What is a differential equation?

A differential equation is a mathematical equation that relates a function to its derivatives, showing how the function changes over time or space.

What are the types of differential equations?

Differential equations are classified into ordinary differential equations (ODEs) and partial differential equations (PDEs) based on the number of independent variables involved.

What is the order of a differential equation?

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

What are some different types of differential equations?

Separation of variables homogeneous differential equation linear differential equation.

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Continuity and Differentiability

Continuity and differentiability are key concepts in calculus. Continuity means a function has no breaks in its graph.

Explore The Different Methods of Differentiation

Differentiation is a fundamental concept in calculus that deals with how things change. Specifically, it's about understanding

Partial Differential Equations

Partial Differential Equations (PDEs) are mathematical equations that involve multiple independent variables, their partial derivatives, and an unknown function or we can say that an equation which involves partial derivatives of a function of two or more independent variables is called Partial Differential Equation.

Differentiation

Differentiation is a mathematical process that measures how a function changes as its input changes.

Binomial Theorem

The Binomial Theorem is an essential tool for simplifying long expressions that follow the pattern (a + b)n. By providing...

Understanding The Conjugate Of A Complex Number

A complex conjugate is a concept in complex number theory where for any given complex number, a conjugate exists that reverses the sign of the imaginary part while keeping the real part unchanged.

Domain and Range of a Relation

The domain is all the possible input values, while the range is all the possible output values. Knowing...

Differential Equations

An equation that involves independent and dependent variables and the derivatives of the dependent variables is called a Differential Equation (D. E).

A Differential Equation (D. E) is said to be ordinary, if the differential coefficients have reference to a single independent variable only e.g. $\frac{d ^{2} y}{d x ^{2}} - \frac{2 d y}{d x} + cos x = 0$ and it is said to be partial if there are two or more independent variables. e.g. $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0$ is a partial differential equation. We are concerned with ordinary differential equations only.

1.0Order Of Differential Equations

The order of a differential equation (D.E) is the order of the highest differential coefficient occurring in it.

Example:

$\frac{d ^{2} y}{dx ^{2}} + \frac{dy}{dx} + 1 = 0$ , Order is 2

$\frac{d ^{3} y}{d x ^{3}} + x = 0$ , Order is 3.

2.0Degree of Differential Equations

The highest exponent of the highest order differential coefficient, when the differential equation is expressed as a polynomial in all the differential coefficients, is called degree of the differential equation.

Thus, the differential equation:

$a_{0, 0} + \sum_{i = 1}^{n_{1}} a_{i, 1} (\frac{d ^{m} y}{d x ^{m}})^{i} + \sum_{i = 1}^{n_{2}} a_{i, 2} (\frac{d ^{m - 1} y}{d x ^{m - 1}})^{i} + \dots = 0$ is of order m & degree $n_{1} .$

where $a_{i, j}$ is some expression in x, y and $a_{n_{1}, 1} \neq = 0$

Example: Find degree of $\frac{d ^{2} y}{d x ^{2}} = 3 \frac{d y}{d x} + 4$

Solution: $(\frac{d ^{2} y}{d x ^{2}})^{3} = \frac{d y}{d x} + 4$

Degree = 3

3.0Solution Of Differential Equations

The solution to a differential equation is a relation between the variables that does not include derivatives but satisfies the given differential equation. In other words, it is a function from which the given differential equation can be derived.

Thus, the solution of $\frac{d y}{d x} = e^{x}$ could be obtained by simply integrating both sides, i.e., $y = e^{x} + c$ and that of, $\frac{d y}{d x} = p x + q i sy = \frac{p x ^{2}}{2} + q x + c$ , where c is an arbitrary constant.

(i) A general solution, or integral, of a differential equation is a relation between the variables that does not include derivatives. This relation contains as many arbitrary constants as the order of the differential equation.

For example, a general solution of the differential equation $\frac{d ^{2} x}{d t ^{2}} = - 4 x i s x = A cos 2 t + B sin 2 t$

where A and B are the arbitrary constants.

(ii) Particular solution or integral is that solution of the differential equation which is obtained from the general solution by assigning particular values to the arbitrary constant in the general solution.

