# Applications of First Order Differential Equation

## Applications of First Order Differential Equation

Growth and Decay: Let N(t) denote the amount of substance (or population) that is either growing or decaying. If we assume that $$\frac{dN}{dt}$$, the time rate of change of this amount of substance, is proportional to the amount of substance present, then $$\frac{dN}{dt}=kN$$ or $$\frac{dN}{dt}-kN=0…….(a)$$.

Where, k is the constant of proportionality. We are assuming that N(t)  is a differentiable, hence continuous function of time.

Example: The population of a village increases at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20000 in year 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Solution: Let P be the population at time t, then $$\frac{dy}{dt}\propto y$$

$$\Rightarrow \frac{dy}{dt}=ky$$, where k is constant

$$\Rightarrow \frac{dy}{y}=k\,dt$$

On integrating both sides, we get

$$\Rightarrow \int{{}}\frac{dy}{y}=\int{{}}k\,dt$$

Log y = kt + C …..(1)

In the year 1999, t = 0, y = 20000

∴ From Eq. (i)

log2000 = k(0) + C

log2000 = C … (ii)

In the year 2004, t = 5, y = 25000, so from Eq. (i),

log2500 = k(5) + C

[Using Eq. (ii)]

log2500 = k (5) + log2000

log2500 – log2000 = k (5)

$$\Rightarrow \,5k=\log \left( \frac{25000}{20000} \right)=\log \left( \frac{5}{4} \right)$$

$$\Rightarrow \,k=\frac{1}{5}\log \frac{5}{4}$$

For year 2009, t = 10 yr

Now, substituting the values of t, k and C in Eq. (i), we get

$$\log y=10\times \frac{1}{5}\log \left( \frac{5}{4} \right)+\log \left( 20000 \right)$$

$$\Rightarrow \,\,\log y=\log \left[ 20000\times {{\left( \frac{5}{4} \right)}^{2}} \right]\,\,\,\left( \because \,\log m+n=\log mn \right)$$

$$y=20000\times \frac{5}{4}\times \frac{5}{4}$$

y = 31250

Hence, the population of the village in 2009 will be 31250.