Respuesta :
Answer:
[tex]T(t)=5e^{-t}+27e^{-0.2t}[/tex]
Step-by-step explanation:
QUESTION
The rate of change of the temperature T(t) of a body is still governed by
[tex]\frac{dT}{dt}=-k(T-A), T(0)=T_0[/tex] when the ambient temperature A(t) varies with time. Suppose the body is known to have k = 0.2 and initially is at 32°C; suppose also that [tex]A(t) = 20e^{-t}[/tex]. Find the temperature T(t).
SOLUTION
[tex]\frac{dT}{dt}=-k(T-A), T(0)=T_0, A(t) = 20e^{-t}, k=0.2[/tex]
[tex]\frac{dT}{dt}=-0.2T+20(0.2)e^{-t}\\\frac{dT}{dt}+0.2T=4e^{-t}\\\text{Integrating factor}=e^{0.2t}\\\frac{dTe^{0.2t}}{dt}=4e^{-t}e^{0.2t}\\dTe^{0.2t}=4e^{-t}e^{0.2t}dt\\\int d[Te^{0.2t}]=4\int e^{-t(1-0.2)}dt\\Te^{0.2t}=4\int e^{-0.8t}dt\\Te^{0.2t}=\frac{4}{-0.8} e^{-0.8t}+C, \text{C a constant of integration}\\Te^{0.2t}=-5 e^{-0.8t}+C\\T(t)=5 e^{-0.8t}e^{-0.2t}+Ce^{-0.2t}\\T(t)=5e^{-t}+Ce^{-0.2t}\\When t=0, T_0=32\\32=5+C\\C=27[/tex]
Therefore:
[tex]T(t)=5e^{-t}+27e^{-0.2t}[/tex]
