Differential equation - Newton's law of cooling

Question

Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. Let $y(t)$ be the temperature of the water in a pot at the time t minutes. When the water boils the pot is put outside where the temperature is $-20^o$ (sorry about that all Americans).

The temperature $y(t)$ corresponds to the differential equation on the form $y'(t) = k(y(t)+20)$. We also know that the temperature is $40^o$ after 10 minutes.

a) Solve the differential equation. You should use the substitution $u(t) = y(t) + 20 $.

The problem here is that I don't know what they mean by using a substitution.

Answer 1

We have that

• $y'(t) = k(y(t)+20)$

let

• $u(t)=y(t)+20 \implies u'(t)=y'(t)$

then we need to solve by separation of variables

• $u'(t) = ku(t)\iff \frac{du}{u}=k\,dt\iff\ln u=kt+c_1 \iff e^{\ln u}=e^{kt+c_1}\iff u=e^{c_1}e^{kt}\\\iff u(t)=ce^{kt}\iff y(t)=ce^{kt}-20$

and from the initial condition

• $y(0)=ce^{k0}-20=c-20=100 \implies c=120$

and the condition after 10 minute

• $y(10)=120e^{10k}-20=40$

from which we can determine $k$ and solve the problem.

Answer 2

If $y(t)$ is a solution, let $u(t)=y(t)+20$. Then$$u'(t)=y'(t)=k\bigl(y(t)+20\bigr)=ku(t).$$Can you solve the equation $u'(t)=ku(t)$? I hope so. Solve it and then take $y(t)=u(t)-20$.

Answer 3

Without using substitution observe that $$ \frac{1}{y(t)+20}\frac{dy}{dt}=k\implies\int\frac{1}{y(t)+20}\frac{dy}{dt}\,dt=\int k \,dt $$ Integrate both sides w.r.t to $t$ to get that $\log(y(t)+20)=kt+c$ for some $c$.