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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 November 11th, 2017, 02:00 PM #1 Senior Member     Joined: Nov 2015 From: United States of America Posts: 181 Thanks: 21 Math Focus: Calculus and Physics thermo An object is placed in an environment where temperature is kept at a constant temperature $T_o$. The temperature of the object is initially $T_o$. Now, the object is directly heated by an infra red light beam. Assuming that the temperature of the environment remains $T_o$, the temperature of the object at time t, T(t), obeys the following differential equation... $\frac{dT}{dt} = -k(T-T_o) + q$ where k and q are positive constant. The first term in the right hand side represents cooling by the environment and the second term heating by the beam. Find T(t). What is the final temperature when time goes to infinity? I am going to run through how I did this, but it is incorrect. I have a solution for this, but the limits of the integration through the differential equation keep changing to meaningless variables not introduced in the problem or solution which makes for a very confused student. Anyhow... $T-T_o$ is the change in temperature, so I will call that $\tau$ $T-T_o = \tau \Rightarrow \frac{d\tau}{dt} = -k\tau + q$ This looks like a linear differential equation in the form of $\frac{dy}{dx} + P(x)y = Q(x)$ I will now find my integrating factor. $\mu(x) = e^{\int{kdt}} = e^{kt}$ $\frac{d\tau}{dt}*e^{kt} + k\tau*e^{kt} = qe^{kt}$ Then I make not of the product rule going here... $\frac{d}{dt} \tau e^{kt} = qe^{kt}$ $\Rightarrow$ $\tau e^{kt} = \frac{q}{k}e^{kt}$ $\Rightarrow$ $\tau = T-T_o = \frac{q}{k}$ $T = \frac{q}{k} + T_o$ My question is, why did the solution I provided below use the variable "s" in one of the integrals? Also, why is there a negative in front of the integrating factor? If I had the correct limits of integration, I might have a chance of solving this correctly, but I don't see where these limits of integration are coming from. Can anyone help me out? Correct solution:
 November 11th, 2017, 02:00 PM #2 Senior Member     Joined: Nov 2015 From: United States of America Posts: 181 Thanks: 21 Math Focus: Calculus and Physics Sorry about the poor choice in the subject of this post. It was just a place holder, but I hit submit without changing it.
 November 12th, 2017, 01:21 PM #3 Senior Member     Joined: Nov 2015 From: United States of America Posts: 181 Thanks: 21 Math Focus: Calculus and Physics Did it a different way and arrived at the same solution. Finally! Thanks from romsek

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