Question

Prove Theorem 6.22 by solving the initial-value problem $\frac{dT}{dt}=k(A-T)$with T(0) = T0, where k and A are constants.

Short answer

$T\left(t\right)=A-(A-T_0)e^{-kt}$

Step-by-step solution

Step 1. Given information

$\frac{dT}{dt}=k(A-T)$

Step 2. Integrating both the sides

$\int \frac{dT}{(A-T)}=\int kdt$

$-\ln (A-T)=Kt+C1$

Therefore,

$\ln (A-T)=-Kt-C1$

Since, $e^{-C1}=C$

Therefore,

localid="1649156311528" $A-T=C{e}^{-kt}$

localid="1649156328670" $T=A-C{e}^{-kt}$

Step 3. Using initial conditions

Using initial conditions, we get,

$C=A-T_0$

Substitute the value of C,

$T\left(t\right)=A-(A-T_0)e^{-kt}$