Answer
$T=C_1e^{-kt}+\alpha(t-k^{-1})+C_2$
Work Step by Step
We are given:
$T(t)=\alpha (t-k^{-1})+c_1+c_2e^{-kt}$
Newton’s law of cooling:
$\frac{dT}{dt}=k(T_0-T)$
Multiply both sides by $\frac{d}{dt}$:
$\frac{d^2T}{dt^2}=k\frac{d}{dt}T_0-k\frac{dT}{dt}$
where $k$ is proportionality constant
and $T_0$ is the environment's temperature
since $\frac{d}{dt}T_0 = \alpha$
$\rightarrow \frac{d^2}{t^2}T+k\frac{dT}{dt}$
Set $\frac{dT}{dt}=v(t)$
then the equation becomes:
$\frac{dv}{dt}+ku=k\alpha$
Intergrating factor:
$I=e^{\int kdt}=e^{kt}$
The general solution is:
$v=I^{-1}(C_1+\int Ik \alpha)$
$v=C_1e^{-kt}+ \alpha$
Intergrate both sides:
$T=-\frac{C_1}{k}e^{-kt}+\alpha t+C_2$
Since $C_1$ and $C_2$ are constants, assume $-\frac{C_1}{k}=C_1$ and $C_2=C_2-\frac{\alpha}{k}$
We get: $T=C_1e^{-kt}+\alpha(t-k^{-1})+C_2$
