Differential Equation

A differential equation is a mathematical tool or equation that relates or equates a polynomial types function with its derivatives. Differential equations play a crucial role in various fields such as mathematics, physics, engineering, economics, biology, etc. Differential equations describe the relationship between changing quantities and the rates at which these quantities change.

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In this article, we will cover the concept of types of differential equations, order, and degree of differential equations. This concept falls under the broader category of differential equations, which is a crucial Chapter in class 12 Mathematics. 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 ten questions have been asked on this concept, including two in 2014, three in 2017, three in 2019, one in 2021, and one in 2023.

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.
The following relations are some of the examples of differential equations:
(i) $\frac{d y}{d x}=\sin 2 x+\cos x$
(ii) $\mathrm{k} \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right]^{3 / 2}$

Ordinary Differential equation
A differential equation that involves derivatives with respect to a single independent variable is known as an ordinary differential equation.
For example
(i) $\frac{\mathrm{dy}}{\mathrm{dx}}+x y=\sin \mathrm{x}$
(ii) $\frac{\mathrm{d}^3 \mathrm{y}}{\mathrm{dx}^3}+2 \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=\mathrm{e}^{\mathrm{x}}$

Order of a Differential Equation

The order of a differential equation is the highest order of any derivative of the dependent variable with respect to the independent variable involved in the given differential equation.

Some examples of different orders of the differential equation are given:

(i) $\frac{\mathrm{dy}}{\mathrm{dx}}=\sin 2 \mathrm{x}+\cos \mathrm{x}$ is of order 1
(ii) $\frac{d^2 y}{d x^2}+y=0$ is of order 2
(iii) $\frac{d^3 y}{d x^3}+2\left(\frac{d^2 y}{d x^2}\right)^5+y=e^x$ is of order 3

Types of the order of the differential equation

(i) First-order differential equation: The differential equation with a degree of 1. All the linear equations in the form of derivatives are in the first order. It has only the first derivative such as dy/dx, where x and y are the two variables.

(ii) Second-order differential equation: The differential equation with a degree of 2. All the second-order derivatives are in the form of the second order.

Degree of differential equation

The degree of differential equation is the degree (power) of the highest order derivative present in the equation after the differential equation has been made free from the radicals and fractions as far as the derivatives are concerned.

Consider the following example:

$\begin{aligned} & \frac{d^3 y}{d x^3}+2\left(\frac{d^2 y}{d x^2}\right)^2+\frac{d y}{d x}=x y^2 \\ & \frac{d^3 y}{d x^3}+\sin \frac{d y}{d x}=0\end{aligned}$

Observe that equation (i) is a polynomial equation in derivatives: y''', y'', and y' in this case: these terms occur only with some whole number power. So for this equation, the degree of differential equation can be defined.

But equation (ii) is not a polynomial in y’, as sin(y') term is present, so the degree of this equation is not defined.

Now, consider another example,
$
\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{2}}=\left(y+\left(\frac{d y}{d x}\right)^4\right)^{\frac{1}{3}}
$

Here, derivatives have no integral power. We can rewrite the equation as
$
\left(\frac{d^2 y}{d x^2}\right)^3=\left(y+\left(\frac{d y}{d x}\right)^4\right)^2
$

After the expansion of these brackets, this will become a polynomial equation of derivatives. So, its degree is defined.

So, the degree of (i) is 1, as the power of the highest order derivative ( $y^{\prime \prime \prime}$ ) is 1.

And the degree of
$
\left(\frac{d^2 y}{d x^2}\right)^3=\left(y+\left(\frac{d y}{d x}\right)^4\right)^2
$
is 3 (the power of the highest order derivative, which is $\left.y^{\prime \prime}\right)$.

Applications of differential equation

Differential equations are used in many fields, including biology, economics, physics, chemistry, and engineering.

1) Differential equations are 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.

Note :
- From now on, we will use the term 'differential equation' or 'DE' for 'ordinary differential equation'.
- We will use the following notations for derivatives: $\frac{d y}{d x}=y^{\prime}, \frac{d^2 y}{d x^2}=y^{\prime \prime}, \frac{d^3 y}{d x^3}=y^{\prime \prime \prime}$
- For derivatives of higher order, we use the notation $\mathrm{y}_{\mathrm{n}}$ for $\mathrm{n}^{\text {th }}$ order derivative $\frac{d^n y}{d x^n}$.
- Order and degree (if defined) of a differential equation are always positive integers.

Recommended Video Based on Differential Equations

Solved Examples Based On Differential Equations

Example 1: The order and the degree of the differential equation of all ellipses with the centre at the origin, the major axis along the $\mathrm{x}$-axis and eccentricity $\frac{\sqrt{3}}{2}$ are, respectively : [JEE Main 2017]

Solution:
Let the equation of ellipse is
$
\begin{aligned}
& \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { but } 1-\frac{b^2}{a^2}=\frac{3}{4} \\
& \therefore \frac{b^2}{a^2}=\frac{1}{4} \\
& \therefore a^2=4 b^2 \\
& \therefore \frac{x^2}{4 b^2}+\frac{y^2}{b^2}=1 \\
& \therefore x^2+4 y^2=4 b^2 \\
& \therefore 2 x+8 y \cdot \frac{d y}{d x}=0 \\
& \therefore x+4 y \cdot \frac{d y}{d x}=0 \\
& \text { degree }=1
\end{aligned}
$

Hence, the order is 1 and the degree is also 1.

