# The Application of Differential Equations in Business

In the following example we shall discuss the application of simple differential equation in business.

If $P$ is the principal amount and $r$ is the rate of interest, then the differential equation relating to the time $t$, $P$ and $r$ is
$\frac{{dP}}{{dt}} = rP$

Example:

How long will it take for Rs.10000 to double if it is compounded continuously at 4 percent per annum?

Here $P = 10000$, $r = \frac{4}{{100}} = 0.04$

The differential equation governing the given problem is
$\frac{{dP}}{{dt}} = rP\,\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Since the rate $r$ is fixed, putting the values of $r$ in equation (i), we have
$\frac{{dP}}{{dt}} = 0.04P\,\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

Separating the variables in equation (ii), we have
$\frac{{dP}}{P} = 0.04dt\,\,\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right)$

Since the time during which $P$ is doubled is to be calculated, integrating the left hand side of equation (iii) from 10000 to 20000 and integrating its right hand side from 0 to $t$, we have
$\begin{gathered} \int\limits_{10000}^{20000} {\frac{{dP}}{P} = 0.04\int\limits_0^t {dt} } \\ \Rightarrow \left| {\ln P} \right|_{10000}^{20000} = 0.04\left| t \right|_0^t \\ \Rightarrow \ln 20000 – \ln 10000 = 0.04\left( {t – 0} \right) \\ \Rightarrow \ln \frac{{20000}}{{10000}} = 0.04t \\ \Rightarrow t = \frac{1}{{0.04}}\ln 2 \\ \end{gathered}$
$\Rightarrow t = 17.328$ years, approximately.