For example, $x = 10 cos 2 t + 5 s in 2 t$ is a particular solution of differential equation $\frac{d ^{2} x}{d t ^{2}} = - 4 x$ .

4.0Elementary Types of First Order and First-Degree Differential Equations

Separation of Variables

Some differential equations can be solved by the method of separation of variables (or “variable separable”).

This method is only possible, if we can express the differential equation in the form

$A (x) d x + B (y) d y = 0$

where A(x) is a function of '' only and B(y) is a function of $’ y’$ only.

A general solution of this is given by,

$\int A (x) d x + \int B (y) d y = c$

where $’C’$ is the arbitrary constant.

Homogeneous Differential Equations

A function f(x, y) is said to be a homogeneous function of degree n, if the substitution

$x = λ x, y = λ y, λ > 0$ produces the equality

$f_{(λ^{x}, λ^{y})} = λ^{n} f (x, y)$

The degree of homogeneity $’n’$ can be any real number.

A differential equation of the form $\frac{d y}{d x} = \frac{f ( x , y )}{g ( x , y )}$

Where f(x, y) and g(x, y) are homogeneous functions of x, y and of the same degree are said to be homogeneous. Such equations can be solved by substituting y = vx.

Equation Reducible to Homogeneous form

The equation of the form $\frac{d y}{d x} = \frac{a _{1} x + b _{1} y + c _{1}}{a _{2} x + b _{2} y + c _{2}} Wh ere \frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ can be reduced to homogeneous form by changing the variable x, y to u, v as x = u + h, y = v + k.

Also Read Homogeneous Differential Equation

Linear Differential Equations

A differential equation is considered linear if the dependent variable and its derivatives appear only to the first power and are not multiplied by each other.

The n^th order linear differential equation is of the form:

$a_{0} (x) \frac{d ^{n} y}{d x ^{n}} + a_{1} (x) \frac{d ^{n - 1} y}{d x ^{n - 1}} + \dots\dots .. + a_{n} (x) y = φ (x), w h ere a_{0} (x), a_{1} (x) \dots . a_{n} (x)$ are called the coefficients of the differential equation.

Note that a linear differential equation is always of the first degree. However, not every differential equation of the first degree. e.g. the differential equation $\frac{d ^{2} y}{d x ^{2}} + (\frac{d y}{d x})^{3} + y^{2} = 0$ is not linear, though its degree is 1.

(i) Linear Differential Equations of First Order:

The most general form of a linear differential equation of first order is $\frac{d y}{d x} + P y = Q$ , where $P_{&} Q$

are functions of x.

To solve such an equation multiply both sides by $e^{\int P d x}$ .

So that we get $e^{\int P d x} [\frac{d y}{d x} + P y] = Q e^{\int P d x}$ ...(i)

$\frac{d}{d x} (e^{\int P d x} \cdot y) = Q e^{\int P d x}$ ...(ii)

On integrating equation (ii), we get

$y e^{\int P d x} = \int Q e^{\int P d x} d x + c$

This is the required general solution.

Also Read: First Order Differential Equation

(ii) Equation Reducible to Linear Form:

The equation of the form $\frac{d y}{d x} + P y = Q y^{n}$ , where P and Q are functions of x, is called Bernoulli’s equation.

On dividing by $y^{n}$ , we get $y^{- n} \frac{d y}{d x} + P y^{- n + 1} = Q$

Let $y^{- n + 1} = t, so t ha t (- n + 1) y^{- n} \frac{d y}{d x} = \frac{d t}{d x}$

then equation becomes $\frac{d t}{d x} + P (1 - n) t = Q (1 - n)$

which is linear with t as a dependent variable.

5.0Solved Examples of Differential Equations

Example 1: Find the order and degree of the following differential equation: $\frac{d ^{2} y}{d x ^{2}} = sin (\frac{d y}{d x})$

Solution: The given differential equation has order 2. Since the given differential equation is not a polynomial in the differential coefficients, the degree of the equation is not defined.

