JEE Important Chapter - Differential Equations

A differential equation is a mathematical equation that connects the derivatives of a function. The functions are usually used to represent physical quantities, the derivatives are used to represent their rates of change, and the equation is used to define a relationship between the two.

JEE Main Maths Chapter-wise Solutions 2022-23 

Important Topics of Differential Equation

• Differential Equations

• Differential Equation And Its Types

• Partial Differential Equations

• Partial Derivative

• Linear Differential Equations

• Homogeneous Differential Equations

Important Concepts of Differential Equation

Definition

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

$\dfrac{dy}{dx}=f(x)$, Where ‘x’ is an independent variable and ‘y’ is a dependent variable.

Differential Equations Formulas

$\dfrac{dy}{dx} = \sin x$

$\dfrac{d^2y}{dx^2} + k^2y = 0$

$\dfrac{d^2y}{dt^2} + \dfrac{d^2x}{dt^2} = x$

$\dfrac{d^3y}{dx^3} + x\left(\dfrac{dy}{dx}\right) - 4xy = 0$

$\dfrac{rdr}{d\theta} + \cos \theta = 5$

Order of Differential Equations

The highest order of the derivative appearing in a differential equation is the order of the equation.

Example: $2x + \dfrac{dy}{dx} = 1$ order of a differential equation is equal to 1.

$\dfrac{d^{2} y}{dx} + \cos x = \dfrac{1}{2}$ order of a differential equation is equal to 2.

$\dfrac{d^{3} y}{dx} + 2\dfrac{d^{2} y}{dx} + x^{3} + 6 = 0$ order of a differential equation is equal to 3.

First Order Differential Equation

A first-order differential equation is one where f(x, y) is a function of two variables defined in the XY-plane. All linear equations are in the first-order form of derivatives. It only has the first derivative, as in $\dfrac{dy}{dx}$. As you can see in the first example, it's a one-degree first-order differential equation.

$\dfrac{dy}{dx} = f(x, y) = y^\prime$

Second-Order Differential Equation

A second-order linear differential equation is an equation of the form that is linear in y and its derivatives. 

$\dfrac{d}{dx} \left(\dfrac{dy}{dx}\right) = \dfrac{d^2y}{dx^2} = f^{\prime \prime}(x) = y^{\prime \prime}$

Degree of Differential Equations

The highest power of the highest order derivative involved in the given differential equation is the degree of the differential equation.

Example: $\dfrac{dy}{dx})^{2}  +\dfrac{d^{2}y}{dx} + 1 = 0$ The highest degree derivative in this equation has a power of one. As a result, the order of the differential equation is equal to 1.

$\dfrac{dy}{dx} + 1 = 0$, degree of a differential equation is equal to 1

$(y^{\prime \prime \prime})3 + 2y^{\prime \prime} + 10y^\prime - 4 = 0$,  degree of a differential equation is equal to 3

Note: It is not always necessary for the degree and order of a differential equation to be equal, but they must both be positive.

Types of Differential Equations

• Homogeneous Differential Equations

• Nonhomogeneous Differential Equations

• Ordinary differential equations

• Partial differential equations

• Nonlinear differential equations

• Linear Differential Equations

Homogeneous Differential Equation

A homogeneous differential equation is a differential equation in which all of the terms have the same degree. 

$P(x,y)dx + Q(x,y)dy = 0$ is a general representation, where P(x,y) and Q(x,y) are homogeneous functions of the same degree.

Examples: $y + 2x\left(\dfrac{dy}{dx}\right) = 0$ is a homogeneous differential equation of degree equal to 1.

$x^2 + y^4\left(\dfrac{dy}{dx}\right) = 0$ is a homogeneous differential equation of degree equal to 4.

Non-Homogeneous Differential Equation

A homogeneous differential equation is a differential equation in which all of the terms have different degrees.

Example: $y\dfrac{dy}{dx} + y^2 + 2x = 0$ is a non-homogeneous differential equation.

Linear Differential Equations

A linear differential equation is a differential equation of the form $\dfrac{dy}{dx} + Py = Q$. 

Where P and Q are numeric constants or functions in x. It is made up of a y and a y derivative. The differential equation is called the first-order linear differential equation because it is a first-order differentiation.

General solution of differential equation is, $y \times (I.F) = \int{Q(I.F)}dx + c$ where, I.F (integrating factor) $= e^{\int p dx}$

Ordinary Differential Equation 

The function and its derivatives are involved in an ordinary differential equation. It only has one independent variable and one or more derivatives with respect to the variable. The ordinary differential equation can be homogeneous or nonhomogeneous. The general form of an $n^{th}$ order ordinary differential equation is as follows:

Example: $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} = y \sin x$

Partial Differential Equations

Differential equations that have derivatives with respect to more than one independent variable are called Partial differential equations.

