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Math Help - differentiation under the integral sign

  1. #1
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    differentiation under the integral sign

    I have this problem:

    With the aid of differentiation under the integral sign, show that the function

    y=integral from 0 to x of h(t)sin(x-t)dt

    satisfies the differential equation

    y''+y=h(t).

    Ok. I have the following equation to help me:
    the derivative of the integral from a(x) to b(x) of f(x,t)dt is equal to
    the integral from a(x) to b(x) of the partial derivative of x in f(x,t)dt + f(x,b(x))*b'(x)-f(x,a(x))*a'(x).

    So I note that
    a(x)=0
    b(x)=x
    a'(x)=0
    b'(x)=1
    f(x,t)=h(t)sin(x-t)
    the partial derivative of x of f(x,t)=h(t)cos(x-t)
    f(x,b(x))=0
    f(x,a(x))=h(0)sin(x)

    Thus, the derivative of the integral from a(x) to b(x) of f(x,t)dt is equal to
    the integral from 0 to x of h(t)cos(x-t)dt.

    This is where I get stuck. I'm pretty sure that if you plug in 0 and x into the above you get the second derivative, but I'm not positive.

    Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by James234 View Post
    I have this problem:

    With the aid of differentiation under the integral sign, show that the function

    y=integral from 0 to x of h(t)sin(x-t)dt
    y=\int_0^x h(t)\sin(x-t)dt

    y=\int_0^x h(t)(\sin x\cos t-\sin t\cos x ) dt
    y=\int_0^x h(t)\sin x\cos t-\int_0^x h(t)\cos x\sin t dt
    y=\sin x\int_0^x h(t)\cos t dt-\cos x \int_0^x h(t)\sin t dt
    y'=\cos x\int_0^x h(t)\cos tdt + h(x)\sin x \cos x+\sin x \int_0^x h(t)\sin t dt-h(x)\sin x \cos x
    y'=\cos x\int_0^x h(t)\cos tdt+\sin x\int_0^x h(t)\sin tdt
    y''=-\sin x\int_0^x h(t)\cos tdt+h(x)\cos^2 x+\cos x\int_0^x h(t)\sin tdt+h(x)\sin^2 x
    y''+y=h(x)\cos^2 x+h(x)\sin^2 x=h(x)(\cos^2 x+\sin^2 x)=h(x)(1)=h(x)
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  3. #3
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    Thank you for the response. Makes a lot of sense.

    Do you know the initial conditions that this function has to follow to justify this differentiation?
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