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Math Help - Variation of Parameters

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    Variation of Parameters

    The question to this problem is t^2y'' - 2y = 3t^2 - 1, t > 0; y1(t) = t^2, y2(t) = t^-1
    Where y1 and y2 satisfy the corresponding homogeneous equation. Then the particular solution to that problem would be:
    At^2 + Bt + C right.
    Last edited by fudawala; July 4th 2007 at 09:16 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fudawala View Post
    The question to this problem is t^2y'' - 2y = 3t^2 - 1, t > 0; y1(t) = t^2, y2(t) = t^-1
    Where y1 and y2 satisfy the corresponding homogeneous equation. Then the particular solution to that problem would be:
    At^2 + Bt + C right.
    No, you do not guess solutions with Variation of Parameters as you do with Undetermined coefficients.

    Here's the method of Variation of Parameters:

    Given a differential equation of the form: y'' + p(t)y' + q(t) = f(t) having homogeneous solutions y_1(t) and y_2(t), we assume, by the method of Variation of Parameters, that a particular solution exists of the form:

    y_p = A(t)y_1(t) + B(t)y_2(t)

    where A(t) and B(t) are found from the system of equations:

    A'(t)y_1 + B'(t)y_2 = 0 ...............(1)
    A'(t)y_1' + B'(t)y_2' = f(t) ...............(2)

    since y_1 and y_2 are knowns, we can find the values of A'(t) and B'(t) by Cramer's Rule, and then integrate them to get A(t) and B(t)

    then as always, our general solution y_g(t) is given by:

    y_g(t) = y_1(t) + y_2(t) + y_p(t)


    Most textbooks, including your own, condenses this whole method into a single formula involving an integral, look it up if you can't bother with Cramer's rule
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