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Math Help - Differential Equations... find the general sum

  1. #1
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    Differential Equations... find the general sum

    dy/dt + 3y = t + e^(-2t)

    Find the general solution of the given differential equation and use it to determine how solutions behave at t --> inf

    So we did this problem in class and I'm a little confused about how the TA integrated it.

    (e^(3t)y)' = te^(3t) + e^t
    e^(3t)y = INT(te^(3t) + e^t)dt

    u = t
    du = dt
    dv = e^(3t)dt
    v = (1/3)e^(3t)

    e^(3t)y = (1/3)te^(3t) - (1/9)e^(3t) + e^t + C
    y = (1/3)t - (1/9) + e^(-2t) + Ce^(-3t)

    If someone could please explain how they got that after integration. Thank you.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by pakman View Post
    dy/dt + 3y = t + e^(-2t)

    Find the general solution of the given differential equation and use it to determine how solutions behave at t --> inf

    So we did this problem in class and I'm a little confused about how the TA integrated it.

    (e^(3t)y)' = te^(3t) + e^t
    e^(3t)y = INT(te^(3t) + e^t)dt

    u = t
    du = dt
    dv = e^(3t)dt
    v = (1/3)e^(3t)

    e^(3t)y = (1/3)te^(3t) - (1/9)e^(3t) + e^t + C
    y = (1/3)t - (1/9) + e^(-2t) + Ce^(-3t)

    If someone could please explain how they got that after integration. Thank you.

    which are you confused about? the integration or the y = ... or both.
    The TA did integration by parts to find the integral, and to go from
    e^(3t)y = (1/3)te^(3t) - (1/9)e^(3t) + e^t + C
    to
    y = (1/3)t - (1/9) + e^(-2t) + Ce^(-3t)
    he divided both sides by e^(3t)

    which parts exactly do you want explained? do you know the method for integration by parts?
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  3. #3
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    Quote Originally Posted by pakman View Post
    dy/dt + 3y = t + e^(-2t)

    Find the general solution of the given differential equation and use it to determine how solutions behave at t --> inf

    So we did this problem in class and I'm a little confused about how the TA integrated it.

    (e^(3t)y)' = te^(3t) + e^t
    e^(3t)y = INT(te^(3t) + e^t)dt

    u = t
    du = dt
    dv = e^(3t)dt
    v = (1/3)e^(3t)

    e^(3t)y = (1/3)te^(3t) - (1/9)e^(3t) + e^t + C
    y = (1/3)t - (1/9) + e^(-2t) + Ce^(-3t)

    If someone could please explain how they got that after integration. Thank you.
    It is of the form,

    y'+P(t)y=Q(t) where P,Q are continous almost everywhere.

    The integrating factor,

    R(t)=exp (INT P(t) dt)=exp (3t) = e^{3t}

    Thus, the solution is,

    y= 1/R(t) *INT [R(t)*Q(t)] dt

    =e^{-3t} * INT [e^{3t}*(e^{-2t}+t)] dt

    =e^{-3t}*INT [t*e^{3t}+e^t ]dt

    Now find the integral,

    INT [t*e^{3t}+e^t] = INT t*e^{3t} dt +INT e^t dt

    Integration by parts,

    (1/3)te^{3t}-(1/9)e^{3t}+e^t+C

    Thus,

    y=e^{-3t} [ (1/3)te^{3t}-(1/9)e^{3t}+e^t+C]

    =(1/3)t-(1/9)+e^{-2t}+C=(1/3)t+e^{-2t}+C

    Or something like that, I am too lazy to check.
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