Example: $\dfrac{\partial u}{\partial x}+\dfrac{\partial u}{\partial t}=0$

Differential Equations Solutions

• The solution of a differential equation is a function that satisfies its solution. 

• A general solution is one that contains as many arbitrary constants as the order of the differential equation. 

• A particular solution is a solution that is free of arbitrary constants. 

• The solution to the differential equation can be found using two methods.

1. Variable Separable method

2. Integrating factor

Variable Separable Differential Equation

The first-order differential equation is of the form $\dfrac{dy}{dx} = F(x,y)$

Where F(x, y) can be written as h(x)g(y), with h(x) being a function of x and g(x) being a function of y. The equation is then known as a variable separable differential equation. $\dfrac{dy}{dx} = h(x)g$ is the differential equation's form (x).

Integrating Factor

When the differential equation is of the form $\dfrac{dy}{dx} + p(x)y = q(x)$, the integrating factor method is used. $y^\prime + P(x)y = Q$ is a first-order differential equation (x). An equation containing partial or ordinary derivatives of an unknown function is referred to as a higher-order differential equation. It is possible to represent it in any order.

Solution of Differential Equations - How to Solve Differential Equations?

• There are an infinite number of solutions to the differential equation. Because the process of solving a differential equation is referred to as integrating a differential equation. 

• A differential equation solution is an expression for the dependent variable that satisfies the differential equation in terms of the independent variable.

• The general solution is one that contains as many arbitrary constants as possible. The resulting solution is called a Particular Solution when the arbitrary constants in the general solution of the differential equation are given particular values. 

• A first-order differential equation is obtained by eliminating one arbitrary constant.

• A second-order differential equation is obtained by eliminating two arbitrary constants, and so on.

• Differential equations are equations that have an independent variable, a dependent variable, and the derivative of the dependent variable with respect to the independent variable. 

• Ordinary differential equations have a single independent variable, while partial differential equations have more than one independent variable and their partial derivative.

Methods of Solving Differential Equations

Consider equation $\dfrac{dy}{dx} = x^2y + 2y$

Step 1: Divide the above differential equation w.r.t y.

$\dfrac{1}{y}\left(\dfrac{dy}{dx}\right) = \dfrac{1}{y}\times y(x^2 + 2)$

$\dfrac{1}{y}\left(\dfrac{dy}{dx}\right) =(x^2 + 2)$

Both y and x are variables, so we rewrite the equation as

$\dfrac{dy}{y} = (x^2 + 2)dx$

Step 2: Now L.H.S. should be integrated both in terms of y and x.

$\int{\dfrac{1}{y}dx} = \int{(x^2 + 2)}dx$

Step 3: On integrating, we get:

$\text{log }y = \dfrac{x^3}{3} + 2x + c$

Applications of Differential Equations

Differential equations are used in various fields, including applied mathematics, science, and engineering. Apart from technical applications, they are also used to solve a variety of real-world issues. In real life, ordinary differential equations are used to calculate the movement or flow of electricity, the motion of an object to and fro, such as a pendulum, and to explain thermodynamics concepts. In medical terms, they're also used to visualize the progression of diseases in graphical form. Differential equations can be used to describe mathematical models such as population growth or radioactive decay.

Important Notes

For derivatives, use the following notations. 

$\dfrac{dy}{dx} = y^\prime$

$\dfrac{d^2y}{dx^2} = y^{\prime\prime}$

$\dfrac{d^3y}{dx^3} = y^{\prime\prime\prime}$

Examples of Differential Equations

Example 1: Form the differential equation of the family of curves, y = mx.

Ans: Given: $y = mx$ 

Differentiating both sides of the equation we get,

$\dfrac{dy}{dx} = m$, because m is a constant 

Substitute the value of m in the given equation, we get

$y = \dfrac{dy}{dx}x$

$y - \dfrac{dy}{dx}x = 0$

Or

$y - y^\prime x = 0$

Example 2: Check whether the differential equation, $(x+y)\dfrac{dy}{dx} = x - 2y$ is homogeneous. 

Ans: Given $(x+y)\dfrac{dy}{dx} = x - 2y$ is a homogeneous equation,

The given equation can be written as $\dfrac{dy}{dx} = \dfrac{x-2y}{x+y}$

As it is an homogeneous equation it can be written as,

$F(x,y) = \dfrac{x-2y}{x+y}$

Let a be a constant, so

$F(ax,ay) = \dfrac{ax-2ay}{ax+ay}$

$F(ax,ay) = \dfrac{x-2y}{x+y}a^0$

$F(ax,ay) = a^0 F(x,y)$

Since this function is homogeneous, so is the differential equation.