Example 2: Solve the differential equation $x y \frac{d y}{d x} = \frac{1 + y ^{2}}{1 + x ^{2}} (1 + x + x^{2})$

Solution: Differential equation can be rewritten as

$x y = (1 + y^{2}) (1 + \frac{x}{1 + x ^{2}}) \Rightarrow \frac{y}{1 + y ^{2}} d y = (\frac{1}{x} + \frac{1}{1 + x ^{2}}) d x$

Integrating, we get

$\frac{1}{2} ln (1 + y^{2}) = ln x + tan^{- 1} x + ln c \Rightarrow 1 + y^{2} = c x e^{t a n^{- 1} x}$

Example 3: Solve $\frac{d y}{d x} = \frac{x - 2 y + 1}{2 x + 2 y + 3}$

Solution:

$\frac{d y}{d x} = \frac{x - 2 y + 1}{2 x + 2 y + 3}$

$⊨^{2 x d y + 2 y d y + 3 d y} = x d x - 2 y d x + d x$

$\Rightarrow (2 y + 3) d y = (x + 1) d x - 2 (x d y + y d x)$

On integrating, we get

$\Rightarrow \int (2 y + 3) d y = \int (x + 1) d x - \int 2 d (x y)$

Solving : $y^{2} + 3 y = \frac{x ^{2}}{2} + x - 2 x y + c$

Example 4: Find the degree of homogeneity of function

(i) $f (x, y) = x^{2} + y^{2}$

(ii) $f (x, y) = (x^{\frac{3}{2}} + y^{\frac{3}{2}}) / (x + y)$

(iii) $f (x, y) = sin (\frac{x}{y})$

Solution:

(i) $f (λ x, λ y) = λ^{2} x^{2} + λ^{2} y^{2} = λ^{2} (x^{2} + y^{2}) = λ^{2} f (x, y)$

degree of homogeneity ⟶ 2

(ii) $f (λ x, λ y) = \frac{λ ^{3/2} x ^{3/2} + λ ^{3/2} y ^{3/2}}{λ x + λ y} f (λ x, λ y) = λ^{\frac{1}{2}} f (x, y)$

degree of homogeneity ⟶ 1/2

(iii) $f (λ x, λ y) = sin (\frac{λ x}{λ y}) = λ^{\circ} sin (\frac{x}{y}) = λ^{\circ} f (x, y)$

degree of homogeneity ⟶ 0

Example 5: The solution of the given differential equation $\frac{d y}{d x} = \frac{s i n y + x}{s i n 2 y - x c o s y} s$ -

(A) $sin^{2} y = x sin y + \frac{x ^{2}}{2} + c$ (B) $sin^{2} y = x sin y - x + c$

(C) $sin^{2} y = x + sin y + x^{2} + c$ (D) $sin^{2} y = x - sin y + \frac{x ^{2}}{2} + c$

Ans. (A)

Solution:

Here, $\frac{d y}{d x} = \frac{s i n y + x}{s i n 2 y - x c o s y}$

$\Rightarrow (cos y) \frac{d y}{d x} = \frac{s i n y + x}{2 s i n y - x} (p u t sin y = t)$

$\Rightarrow \frac{d t}{d x} = \frac{t + x}{2 t - x}$ (put t=v x)

$\frac{x d v}{d x} + v = \frac{v + 1}{2 v - 1}$

$∴ \frac{x d v}{d x} = \frac{v + 1}{2 v - 1} - v = \frac{v + 1 - 2 v ^{2} + v}{2 v - 1}$

or $\frac{2 v - 1}{- 2 v ^{2} + 2 v + 1} d v = \frac{d x}{x}$

on solving, we get

$sin^{2} y = x sin y + + c$ .

