# Thread: Model with First Order Differential Equation

1. ## 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.

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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: $u(t) = ce^{-kt} + T_0 + \frac{kT_1}{k^2 + w^2}(k\cos{wt} + w\sin{wt})$

2. Originally Posted by scherz0
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: $u(t) = ce^{-kt} + T_0 + \frac{kT_1}{k^2 + w^2}(k\cos{wt} + w\sin{wt})$

3. Thanks for your reply, ackbeet. I've since corrected my solution, but it still differs from the given answer.

4. Can you show your work, please? How did you integrate?

5. 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.

6. Originally Posted by scherz0
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.

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: $u(t) = ce^{-kt} + T_0 + \frac{kT_1}{k^2 + w^2}(k\cos{wt} + w\sin{wt})$

8. Originally Posted by scherz0
$\Rightarrow (e^{kt}u)=k\left(\int T_0e^{kt} dt + \int T_1\cos{wt} dt \right)$
$\Rightarrow (e^{kt}u)=k\left(\int T_0 e^{kt}\,dt + \int T_1 e^{kt}\cos(wt)\, dt \right).$
You failed to distribute the exponential $e^{kt}$ into your second integrand. Carry that correction through and see what happens.