Example 2: Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be a solution curve of the differential equation. $\left(1-x^2 y^2\right) d x=y d x+x d y$ If the line $\mathrm{x}=1$ intersects the curve $\mathrm{y}=\mathrm{y}(\mathrm{x})$ at $\mathrm{y}=2$ and the line $\mathrm{x}=2$ intersects the curve $\mathrm{y}=\mathrm{y}(\mathrm{x})$ at $\mathrm{y}=\alpha$, then a value of $\alpha$ is: [JEE MAIN 2023]

Solution:

$
\begin{aligned}
& \left(1-x^2 y^2\right) d x=y d x+x d y, y(1)=2 \\
& y(2)=\infty \\
& d x=\frac{d(x y)}{1-(x y)^2} \\
& \int d x=\int \frac{d(x y)}{1-(x y)^2} \\
& x=\frac{1}{2} \ln \left|\frac{1+x y}{1-x y}\right|+C
\end{aligned}
$

Put $x=1$ and $y=2$ :
$
\begin{gathered}
1=\frac{1}{2} \ln \left|\frac{1+2}{1-2}\right|+C \\
C=1-\frac{1}{2} \ln 3 \\
2=\frac{1}{2} \ln \left|\frac{1+2 \alpha}{1-2 \alpha}\right|+1-\frac{1}{2} \ln 3 \\
1+\frac{1}{2} \ln 3=\frac{1}{2}\left|\frac{1+2 \alpha}{1-2 \alpha}\right| \\
2+\ln 3=\left|\frac{1+2 \alpha}{1-2 \alpha}\right|
\end{gathered}
$
$
\begin{aligned}
& 2=\frac{1}{2} \ln \left|\frac{1+2 \alpha}{1-2 \alpha}\right|+1-\frac{1}{2} \ln 3 \\
& 1+\frac{1}{2} \ln 3=\frac{1}{2}\left|\frac{1+2 \alpha}{1-2 \alpha}\right| \\
& 2+\ln 3=\left|\frac{1+2 \alpha}{1-2 \alpha}\right|
\end{aligned}
$

Nowputx $=2:\left|\frac{1+2 \alpha}{1-2 \alpha}\right|=3 \mathrm{e}^2$
$
\begin{aligned}
& \frac{1+2 \alpha}{1-2 \alpha}=3 \mathrm{e}^2,-3 \mathrm{e}^2 \\
& \frac{1+2 \alpha}{1-2 \alpha}=3 \mathrm{e}^2 \Rightarrow \alpha=\frac{3 \mathrm{e}^2-1}{2\left(3 \mathrm{e}^2+1\right)}
\end{aligned}
$

And $\frac{1+2 \alpha}{1-2 \alpha}=-3 \mathrm{e}^2 \Rightarrow \alpha=\frac{3 \mathrm{e}^2+1}{2\left(3 \mathrm{e}^2-1\right)}$
Hence, the answer is $\frac{1+3 e^2}{2\left(3 e^2-1\right)}$

Example 3: Let $\mathrm{f}: R \rightarrow R$ be a function such that $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)_2$ $x \in R$. Then $\mathrm{f}(2)$ equals : [JEE MAIN 2019]

Solution:
Given $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)$
$\Rightarrow f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2)$
$\Rightarrow f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$.
$\Rightarrow f^{\prime \prime \prime}(x)=6$
put $x=1$ in equation (1)
$
\Rightarrow f^{\prime}(1)=3+2 f^{\prime}(1)+f^{\prime \prime}(2)
$
put $\mathrm{x}=2$ in equation (2)
$
\Rightarrow f^{\prime \prime}(2)=12+2 f^{\prime}(1) \text {. }
$
from eqn (4) and (5)
$-3-f^{\prime}(1)=12+2 f^{\prime}(1)$
$\Rightarrow 3 f^{\prime}(1)=-15$
$\Rightarrow f^{\prime}(1)=-5 \quad$ and $\quad f^{\prime \prime}(2)=2$
Now,
put $x=3$ in eqn (3)
$f^{\prime \prime \prime}(3)=6$
$\therefore f(x)=x^3-5 x^2+2 x+6$
$\therefore f(2)=8-20+4+6=-2$
Hence, the answer is -2.

Example 4: 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:

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 5: 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 $\frac{k}{\log _e\left(\frac{6}{5}\right)}$ 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} \ell \mathrm{n}\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 \ell \ln 2}{\ln \left(\frac{6}{5}\right)} \\
& \Rightarrow \mathrm{k}=2 \ln 2 \\
& \left(\frac{\mathrm{k}}{\log _e 2}\right)^2=\left(\frac{2 \ln 2}{\log 2}\right)^2=4 \\
& \ln 2 a \text { and } \log _e 2 \text { is same thing }
\end{aligned}
$

Hence, the answer is 4.

Summary

Differential equations are powerful tools for modelling and understanding the behaviour of real-world dynamic systems. By translating real-life events into differential equations, we can predict future behaviour and gain insight into the underlying mechanisms. The example of the increasing population shows that a simple differential equation can explain a complex world phenomenon.