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Math Help - differentiating through the integral

  1. #1
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    differentiating through the integral

    The problem is as follows:
    Solve for f(x), given

    f(x) = x + \int^{x}_{0}{(x-2t)f(t) dt}


    What I did:
    1. Differentiate both sides wrt x, and partially differentiate through the integral.

    f'(x) = 1 + \frac{d}{dx} \int^{x}_{0}{(x-2t)f(t) dt}
    f'(x) = 1 + \int^{x}_{0}\frac{\partial}{\partial x}{(x-2t)f(t) dt}
    f'(x) = 1 + (x-2x)f(x) + \int^{x}_{0}{f(t) dt}
    f'(x) = 1 - xf(x) + \int^{x}_{0}{f(t) dt}

    2. Differentiate both sides wrt x again.
    f''(x) = - xf'(x) - f(x) + \frac{d}{dx}\int^{x}_{0}{f(t) dt}
    f''(x) = - xf'(x) - f(x) + f(x)
    f''(x) = - xf'(x)

    This is an ODE, yes, but how would I solve for this? (i.e. is there a less messy way to calculate this other than using series solutions, which I hate with a passion?)
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  2. #2
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    Quote Originally Posted by compliant View Post
    The problem is as follows:
    Solve for f(x), given

    f(x) = x + \int^{x}_{0}{(x-2t)f(t) dt}


    What I did:
    1. Differentiate both sides wrt x, and partially differentiate through the integral.

    f'(x) = 1 + \frac{d}{dx} \int^{x}_{0}{(x-2t)f(t) dt}
    f'(x) = 1 + \int^{x}_{0}\frac{\partial}{\partial x}{(x-2t)f(t) dt}
    f'(x) = 1 + (x-2x)f(x) + \int^{x}_{0}{f(t) dt}
    f'(x) = 1 - xf(x) + \int^{x}_{0}{f(t) dt}

    2. Differentiate both sides wrt x again.
    f''(x) = - xf'(x) - f(x) + \frac{d}{dx}\int^{x}_{0}{f(t) dt}
    f''(x) = - xf'(x) - f(x) + f(x)
    f''(x) = - xf'(x)

    This is an ODE, yes, but how would I solve for this? (i.e. is there a less messy way to calculate this other than using series solutions, which I hate with a passion?)
    Let y = f(x). Then you have

    \frac{d^2 y}{dx^2} = - x \frac{dy}{dx}.

    Substitute \frac{dy}{dx} = p .... (1):

    \frac{dp}{dx} = - x p .... (2)

    DE (2) is seperable. Solve it for p as a function of x. Then substitute your expression for p into DE (1) and solve for y as a function of x.
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