Example 6: Solve $\frac{d y}{d x} = \frac{x + 2 y + 3}{2 x + 3 y + 4}$

Solution:

Equation reducible to homogeneous form

Put $x = X + h, y = Y + k 0$

We have $\frac{d Y}{d X} = \frac{X + 2 Y + ( h + 2 k + 3 )}{2 X + 3 Y + ( 2 h + 3 k + 4 )}$

To determine h and k, we write

$h + 2 k + 3 = 0, 2 h + 3 k + 4 = 0 \Rightarrow h = 1, k = - 2$

So that $\frac{d Y}{d X} = \frac{X + 2 Y}{2 X + 3 Y}$

Putting $Y = V X$ , we get

$V + X \frac{d V}{d X} = \frac{1 + 2 V}{2 + 3 V} \Rightarrow \frac{2 + 3 V}{3 V ^{2} - 1} d V = - (\frac{d X}{X})$

$\Rightarrow [\frac{2 + 3}{2 ( 3 V - 1 )} - \frac{2 - 3}{2 ( 3 V + 1 )}] d V = - (\frac{d X}{X})$

$\Rightarrow \frac{2 + 3}{2 3} lo g (3 V - 1) - \frac{2 - 3}{2 3} lo g (3 V + 1) = (- lo g X + c)$

$∣^{\frac{2 + 3}{2 3}} lo g (3 Y - X) - \frac{2 - 3}{2 3} lo g (3 Y + X) = A$

where A is another

constant and

$X = x - 1, Y = y + 2 .$

Example 7: Which of the following equations is a linear differential equation?

(A) $\frac{d y}{d x} + x y^{2} = 1$ (B) $x^{2} + y = e^{x}$

(C) $\frac{d y}{d x} + 3 y = x y^{2}$ (D) $x \frac{d y}{d x} + y^{2} = sin x$

Ans. (B)

Solution:

Clearly answer is (B)

Example 8: Solve the differential equation $\frac{d y}{d x} + \frac{y}{x} = x^{2} y^{6}$ .

Solution:

The provided differential equation can be written as $\frac{1}{y ^{6}} \frac{d y}{d x} + \frac{1}{x y ^{5}} = x^{2}$

Putting y^{-5}=v so that

$- 5 y^{- 6} \frac{d y}{d x} = \frac{d v}{d x} or y^{- 6} \frac{d y}{d x} = - \frac{1}{5} \frac{d v}{d x}$

we get

$- \frac{1}{5} \frac{d v}{d x} + \frac{1}{x} v = x^{2} \Rightarrow \frac{d v}{d x} - \frac{5}{x} v = - 5 x^{2}$ ...(i)

This is the standard form of the linear differential equation having integrating factor

I.F = $e^{\int - \frac{5}{x} d x} = e^{- 5 l o g x} = \frac{1}{x ^{5}}$

Multiplying both sides of (i) by I.F. and integrating w.r.t. x

We get $v. \frac{1}{x ^{5}} = \int - 5 x^{2} \cdot \frac{1}{x ^{5}} d x$

$\Rightarrow \frac{v}{x ^{5}} = \frac{5}{2} x^{- 2} + c$

$\Rightarrow y^{- 5} x^{- 5} = \frac{5}{2} x^{- 2} + c$

which is the required solution.

Also Read Differential Equation previous year questions with solutions

6.0Practice Question on Differential Equations

1. Find the order and degree of differential equation $[\frac{d ^{2} x}{d t ^{2}}]^{3} + [\frac{d x}{d t}]^{4} - x t = 0$

(A) order : 3 ; degree : 2 (B) order : 2 ; degree : 4

(C) order : 2 ; degree : 3 (D) order : 4 ; degree : 2

2. Obtain the differential equation of the family of circles $x^{2} + y^{2} + 2 gx + 2 f y + c = 0$ ; where $g, f & c$

are arbitrary constants.

