Question

Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k > 0 is a constant. Suppose that A = A0 cos(ωt) for some constants A0 and ω. That is, the ambient temperature oscillates (for example night and day temperatures). a) Find the general solution. b) In the long term, will the initial conditions make much of a difference? Why or why not?

Answer 1

Answer:

a)$X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

b)Does not affect the long term.

Explanation:

Given that

$\dfrac{dx}{dt}=-k(x-A)$

A = A0 cos(ωt)

$\dfrac{dx}{dt}=-k(x-A_o cos(\omega t))$

$\dfrac{dx}{dt}+kx=kA_o cos(\omega t)$

This is linear equation so integration factor ,I

$I=e^{\int kdt}$

$I=e^{kt}$

Now by using linear equation property

$e^{kt} X=\int e^{kt} kA_o cos(\omega t) dt +C$

$e^{kt} X= kA_o \dfrac{e^{kt}}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+C$

$X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

b)

at t= 0

$X(0)=\dfrac{k^2A_o}{\omega^2+k^2}+C$

$X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+e^{-kt}\times \left ( X(0)-\dfrac{k^2A_o}{\omega^2+k^2} \right )$

So the initial condition does not affect the long term.