# 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.