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Math Help - Model with First Order Differential Equation

  1. #1
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    Model with First Order Differential Equation

    Dear all,

    I'm having some trouble with the following question involving modelling with a first-order differential equation. My solution is below but it differs from the given answer. I also can't see where I might have erred.

    Thank you.

    ----

    18. Consider an insulated box (a building, perhaps) with internal temperature u(t). According to Newton's law of cooling, u satisfies the differential equation

    \cfrac{du}{dt} = -k[u - T(t)],

    where T(t) is the external temperature. Suppose that T(t) varies sinusoidally; assume that T(t) = T_0 + T_1cos(wt)

    a) Solve the differential equation above and express u(t) in terms of t,k,T_0,T_1, w.

    My solution:

    The above ODE is a linear equation. Rearrangement gives:
    \dfrac{du}{dt} + ku = k(T_0 + T_1\cos{wt}).

    This linear ODE has integrating factor  e^{\int k dt} = e^{kt} .

    Therefore,  \dfrac{d}{dt}(e^{kt}u) = e^{kt}k(T_0 + T_1\cos{wt})

    \Rightarrow (e^{kt}u)=k\left(\int T_0e^{kt} dt + \int T_1\cos{wt} dt \right)

    \Rightarrow (e^{kt}u)=k\left(T_0\frac{e^{kt}}{k} + T_1\frac{\sin{wt}}{w} \right) + C, where C has absorbed the product of integration constants from both integrals with  k .

     \Rightarrow u = T_0 - T_1\dfrac{k\sin{wt}}{we^{kt}} + Ce^{-kt}.

    However, the given answer is: u(t) = ce^{-kt} + T_0 + \frac{kT_1}{k^2 + w^2}(k\cos{wt} + w\sin{wt})
    Last edited by scherz0; May 31st 2011 at 09:11 AM.
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  2. #2
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    Quote Originally Posted by scherz0 View Post
    Dear all,

    I'm having some trouble with the following question involving modelling with a first-order differential equation. My solution is below but it differs from the given answer. I also can't see where I might have erred.

    Thank you.

    ----

    18. Consider an insulated box (a building, perhaps) with internal temperature u(t). According to Newton's law of cooling, u satisfies the differential equation

    \cfrac{du}{dt} = -k[u - T(t)],

    where T(t) is the external temperature. Suppose that T(t) varies sinusoidally; assume that T(t) = T_0 + T_1cos(wt)

    a) Solve the differential equation above and express u(t) in terms of t,k,T_0,T_1, w.

    My solution:

    The above ODE is a linear equation. Rearrangement gives:
    \cfrac{du}{dt} + ku = k(T_0 + T_1\cos{wt}).
    This linear ODE has integrating factor  e^{\int k dt} = e^{kt} .
    Therefore,  \frac{d}{dt}(e^{kt}u) = e^{kt}k(T_0 + T_1\cos{wt})
    \Rightarrow (e^{kt}u)=k(\int T_0e^{kt} dt + \int T_1\cos{wt} dt)
    This step is incorrect. The exponential multiplies the trig function integral as well.

     \Rightarrow u = T_0e^{kt} - T_1\frac{\sin{wt}}{wk}.

    However, the given answer is: u(t) = ce^{-kt} + T_0 + \frac{kT_1}{k^2 + w^2}(k\cos{wt} + w\sin{wt})
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  3. #3
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    Thanks for your reply, ackbeet. I've since corrected my solution, but it still differs from the given answer.
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  4. #4
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    Can you show your work, please? How did you integrate?
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  5. #5
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    Thanks for your reply, ackbeet. I've shown my integral evaluations. But I still don't understand the difference between my solution and the given answer.
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  6. #6
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    Quote Originally Posted by scherz0 View Post
    Thanks for your reply, ackbeet. I've shown my integral evaluations. But I still don't understand the difference between my solution and the given answer.
    If you edited the OP as "showing your integral evaluations", I still don't buy your calculations. You're still not doing what I suggested in Post # 2. Please do NOT edit your OP again. Post your workings in a new post in this thread.
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  7. #7
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    Thanks for your reply, Ackbeet.

    You earlier requested that I show how I evaluate the integrals. Here it is below.


     \Rightarrow (e^{kt}u)=k\left(\int T_0e^{kt} dt + \int T_1\cos{wt} dt \right)

    \Rightarrow (e^{kt}u)=k\left(T_0 \int e^{kt} dt + T_1 \int \cos{wt} dt \right) , where T_0, T_1 are constants.

    Since  \int e^{kt} dt = \frac{e^{kt}}{k} and \int \cos{wt} dt = \frac{sin {wt}}{w}

    Therefore  (e^{kt}u)=k\left(T_0\frac{e^{kt}}{k} + T_1\frac{\sin{wt}}{w} \right) + C, where C has absorbed the product of integration constants from both integrals with  k .

     \Rightarrow u = T_0 - T_1\dfrac{k\sin{wt}}{we^{kt}} + Ce^{-kt}.

    However, the given answer is: u(t) = ce^{-kt} + T_0 + \frac{kT_1}{k^2 + w^2}(k\cos{wt} + w\sin{wt})
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  8. #8
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    Quote Originally Posted by scherz0 View Post
    Thanks for your reply, Ackbeet.

    You earlier requested that I show how I evaluate the integrals. Here it is below.


     \Rightarrow (e^{kt}u)=k\left(\int T_0e^{kt} dt + \int T_1\cos{wt} dt \right)
    This line is incorrect. It should be

     \Rightarrow (e^{kt}u)=k\left(\int T_0 e^{kt}\,dt + \int T_1 e^{kt}\cos(wt)\, dt \right).

    You failed to distribute the exponential e^{kt} into your second integrand. Carry that correction through and see what happens.
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