Respuesta :
Answer:
(a). The general solution is
[tex]x(t)=\dfrac{kA_{0}}{k^2+\omega^2}(k\cos(\omega t)+\omega\sin(\omega t))+ce^{-kt}[/tex]
(b). The initial condition does not affect the long term.
Explanation:
Given that,
The equation is
[tex]\dfrac{dx}{dt}=-k(x-A)[/tex]
Where, x = temperature
t = time
A = ambient temperature
(a). We need to calculate the general solution
Using given differential equation,
[tex]\dfrac{dx}{dt}=-k(x-A)[/tex]...(I)
Where, [tex]A = A_{0}\cos(\omega t)[/tex]
Put the value of A in equation (I)
[tex]\dfrac{dx}{dt}=-k(x-A_{0}\cos(\omega t))[/tex]
[tex]\dfrac{dx}{dt}=-kx+kA_{0}\cos(\omega t)[/tex]
[tex]\dfrac{dx}{dt}+kx=kA_{0}\cos(\omega t)[/tex].....(II)
The integrating factor [tex]\mu(t)[/tex] is given by
[tex]\mu (t)=e^{\int{k dt}}[/tex]
[tex]\mu (t)=e^{kt}[/tex]
Now, multiplying the equation (II) by μ(t) and integrating,
[tex]e^{kt}x(t)=\int{k A_{0}e^{kt}\cos(\omega t)}dt+c[/tex]
Where, c= constant
[tex]e^{kt}x(t)=kA_{0}{\dfrac{e^{kt}}{k^2+\omega^2}(k\cos(\omega t)+\omega\sin(\omega t))}+c[/tex]
[tex]x(t)=\dfrac{kA_{0}}{k^2+\omega^2}(k\cos(\omega t)+\omega\sin(\omega t))+ce^{-kt}[/tex]....(III)
(b). We need to find the difference in the long term
Using equation (III)
[tex]x(t)=\dfrac{kA_{0}}{k^2+\omega^2}(k\cos(\omega t)+\omega\sin(\omega t))+ce^{-kt}[/tex]
At t = 0,
[tex]x(0)=\dfrac{k^2A_{0}}{k^2+\omega^2}+c[/tex]
[tex]c=x(0)-\dfrac{k^2A_{0}}{k^2+\omega^2}[/tex]
Now, put the value of c in equation (III)
[tex]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}[/tex]
Now, [tex]\lim_{t \to \infty} x(0) e^{-kt}=0[/tex]
For any x(0) ∈ R
So, the initial condition does not affect the long term.
Hence, (a). The general solution is
[tex]x(t)=\dfrac{kA_{0}}{k^2+\omega^2}(k\cos(\omega t)+\omega\sin(\omega t))+ce^{-kt}[/tex]
(b). The initial condition does not affect the long term.
