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Math Help - Variation of Parameters

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    Variation of Parameters

    Jhevon, Tell me if the particular solutions to these problems are correct:
    Where y1 and y2 satisfy the homogeneous equation. Can you show me it by variation of parameters if you can. But if the integral gets worse than try to show it by Method of Undetermined Coefficient. Tell me if these particular solutions are correct and where is my mistake.

    Problem 14 Page 190:
    t^2y'' - t(t + 2)y' + (t + 2)y = 2t^3, t > 0; y1(t) = t, y2(t) = t*e^t
    The particular solution to this is: At^3 + Bt^2 + Ct + D

    Problem 15 Page 190:
    ty'' - (1 + t)y' + y = t^2*e^2t, t > 0; y1(t) = 1 + t, y2(t) = e^t
    The particular solution to this is: (At^2 + B*t + C)e^2t

    Problem 16 Page 190:
    (1 - t)y'' + ty' - y = 2(t - 1)^2*e^-t, 0 < t < 1; y1(t) = e^t, y2(t) = t
    The particular solution to this is: (At^2 + B*t + C)e^-t
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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
    Jhevon, Tell me if the particular solutions to these problems are correct:
    Where y1 and y2 satisfy the homogeneous equation. Can you show me it by variation of parameters if you can. But if the integral gets worse than try to show it by Method of Undetermined Coefficient. Tell me if these particular solutions are correct and where is my mistake.

    Problem 14 Page 190:
    t^2y'' - t(t + 2)y' + (t + 2)y = 2t^3, t > 0; y1(t) = t, y2(t) = t*e^t
    The particular solution to this is: At^3 + Bt^2 + Ct + D

    Problem 15 Page 190:
    ty'' - (1 + t)y' + y = t^2*e^2t, t > 0; y1(t) = 1 + t, y2(t) = e^t
    The particular solution to this is: (At^2 + B*t + C)e^2t

    Problem 16 Page 190:
    (1 - t)y'' + ty' - y = 2(t - 1)^2*e^-t, 0 < t < 1; y1(t) = e^t, y2(t) = t
    The particular solution to this is: (At^2 + B*t + C)e^-t
    see http://www.mathhelpforum.com/math-he...arameters.html for the method of Variation of Parameters, you do not guess solutions the same way with this method as you do Undetermined coefficients
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