New Notes Blog: `dT + k(T-70)dt = 0 , T(0) = 140` Find the particular solution that satisfies the initial condition

Wednesday, 12 August 2015

$\displaystyle{d}{T}+{k}{\left({T}-{70}\right)}{\left.{d}{t}\right.}={0},{T}{\left({0}\right)}={140}$ Find the particular solution that satisfies the initial condition

Given the differential equation : $\displaystyle{d}{T}+{K}{\left({T}-{70}\right)}{\left.{d}{t}\right.}={0},{T}{\left({0}\right)}={140}$

We have to find a particular solution that satisfies the initial condition.

We can write,

$\displaystyle{d}{T}=-{K}{\left({T}-{70}\right)}{\left.{d}{t}\right.}$

$\frac{dT}{T-70}=-Kdt$

$\int \frac{dT}{T-70}=\int -Kdt$

$\displaystyle{\ln{{\left({T}-{70}\right)}}}=-{K}{t}+{C}$ where C is a constant.

Now,

$T-70=e^{-Kt+C}$

$=e^{-Kt}.e^{C}$

$=C'e^{-Kt}$ where $\displaystyle{{e}}^{{C}}={C}'$ is again a constant.

Hence we have,

$T=70+C'e^{-Kt}$

Applying the initial condition we get,

$\displaystyle{140}={70}+{C}'$ implies $\displaystyle{C}'={70}$

Therefore we have the solution:

$T=70(1+e^{-Kt})$

Given the differential equation : $\displaystyle{d}{T}+{K}{\left({T}-{70}\right)}{\left.{d}{t}\right.}={0},{T}{\left({0}\right)}={140}$

We have to find a particular solution that satisfies the initial condition.

We can write,

$\displaystyle{d}{T}=-{K}{\left({T}-{70}\right)}{\left.{d}{t}\right.}$

$\frac{dT}{T-70}=-Kdt$

$\int \frac{dT}{T-70}=\int -Kdt$

$\displaystyle{\ln{{\left({T}-{70}\right)}}}=-{K}{t}+{C}$ where C is a constant.

Now,

$T-70=e^{-Kt+C}$

$=e^{-Kt}.e^{C}$

$=C'e^{-Kt}$ where $\displaystyle{{e}}^{{C}}={C}'$ is again a constant.

Hence we have,

$T=70+C'e^{-Kt}$

Applying the initial condition we get,

$\displaystyle{140}={70}+{C}'$ implies $\displaystyle{C}'={70}$

Therefore we have the solution:

$T=70(1+e^{-Kt})$

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