Question

Solve the given initial value problem. Give the largest interval lover which the general solution is defined.

$\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathbf{k}\mathbf{(}\mathbf{T}\mathbf{-}{\mathbf{T}}_{\mathbf{m}}\mathbf{)}\mathbf{;}\mathbf{}\mathbf{T}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{T}}_{\mathbf{0}}$, $\mathbf{k}\mathbf{,}{\mathbf{T}}_{\mathbf{m}}$, and T₀ constants

Short answer

The solution for the given initial value is$\mathrm{T}={\mathrm{T}}_{\mathrm{m}}+\left({\mathrm{T}}_0-{\mathrm{T}}_{\mathrm{m}}\right){\mathrm{e}}^{\mathrm{kt}}$.

Step-by-step solution

The given equation in the standard form and determine the integrated factor

Given differential equation is written as

$\frac{dT}{dt}-KT=-T_{m}K$; which is in the standard form.

Here $P\left(t\right)=-Kandf\left(t\right)=-T_{m}K.P,f$ are continuous on .

So, an integrating factor is $e^{\int -kdt}=e^{-kt}$.

Determine the general solution for the given differential equation 

Multiplying the standard form with the integrating factor, we get $e^{-kt}\frac{dT}{dt}-K{e}^{-kt}T=T_m(-k)e^{-kt};$

which is same as $\frac{d}{dt}\left[e^{-kt}T\right]=T_m(-k)e^{-kt}$

On integration, we obtain

$$\begin{gathered} e^{-kt}T=T_m{e}^{-kt}+c \\ T=T_m+c{e}^{kt}; \\ T\left(0\right)=T_0\Rightarrow T_0=T_m+c{e}^0 \\ \Rightarrow c=T_0-T_m for-\infty <t<\infty . \end{gathered}$$

Thus, $T=T_m+\left(T_0-T_m\right)e^{kt}$; is the solution of given initial value problem.