Application of Differential Equation

Differential equations have various applications across various fields due to their ability to model systems and processes. Here, we will discuss the growth and decay problems because this is very helpful in analyzing the growth of bacteria and in the decay of radioactive materials. Differential equations are used in various disciplines, such as biology, economics, physics, chemistry, and engineering.

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This Story also Contains

1. What is a Differential Equation?
2. Applications of differential equation
3. Solved Examples Based On Applications of Differential Equations
4. Summary

In this article, we will cover the applications of differential equations. This concept falls under the broader category of differential equations. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of six questions have been asked on this concept, including one in 2013, one in 2014, one in 2020, one in 2021, and two in 2022.

$$
\begin{aligned}
& x^2+x+1=0 \\
& \sin x+\cos x=0 \\
& x+y=9 \\
& x \frac{d y}{d x}+2 y=0
\end{aligned}
$$
... (1)

What is a Differential Equation?

A differential equation is an equation involving one or more terms and the derivatives of one dependent variable with respect to the other independent variable.

Differential equation: dy/dx = f(x)

Where “x” is an independent variable and “y” is a dependent variable

Example of differential equation: $x \frac{d y}{d x}+2 y=0$
The above-written equation involves variables as well as the derivative of the dependent variable $\mathrm{y}$ with respect to the independent variable $\mathrm{x}$. Therefore, it is a differential equation.

Applications of differential equation

Differential equations are used in a variety of disciplines, such as biology, economics, physics, chemistry, and engineering.

Growth and Decay Problem:

Let the amount of substance (or population) that is either growing or decaying is denoted by N(t). if we assume the time rate of change of this amount of substance, dN / dt, is proportional to the amount of substance present, then

$\frac{d N}{d t}=k N \quad$ or $\quad \frac{d N}{d t}-k N=0$

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

Newton's Law of Cooling:

According to Newton, the cooling of a hot body is proportional to the temperature difference between its temperature T and the temperature T₀ of its surroundings. This statement can be written as:

dT/dt ∝ (T – T₀)…………(1)

This is the form of a linear differential equation.

Introducing a proportionality constant k, the above equation can be written as:

dT/dt = k(T – T₀) …………(2)

Here, T is the temperature of the body and t is the time,

T₀ is the temperature of the surroundings,

dT/dt is the rate of cooling of the body

Some other applications are:

1) They are also used to describe the change in return on investment over time.

2) They are used in the field of medical science for modelling cancer growth or the spread of disease in the body.

3) The movement of electricity can also be described with the help of it.

4) They help economists in finding optimum investment strategies.

5) The motion of waves or a pendulum can also be described using these equations.

Recommended Video Based on Applications of Differential Equations

Solved Examples Based On Applications of Differential Equations

Example 1: In which of the following, a differential equation will not be formed?

(1) The temperature of the body increases at a rate proportional to its instantaneous temperature.

(2) The population of a country increases at a constant rate.

(3) A point moves in a plane such that its distance from its origin is constant.

(4) None of these

Solution:

As we have learned

Application of Homogeneous Differential Equations-

For option (1) $\rightarrow \frac{d T}{d t}=K T$
For option (2) $\rightarrow \frac{d p}{d t}=K$
For option (3) $\rightarrow x^2+y^2=K$

Here, (1) and (2) are differential equations but (3) is not.

Hence, the answer is the option (3).

Example 2: The rate at which a substance cools in moving air is proportional to the difference between the temperature of substance0 and that of the surroundings. If the temperature of the surrounding is 290 K and the substance cools from 370 K to 330 K in 10 min, then after what time from the initial, temperature will be 295 K?

Solution:

As we have learnt,

Temperature Problems -

$\frac{d T}{d t}=-k\left(T-T_m\right)$

- wherein

K is the proportionality constant

T = Temperature of body

$T_m=$ Temperature of Surrounding
Let at time ' $t$ ', temperature is $T$.
So,
$
-\frac{\mathrm{d} T}{\mathrm{~d} t}=k(T-290) \Rightarrow \frac{d T}{T-290}=-k d t
$

On integrating, we get

$\begin{aligned} & \ln (T-290)=k t+c \\ & \text { At } t=0, T=370 \Rightarrow c=\ln 80 \\ & \Rightarrow \ln (T-290)=-k t+\ln (80) \\ & \text { Also, at } t=10, T=330 \Rightarrow \ln 40=-10 k+\ln 80 \\ & \Rightarrow 10 k=\ln 2 \Rightarrow K=\frac{1}{10} \ln 2 \\ & \Rightarrow \ln (T-290)=\left(-\frac{1}{10} \ln 2\right) t+\ln 80 \\ & \text { Now when } T=295 K \text { then } \ln 5=-\frac{t \ln 2}{10}+\ln 80 \\ & \Rightarrow \frac{t \ln 2}{10}=\ln 16 \Rightarrow t=40\end{aligned}$

Hence, the required answer is 40 mins.