Example 3: Find the general solution of the differential equation, $\dfrac{dy}{dx} = \dfrac{x-1}{2+y}$ where $(y \ne 2)$

Ans: Given, $\dfrac{dy}{dx} = \dfrac{x-1}{2+y}$

On simplifying we get,

$(2+y)dy = (x-1)dx$

$\int{(2+y)dy} = \int{(x-1)dx}$

$2y+\dfrac{y^2}{2} = \dfrac{x^2}{2}-x+c$

$4y+y^2-x^2+2x=c$

Solved Problems of Previous Year Question

1. If $x dy = y (dx + y dy)$, $y > 0$ and $y (1) = 1$, then $y (-3)$ is equal to ______.

Ans: Given, $x dy = y (dx + y dy)$

Given equation can be written as,

$x dy = y dx + y^2 dy$

Multiplying with -1 and rearranging we get,

$y dx - x dy = -{y^2} dy$

$\dfrac{y dx - x dy}{y^2} = -dy$

Integrating both the sides we get,

$\dfrac{x}{y} = -y + c$

Given x = 1, and y = 1

$1=-1+c$

$c=2$

Substituting we get,

$\dfrac{x}{y} = -y + 2$

To find y at x = -3, we get,

$-\dfrac{3}{y}=-y+2$

$\Rightarrow -3=-y^{2}+2 y$

$\Rightarrow y^{2}-2 y-3=0$

$\Rightarrow y^{2}-3 y+y-3=0$

$\Rightarrow y(y-3)+(y-3)=0$

$\Rightarrow(y+1)(y-3)=0$

$\Rightarrow y=-1$ or $y=3$

2. If ${{dy} \over {dx}} = {{{2^{x + y}} - {2^x}} \over {{2^y}}}$, y(0) = 1, then y(1) is equal to ______.

Ans: Given $\dfrac{d y}{dx}=\dfrac{2^{x+y}-2^{x}}{2^{y}}$

The given equation can be written as,

$2^{y} \dfrac{d y}{dx}=2^{x}\left(2^{y}-1\right)$

Integrating on both the sides, we get

$\int \dfrac{2^{y}}{2^{y}-1} d y=\int 2^{x} dx$

$\dfrac{\ln \left(2^{y}-1\right)}{\ln 2}=\dfrac{2^{x}}{\ln 2}+c$

$\Rightarrow \log _{2}\left(2^{y}-1\right)=2^{x} \log _{2} e+c$ —(1)

As $y(0)=1$

$\Rightarrow 0=\log _{2} e+c$

$C=-\log _{2} e$

Substitute in equation (1), we get

$\Rightarrow \log _{2}\left(2^{y}-1\right)=\left(2^{x}-1\right) \log _{2} e$

Put $x=1$, $\log _{2}\left(2^{y}-1\right)=\log _{2} e$

$2^{y}=e+1$

Hence, $y=\log _{2}(e+1)$

3. If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is $\dfrac{d y}{dx}=\dfrac{(x-2)^{2}+y+4}{x-2}$, then this curve also passes through the point ______.

Ans: Given, $\dfrac{d y}{dx}=\dfrac{(x-2)^{2}+y+4}{x-2}$ and as the curve passes through the origin we can say that $y(0)=0$

$\dfrac{d y}{dx}-\dfrac{y}{x-2}=(x-2)+\dfrac{4}{x-2}$

$\Rightarrow I.F=e^{-\int \frac{1}{x-2} dx}=\dfrac{1}{x-2}$

Hence the solution of D.E is

$\Rightarrow y \cdot \dfrac{1}{x-2} = \int \dfrac{1}{x-2}\left((x-2) + \dfrac{4}{x-2}\right) \cdot dx$

$\Rightarrow \dfrac{y}{x-2}=x-\dfrac{4}{x-2}+c$

Now, at $x=0, y=0 \Rightarrow c=-2$

On substituting the value of c in the above equation we get,

$y=x(x-2)-4-2(x-2)$

$\Rightarrow y=x^{2}-4 x$

Hence the curve passes through $(5,5)$

Practice Question

1. Solve, $x(\dfrac{dy}{dx}) + y =y^{2logx}$

Ans: $\dfrac{1}{xy} = \dfrac{1}{x}logx + \dfrac{1}{x} + c$

2. Solve the differential equation: $\dfrac{dy}{dx} = (e^x + 1)y$

Ans: $log|y| =ex + x + c$

3. Solve $\left(x tan \dfrac{y}{x}-ysec^2 \dfrac{y}{x}\right)dx-x\sec^{2} \dfrac{y}{x}dy=0$

Ans: $\dfrac{xtany}{x} = c$

Conclusion

We can conclude that a differential equation is defined as an equation that contains the derivative or derivatives of the dependent variable with respect to the independent variable. We also discussed differential equation formulas and various methods for solving differential equations using basic differential equation formulas. Students can carefully read through the Concepts, Definitions, and Questions in the PDFs also, which are also free to download and understand the concepts used to solve these questions which is extremely beneficial in the exams.

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