(A) $[1 + (y^{'})^{2}] \cdot y^{'''} - 3 y^{'} (y^{''})^{2} = 0$

(B) $[1 + (y^{'})^{2}] \cdot y^{'''} + (y^{''} + 2 y^{'}) (y^{''})^{2} = 0$

(C) $[1 + (y^{'})^{2}] \cdot y^{'''} + 3 y^{'} (y^{''})^{2} = 0$

(D) $[1 - (y^{'})^{2}] \cdot y^{'''} + 3 y^{'} (y^{''})^{2} = 0$

3. Solution of the differential equation $\frac{d y}{d x} = sin (x + y) + cos (x + y)$ is

(A) $tan (\frac{x + y}{2}) = a e^{x} - 1$

(B) $tan (\frac{x - y}{2}) = a e^{x} - 1$

(C) $tan (\frac{x + y}{2}) = a e^{- x} - 1$

(D) $tan (\frac{x + y}{2}) = a e^{x} + 1$

4. Solution of the differential equation $\frac{x d x + y d y}{x ^{2} + y ^{2}} = \frac{x d y - y d x}{x ^{2}}$ is

(A) $x^{2} - y^{2} = \frac{y}{x} + c$

(B) $x^{2} + y^{2} = - \frac{y}{x} + c$

(C) $x^{2} + y^{2} = \frac{x}{y} + c$

(D) $x^{2} + y^{2} = \frac{y}{x} + c$

Answer Sheet:

Question	1	2	3	4
Answer	C	C	A	D

Also Read

Divergence and Curl	Differentiability	Applications of Derivatives
Divergence Theorem	Differential Calculus	Partial Differential Equations

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I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Continuity and Differentiability

Continuity and differentiability are key concepts in calculus. Continuity means a function has no breaks in its graph.

Explore The Different Methods of Differentiation

Differentiation is a fundamental concept in calculus that deals with how things change. Specifically, it's about understanding

Partial Differential Equations

Partial Differential Equations (PDEs) are mathematical equations that involve multiple independent variables, their partial derivatives, and an unknown function or we can say that an equation which involves partial derivatives of a function of two or more independent variables is called Partial Differential Equation.

Differentiation

Differentiation is a mathematical process that measures how a function changes as its input changes.

Binomial Theorem

The Binomial Theorem is an essential tool for simplifying long expressions that follow the pattern (a + b)n. By providing...

Understanding The Conjugate Of A Complex Number

A complex conjugate is a concept in complex number theory where for any given complex number, a conjugate exists that reverses the sign of the imaginary part while keeping the real part unchanged.

Domain and Range of a Relation

The domain is all the possible input values, while the range is all the possible output values. Knowing...

Differential Equations

An equation that involves independent and dependent variables and the derivatives of the dependent variables is called a Differential Equation (D. E).

A Differential Equation (D. E) is said to be ordinary, if the differential coefficients have reference to a single independent variable only e.g. $\frac{d ^{2} y}{d x ^{2}} - \frac{2 d y}{d x} + cos x = 0$ and it is said to be partial if there are two or more independent variables. e.g. $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0$ is a partial differential equation. We are concerned with ordinary differential equations only.

1.0Order Of Differential Equations

The order of a differential equation (D.E) is the order of the highest differential coefficient occurring in it.

Example:

$\frac{d ^{2} y}{dx ^{2}} + \frac{dy}{dx} + 1 = 0$ , Order is 2

$\frac{d ^{3} y}{d x ^{3}} + x = 0$ , Order is 3.

2.0Degree of Differential Equations

The highest exponent of the highest order differential coefficient, when the differential equation is expressed as a polynomial in all the differential coefficients, is called degree of the differential equation.

Thus, the differential equation:

$a_{0, 0} + \sum_{i = 1}^{n_{1}} a_{i, 1} (\frac{d ^{m} y}{d x ^{m}})^{i} + \sum_{i = 1}^{n_{2}} a_{i, 2} (\frac{d ^{m - 1} y}{d x ^{m - 1}})^{i} + \dots = 0$ is of order m & degree $n_{1} .$

where $a_{i, j}$ is some expression in x, y and $a_{n_{1}, 1} \neq = 0$

Example: Find degree of $\frac{d ^{2} y}{d x ^{2}} = 3 \frac{d y}{d x} + 4$

Solution: $(\frac{d ^{2} y}{d x ^{2}})^{3} = \frac{d y}{d x} + 4$

Degree = 3

3.0Solution Of Differential Equations

The solution to a differential equation is a relation between the variables that does not include derivatives but satisfies the given differential equation. In other words, it is a function from which the given differential equation can be derived.

Thus, the solution of $\frac{d y}{d x} = e^{x}$ could be obtained by simply integrating both sides, i.e., $y = e^{x} + c$ and that of, $\frac{d y}{d x} = p x + q i sy = \frac{p x ^{2}}{2} + q x + c$ , where c is an arbitrary constant.

(i) A general solution, or integral, of a differential equation is a relation between the variables that does not include derivatives. This relation contains as many arbitrary constants as the order of the differential equation.

For example, a general solution of the differential equation $\frac{d ^{2} x}{d t ^{2}} = - 4 x i s x = A cos 2 t + B sin 2 t$

where A and B are the arbitrary constants.

(ii) Particular solution or integral is that solution of the differential equation which is obtained from the general solution by assigning particular values to the arbitrary constant in the general solution.

For example, $x = 10 cos 2 t + 5 s in 2 t$ is a particular solution of differential equation $\frac{d ^{2} x}{d t ^{2}} = - 4 x$ .

4.0Elementary Types of First Order and First-Degree Differential Equations

Separation of Variables

Some differential equations can be solved by the method of separation of variables (or “variable separable”).

This method is only possible, if we can express the differential equation in the form

$A (x) d x + B (y) d y = 0$

where A(x) is a function of '' only and B(y) is a function of $’ y’$ only.

A general solution of this is given by,

$\int A (x) d x + \int B (y) d y = c$

where $’C’$ is the arbitrary constant.

Homogeneous Differential Equations

A function f(x, y) is said to be a homogeneous function of degree n, if the substitution

$x = λ x, y = λ y, λ > 0$ produces the equality

$f_{(λ^{x}, λ^{y})} = λ^{n} f (x, y)$

The degree of homogeneity $’n’$ can be any real number.

A differential equation of the form $\frac{d y}{d x} = \frac{f ( x , y )}{g ( x , y )}$

Where f(x, y) and g(x, y) are homogeneous functions of x, y and of the same degree are said to be homogeneous. Such equations can be solved by substituting y = vx.

Equation Reducible to Homogeneous form

The equation of the form $\frac{d y}{d x} = \frac{a _{1} x + b _{1} y + c _{1}}{a _{2} x + b _{2} y + c _{2}} Wh ere \frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ can be reduced to homogeneous form by changing the variable x, y to u, v as x = u + h, y = v + k.

Also Read Homogeneous Differential Equation

Linear Differential Equations

A differential equation is considered linear if the dependent variable and its derivatives appear only to the first power and are not multiplied by each other.

The n^th order linear differential equation is of the form:

$a_{0} (x) \frac{d ^{n} y}{d x ^{n}} + a_{1} (x) \frac{d ^{n - 1} y}{d x ^{n - 1}} + \dots\dots .. + a_{n} (x) y = φ (x), w h ere a_{0} (x), a_{1} (x) \dots . a_{n} (x)$ are called the coefficients of the differential equation.

Note that a linear differential equation is always of the first degree. However, not every differential equation of the first degree. e.g. the differential equation $\frac{d ^{2} y}{d x ^{2}} + (\frac{d y}{d x})^{3} + y^{2} = 0$ is not linear, though its degree is 1.

(i) Linear Differential Equations of First Order:

The most general form of a linear differential equation of first order is $\frac{d y}{d x} + P y = Q$ , where $P_{&} Q$

are functions of x.

To solve such an equation multiply both sides by $e^{\int P d x}$ .

So that we get $e^{\int P d x} [\frac{d y}{d x} + P y] = Q e^{\int P d x}$ ...(i)

$\frac{d}{d x} (e^{\int P d x} \cdot y) = Q e^{\int P d x}$ ...(ii)

On integrating equation (ii), we get

$y e^{\int P d x} = \int Q e^{\int P d x} d x + c$

This is the required general solution.

Also Read: First Order Differential Equation

(ii) Equation Reducible to Linear Form:

The equation of the form $\frac{d y}{d x} + P y = Q y^{n}$ , where P and Q are functions of x, is called Bernoulli’s equation.

On dividing by $y^{n}$ , we get $y^{- n} \frac{d y}{d x} + P y^{- n + 1} = Q$

Let $y^{- n + 1} = t, so t ha t (- n + 1) y^{- n} \frac{d y}{d x} = \frac{d t}{d x}$

then equation becomes $\frac{d t}{d x} + P (1 - n) t = Q (1 - n)$

which is linear with t as a dependent variable.

5.0Solved Examples of Differential Equations

Example 1: Find the order and degree of the following differential equation: $\frac{d ^{2} y}{d x ^{2}} = sin (\frac{d y}{d x})$

Solution: The given differential equation has order 2. Since the given differential equation is not a polynomial in the differential coefficients, the degree of the equation is not defined.

Example 2: Solve the differential equation $x y \frac{d y}{d x} = \frac{1 + y ^{2}}{1 + x ^{2}} (1 + x + x^{2})$

Solution: Differential equation can be rewritten as

$x y = (1 + y^{2}) (1 + \frac{x}{1 + x ^{2}}) \Rightarrow \frac{y}{1 + y ^{2}} d y = (\frac{1}{x} + \frac{1}{1 + x ^{2}}) d x$

Integrating, we get

$\frac{1}{2} ln (1 + y^{2}) = ln x + tan^{- 1} x + ln c \Rightarrow 1 + y^{2} = c x e^{t a n^{- 1} x}$

Example 3: Solve $\frac{d y}{d x} = \frac{x - 2 y + 1}{2 x + 2 y + 3}$

Solution:

$\frac{d y}{d x} = \frac{x - 2 y + 1}{2 x + 2 y + 3}$

$⊨^{2 x d y + 2 y d y + 3 d y} = x d x - 2 y d x + d x$

$\Rightarrow (2 y + 3) d y = (x + 1) d x - 2 (x d y + y d x)$

On integrating, we get

$\Rightarrow \int (2 y + 3) d y = \int (x + 1) d x - \int 2 d (x y)$

Solving : $y^{2} + 3 y = \frac{x ^{2}}{2} + x - 2 x y + c$

Example 4: Find the degree of homogeneity of function

(i) $f (x, y) = x^{2} + y^{2}$

(ii) $f (x, y) = (x^{\frac{3}{2}} + y^{\frac{3}{2}}) / (x + y)$

(iii) $f (x, y) = sin (\frac{x}{y})$

Solution:

(i) $f (λ x, λ y) = λ^{2} x^{2} + λ^{2} y^{2} = λ^{2} (x^{2} + y^{2}) = λ^{2} f (x, y)$

degree of homogeneity ⟶ 2

(ii) $f (λ x, λ y) = \frac{λ ^{3/2} x ^{3/2} + λ ^{3/2} y ^{3/2}}{λ x + λ y} f (λ x, λ y) = λ^{\frac{1}{2}} f (x, y)$

degree of homogeneity ⟶ 1/2

(iii) $f (λ x, λ y) = sin (\frac{λ x}{λ y}) = λ^{\circ} sin (\frac{x}{y}) = λ^{\circ} f (x, y)$

degree of homogeneity ⟶ 0

Example 5: The solution of the given differential equation $\frac{d y}{d x} = \frac{s i n y + x}{s i n 2 y - x c o s y} s$ -

(A) $sin^{2} y = x sin y + \frac{x ^{2}}{2} + c$ (B) $sin^{2} y = x sin y - x + c$

(C) $sin^{2} y = x + sin y + x^{2} + c$ (D) $sin^{2} y = x - sin y + \frac{x ^{2}}{2} + c$

Ans. (A)

Solution:

Here, $\frac{d y}{d x} = \frac{s i n y + x}{s i n 2 y - x c o s y}$

$\Rightarrow (cos y) \frac{d y}{d x} = \frac{s i n y + x}{2 s i n y - x} (p u t sin y = t)$

$\Rightarrow \frac{d t}{d x} = \frac{t + x}{2 t - x}$ (put t=v x)

$\frac{x d v}{d x} + v = \frac{v + 1}{2 v - 1}$

$∴ \frac{x d v}{d x} = \frac{v + 1}{2 v - 1} - v = \frac{v + 1 - 2 v ^{2} + v}{2 v - 1}$

or $\frac{2 v - 1}{- 2 v ^{2} + 2 v + 1} d v = \frac{d x}{x}$

on solving, we get

$sin^{2} y = x sin y + + c$ .