Example 3: The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at the initial time $t=0$. The number of bacteria is increased by $20 \%$ in 2 hours. If the population of bacteria is 2000 after $\frac{k}{l o g}\left(\frac{6}{5}\right)$ hours, then $\left(\frac{k}{\log _e 2}\right)^2$ is equal to:
Solution:

Initial bacteria count = 1000

20% bacteria increased in 2 hours = 1200

$
\begin{aligned}
& \frac{\mathrm{dB}}{\mathrm{dt}}=\lambda \mathrm{B} \\
& \Rightarrow \int_{1000}^{1200} \frac{\mathrm{dB}}{\mathrm{B}}=\lambda \int_0^2 \mathrm{dt} \\
& \Rightarrow \lambda=\frac{1}{2} \ln \left(\frac{6}{5}\right) \\
& \int_{1000}^{2000} \frac{\mathrm{dB}}{\mathrm{B}}=\frac{1}{2} \ln \left(\frac{6}{5}\right) \int_0^{\mathrm{T}} \mathrm{dt} \\
& \Rightarrow \mathrm{T}=\frac{2 \ln 2}{\ln \left(\frac{6}{5}\right)} \\
& \Rightarrow \mathrm{k}=2 \ln 2 \\
& \left(\frac{\mathrm{k}}{\log _{\mathrm{e}} 2}\right)^2=\left(\frac{2 \ln 2}{\log _{\mathrm{e}} 2}\right)^2=4
\end{aligned}
$
$\ln 2 a$ and $\log _e 2$ is same thing

Hence, the answer is 4.

$\text { Example 4: At present, a firm is manufacturing } 2000 \text { items. It is estimated that the rate of change of production } \mathrm{P} \text { w.r.t. additional number of workers } x \text { is given by } \frac{d P}{d x}=100-12 \sqrt{x} \text {. If the firm employs } 25 \text { more workers, then the new level of production of items is: } $

Solution:

$\begin{aligned} & \frac{d P}{d x}=100-12 \sqrt{x} \\ & \Rightarrow \int d P=\int 100 d x-\int 12 \sqrt{x} d x \\ & \Rightarrow P=100 x-\frac{12 x^{3 / 2}}{3 / 2}+C \\ & \Rightarrow P=100 x-8 x^{3 / 2}+C \\ & \text { For } \mathrm{x}=0, \mathrm{P}=2000 \\ & 2000=C \ldots(1)\end{aligned}$
$\begin{aligned} & \text { and for } \mathrm{x}=25 \\ & P^{\prime}=100(25)-8(25)^{3 / 2}+2000=3500\end{aligned}$

Hence, the required answer is 3500.

Example 5: The population of a certain country is known to increase at a rate proportional to the number of people presently living in the country. If after two years, the population has doubled and after four years population is 20,000, then initially population was

Solution:

Let at time $t$, and $N$ be the number of population at that time, So
$
\begin{aligned}
\frac{\mathrm{d} N}{\mathrm{~d} x} & =k N \\
\Rightarrow \frac{\mathrm{d} N}{N} & =k d t
\end{aligned}
$

So, on integrating we get,
$
\ln N=k t+c
$

Let the initial population is $N_0$ at $t=0$
$
\Rightarrow c=\ln N_0 \Rightarrow \ln N-\ln N_0=k t \Rightarrow \frac{N}{N_0}=e^{k t} \Rightarrow N=N_0 e^{k t}
$

It is given, after two years, the population is doubled, So
$
\begin{aligned}
& 2 N_0=N_0 e^{k \cdot 2} \Rightarrow e^{2 k}=2 \Rightarrow 2 k=\ln 2 \Rightarrow k=\frac{1}{2} \ln 2 \\
& \therefore N=N_0 \cdot e^{\frac{1 \ln 2}{2}}
\end{aligned}
$

Given, after four years population is 20,000 , So
$
20000=N_0 \cdot e^{2 \ln 2} \Rightarrow 4 N_0=20000 \Rightarrow N_0=5000
$

Hence, the required answer is 5000.

Summary

We concluded that the differential equation is very important in real life. It helps to simplify complex algorithms and provides a detailed view of how the process occurs at every step. Differential equations play an important role in scientific research and in economics.