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


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.