(t)=a+bt+ \int_0^t (t-s)f(x(s)) ds

    then x(t) is a solution to the initial value problem

    x''(t)=f(x(t)) for t>0, with x(0)=a , \ x'(0)=b.

    ii) Prove the converse of the result in part (i).

    (Hint: You will need to do a change of order of integration in a double integral.)

    Part (i) I differentiated x(t) twice and got the result by applying Leibniz rule and the fundamental theorem of calculus.

    For part (ii) I don't even know where to start. Where would I be using a double integral? I tried proving it just by integrating x''(t) but I wasn't sure if i could do that.

    Any help would be appreciated.

    Thanks.
    Written in term of Laplace-transform the DE is...

    \displaystyle s^{2}\ X(s) + s\ x(0) - x^{'} (0) = \mathcal{L} \{f[x(t)]\}

    ... and from (1), performing the inverse Laplace trasform, You obtain...

    \displaystyle x(t)= (a+b\ t)\ \mathcal {U} (t) + \int_{0}^{t} (t-\tau) f[x(\tau)] \ d \tau (2)

    Kind regards

    \chi \sigma
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