91Ó°ÊÓ

A transformation T from $C^{∞}\mathrm{to}{C}^{∞}$to of the form role="math" localid="1660801324664" $T\left(f\right)=f^{\left(n\right)}+a_{n-1}{f}^{(n-1)}+...+a_{1}f\text{'}+a_{0}f$is called an nth-order linear differential operator.Here $f^{\left(k\right)}$denote the k-th derivative of function f and the coefficients $a_k$are complex scalars.

If T is an nth-order linear differential operator and g is a smooth function, then the equation becomes

$T\left(f\right)=g$

(or)

$f^{\left(n\right)}+a_{n-1}{f}^{(n-1)}+...+a_{1}f\text{'}+a_{0}f=g$

Is called an nth-order linear differential equation (DE).The DE is called homogeneous if g = 0 and inhomogeneous otherwise.

Consider the temperature of a hot cup of coffee can be modelled by the DE

$\mathit{T}\mathbf{\text{'}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{k}\mathbf{\left(}\mathit{T}\mathbf{\right(}\mathit{t}\mathbf{)}\mathbf{-}\mathit{A}\mathbf{)}$

Here k and A be the constants in the differential equation.

The constant is called the continuous growth rate if it is positive and the continuous decay rate if it is negative.

If the constant A is positive then the temperature of the hot cup of coffee modelled by the DE is positive and A is negative then the temperature of the hot cup of coffee modelled by the DE is negative.

Since both the constants k andA be the significant terms in the temperature of the hot cup of coffee modelled by the DE

Consider the temperature of a hot cup of coffee can be modelled by the DE

$\mathit{T}\mathbf{\text{'}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{k}\mathbf{\left(}\mathit{T}\mathbf{\right(}\mathit{t}\mathbf{)}\mathbf{-}\mathit{A}\mathbf{)}$

Here k and A be the constants in the differential equation.

$$\begin{gathered} T\text{'}\left(t\right)=k\left(T\right(t)-A) \\ T\text{'}\left(t\right)=-k\left(T\right(t\left)\right)+kA \\ T\text{'}\left(t\right)=(A-t)k \\ âˆ\mu T\text{'}\left(t\right)=\frac{dT}{dt} \\ \frac{dT}{dt}=(A-T)k \\ \frac{dT}{dt}=-k(T-A) \end{gathered}$$

Integrating on both sides we get,

$$\begin{gathered} \frac{dT}{dt}=-k(T-A) \\ \frac{dT}{(T-A)}=-kdt \\ ∫\frac{dT}{(T-A)}=∫-kdt \\ \ln \left|T-A\right|=-kt+C \\ T-A=e^{-kt}{e}^c \\ T-A=C{e}^{-kt} \\ T\left(t\right)=A+C{e}^{-kt} \\ âˆ\mu A=T-C{e}^{-kt} \end{gathered}$$

Here initial temperature becomes $T_0$and the solution is $A=T_0-C{e}^{-kt}$.