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1 month ago

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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 &g

t; 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?

1 answer:

Maru [3.3K]1 month ago

5 0

Answer:

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

b) D does not influence the long-term results.

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 a linear equation hence the integration factor, I

$I=e^{\int kdt}$

$I=e^{kt}$ Now using the characteristics of linear equations

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

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

b) At t= 0

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

Thus, the initial condition

does not affect the long-term outcome.

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

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