Newton’s law of cooling states that d x d t = − k ( x − A ) where x is the temperature, t is time, A is the ambient temperature, and k >

Question

Newton’s law of cooling states that d x d t = − 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 = A 0 cos ( ω t ) for some constants A 0 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?

in progress 0
Thiên Ân 4 years 2021-07-19T11:58:37+00:00 1 Answers 3 views 0

Answers ( )

  1. thienthi
    0
    2021-07-19T12:00:25+00:00
    Reply

    Answer:

    (a). The general solution is

    x(t)=\dfrac{kA_{0}}{k^2+\omega^2}(k\cos(\omega t)+\omega\sin(\omega t))+ce^{-kt}

    (b). The initial condition does not affect the long term.

    Explanation:

    Given that,

    The equation is

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

    Where, x = temperature

    t = time

    A = ambient temperature  

    (a). We need to calculate the general solution

    Using given differential equation,

    \dfrac{dx}{dt}=-k(x-A)…(I)

    Where, A = A_{0}\cos(\omega t)

    Put the value of A in equation (I)

    \dfrac{dx}{dt}=-k(x-A_{0}\cos(\omega t))

    \dfrac{dx}{dt}=-kx+kA_{0}\cos(\omega t)

    \dfrac{dx}{dt}+kx=kA_{0}\cos(\omega t)…..(II)

    The integrating factor \mu(t) is given by

    \mu (t)=e^{\int{k dt}}

    \mu (t)=e^{kt}

    Now, multiplying the equation (II) by μ(t) and integrating,

    e^{kt}x(t)=\int{k A_{0}e^{kt}\cos(\omega t)}dt+c

    Where, c= constant

    e^{kt}x(t)=kA_{0}{\dfrac{e^{kt}}{k^2+\omega^2}(k\cos(\omega t)+\omega\sin(\omega t))}+c

    x(t)=\dfrac{kA_{0}}{k^2+\omega^2}(k\cos(\omega t)+\omega\sin(\omega t))+ce^{-kt}….(III)

    (b). We need to find the difference in the long term

    Using equation (III)

    x(t)=\dfrac{kA_{0}}{k^2+\omega^2}(k\cos(\omega t)+\omega\sin(\omega t))+ce^{-kt}

    At t = 0,

    x(0)=\dfrac{k^2A_{0}}{k^2+\omega^2}+c

    c=x(0)-\dfrac{k^2A_{0}}{k^2+\omega^2}

    Now, put the value of c in equation (III)

    x(t)=\dfrac{1}{k^2+\omega^2}{k\cos(\omega t)+\omega\sin(\omega t)}+x(0)-\dfrac{k^2A_{0}}{k^2+\omega^2}e^{-kt}

    Now, \lim_{t \to \infty} x(0) e^{-kt}=0

    For any x(0) ∈ R

    So, the initial condition does not affect the long term.

    Hence, (a). The general solution is

    x(t)=\dfrac{kA_{0}}{k^2+\omega^2}(k\cos(\omega t)+\omega\sin(\omega t))+ce^{-kt}

    (b). The initial condition does not affect the long term.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )