Differential equations?

I've worked this out but not too sure if I've done it right.. Would like to see how others would work it out..

Show that the function f(t) = 3/2et2 - 1/2 is a solution to the differential equation y1 - 2ty = t

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  • 10 years ago
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    You do this by separation of variables.

    The equation is :

    dy/dt - 2ty = t

    So

    dy/dt = t (2y +1)

    And thus

    { 1/ (2y+1) }dy = t dt

    (1/2) ln (2y + 1) = (1/2) t^2 + C

    So

    ln (2y + 1) = t^2 + 2C

    (2y + 1) = e^(t^2 + 2C) = e^(t^2) * D where D=e^(2C)

    Rearranging,

    y = (D/2)e^(t^2) - 1/2.

    In particular, when D=3,

    y= f(t) = (3/2)e^(t^2) - 1/2 (shown)

    Regards

    Edem

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