Solving $dy/dt = k(A-y)$ (phrased as rate of change probelm)

Question

The Problem

The rate at which temperature decreases in a house is proportional to the difference between the current house-temperature and the temperature outside.

It's -10 degrees outside, at the start it's 20 degrees inside the house and after an hour it's 17 degrees. After how long is it minus degrees inside?

My solution (attempt)

We know that

$$dy/dt = k(y+10) \\ y(0) = 20 \\ y(1) = 17$$

We do a variable substitution $u(t) = y(t) + 10$ and we now have

$$ du/dt = ku \\ u(0) = 30 \\ u(1) = 27$$

A solution to the equation is $u(t) = e^{kt} + A$

$$u(0) = 30 \rightarrow e^{0k}+A = 1+A = 30 \rightarrow A = 29$$ $$u(1) = 27 \rightarrow e^k + 29 = 27 \rightarrow k = ln(-2) \rightarrow Undefined$$

And that's where I'm stuck. Am I supposed to re-write the difference $y+10$ to something else?

Answer 1

You were close to the solution

$$dy/dt = k(y+10) \,; y(0) = 20 \,; y(1) = 17$$ $$y'=k(y+10) \implies \int \frac {dy}{y+10}=\int kdx=kx+C$$ $$\ln(y+10)=kx+C$$ $$y=e^{kx+c}-10$$ $$\boxed {y=Ce^{kx}-10}$$

Use the initial conditions to calculate k and C

$$C=30 \text{ , and } k=\ln(\frac 9 {10})$$

Answer 2

Hint: Check if your proposed solution satisfied $u'(t) = k\, u(t)$.

Answer 3

Yes, using the change of variable u= y+ 10 then the differential equation becomes du/dt= ku. However, the general solution to that differential equation is NOT "$u(t)= e^{kt}+ A$". You can see that by putting that u into the equation- with $u= e^{ky}+ A$, $du/dt= ke^{kt}$ which is NOT $ku= ke^{kt}+ kA$.

Instead the general solution is $u(t)= Ae^{kt}$. That way, $du/dt= kAe^{kt}$ which is equal to $ku$.

What I would have expected you to do is write the equation as $du/u= kdt$. Integrating both sides gives $ln(u)= kt+ C$ so that, taking the exponential of both sides $u= e^{kt+ C}= e^Ce^{kt}= De^{kt}$ where $D= e^C$. (I am using "D" rather than "A" because you already have a "A" in the problem.)

With $u= y+ 10$, $y= u- 10= De^{kt}- 10$.