Application of Differential Equations in Biology

Differential equations are frequently used in solving the mathematical and physical problems. In the following example we shall discuss application of simple differential equation in biological sciences.

Example:
In a culture bacteria increases at the rate proportional to the number of bacteria present. If bacteria are 400 initially and are doubled in 3 hours, find the number bacteria present 7 hours later.

Let x be the number of bacteria, the rate is \frac{{dx}}{{dt}}. Since the number of bacteria is proportional to the rate, so

\frac{{dx}}{{dt}}  \propto x


If k\,\left( {k > 0}  \right) is the proportionality constant, then

\frac{{dx}}{{dt}}  = kx


Separating the variables, we have

\frac{{dx}}{x}  = kdt\,\,\,\,\,{\text{ -  -  - }}\left( {\text{i}} \right)


Since bacteria are 400 initially and doubled in 3 hours, so we integrate left side of equation (i) from 400 to 800 and integrate its right side from 0 to 3 to find the value of k as follows:

\begin{gathered} \int\limits_{400}^{800} {\frac{{dx}}{x} =  k\int\limits_0^3 {dt} } \\ \Rightarrow \left| {\ln x}  \right|_{400}^{800} = k\left| t \right|_0^3 \\ \Rightarrow \ln 800 - \ln 400 = k\left( {3 -  0} \right) \\ \Rightarrow 3k = \ln \frac{{800}}{{400}} =  \ln 2 \\ \Rightarrow k = \frac{1}{3}\ln 2 \\ \end{gathered}


Putting this value of k in (i), we have

\frac{{dx}}{x}  = \left( {\frac{1}{3}\ln 2} \right)dt\,\,\,\,\,{\text{ -  -  - }}\left(  {{\text{ii}}} \right)


Next to find the number of bacteria present 7 hours later, we integrate the left side of (ii) from 400 to x and its right side from 0 to 7 as follows:

\begin{gathered} \int_{400}^x {\frac{{dx}}{x} = \frac{1}{3}\ln  2\int_0^7 {dt} } \\ \Rightarrow \left| {\ln x} \right|_{400}^x =  \frac{1}{3}\ln 2\left| t \right|_0^7 \\ \Rightarrow \ln x - \ln 400 = \frac{1}{3}\ln  2\left( {7 - 0} \right) \\ \Rightarrow \ln x = \ln 400 + \frac{7}{3}\ln  2 \\ \Rightarrow \ln x = \ln 400 + \ln  {2^{\frac{7}{3}}} \\ \Rightarrow \ln x = \ln \left( {400}  \right){2^{\frac{7}{3}}} \\ \Rightarrow x = \left( {400} \right)\left(  {5.04} \right) = 2016 \\ \end{gathered}


Thus, 2016 is the required number of bacteria after 7 hours.

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