Example 6: Solve $\frac{d y}{d x} = \frac{x + 2 y + 3}{2 x + 3 y + 4}$

Solution:

Equation reducible to homogeneous form

Put $x = X + h, y = Y + k 0$

We have $\frac{d Y}{d X} = \frac{X + 2 Y + ( h + 2 k + 3 )}{2 X + 3 Y + ( 2 h + 3 k + 4 )}$

To determine h and k, we write

$h + 2 k + 3 = 0, 2 h + 3 k + 4 = 0 \Rightarrow h = 1, k = - 2$

So that $\frac{d Y}{d X} = \frac{X + 2 Y}{2 X + 3 Y}$

Putting $Y = V X$ , we get

$V + X \frac{d V}{d X} = \frac{1 + 2 V}{2 + 3 V} \Rightarrow \frac{2 + 3 V}{3 V ^{2} - 1} d V = - (\frac{d X}{X})$

$\Rightarrow [\frac{2 + 3}{2 ( 3 V - 1 )} - \frac{2 - 3}{2 ( 3 V + 1 )}] d V = - (\frac{d X}{X})$

$\Rightarrow \frac{2 + 3}{2 3} lo g (3 V - 1) - \frac{2 - 3}{2 3} lo g (3 V + 1) = (- lo g X + c)$

$∣^{\frac{2 + 3}{2 3}} lo g (3 Y - X) - \frac{2 - 3}{2 3} lo g (3 Y + X) = A$

where A is another

constant and

$X = x - 1, Y = y + 2 .$

Example 7: Which of the following equations is a linear differential equation?

(A) $\frac{d y}{d x} + x y^{2} = 1$ (B) $x^{2} + y = e^{x}$

(C) $\frac{d y}{d x} + 3 y = x y^{2}$ (D) $x \frac{d y}{d x} + y^{2} = sin x$

Ans. (B)

Solution:

Clearly answer is (B)

Example 8: Solve the differential equation $\frac{d y}{d x} + \frac{y}{x} = x^{2} y^{6}$ .

Solution:

The provided differential equation can be written as $\frac{1}{y ^{6}} \frac{d y}{d x} + \frac{1}{x y ^{5}} = x^{2}$

Putting y^{-5}=v so that

$- 5 y^{- 6} \frac{d y}{d x} = \frac{d v}{d x} or y^{- 6} \frac{d y}{d x} = - \frac{1}{5} \frac{d v}{d x}$

we get

$- \frac{1}{5} \frac{d v}{d x} + \frac{1}{x} v = x^{2} \Rightarrow \frac{d v}{d x} - \frac{5}{x} v = - 5 x^{2}$ ...(i)

This is the standard form of the linear differential equation having integrating factor

I.F = $e^{\int - \frac{5}{x} d x} = e^{- 5 l o g x} = \frac{1}{x ^{5}}$

Multiplying both sides of (i) by I.F. and integrating w.r.t. x

We get $v. \frac{1}{x ^{5}} = \int - 5 x^{2} \cdot \frac{1}{x ^{5}} d x$

$\Rightarrow \frac{v}{x ^{5}} = \frac{5}{2} x^{- 2} + c$

$\Rightarrow y^{- 5} x^{- 5} = \frac{5}{2} x^{- 2} + c$

which is the required solution.

Also Read Differential Equation previous year questions with solutions

6.0Practice Question on Differential Equations

1. Find the order and degree of differential equation $[\frac{d ^{2} x}{d t ^{2}}]^{3} + [\frac{d x}{d t}]^{4} - x t = 0$

(A) order : 3 ; degree : 2 (B) order : 2 ; degree : 4

(C) order : 2 ; degree : 3 (D) order : 4 ; degree : 2

2. Obtain the differential equation of the family of circles $x^{2} + y^{2} + 2 gx + 2 f y + c = 0$ ; where $g, f & c$

are arbitrary constants.

(A) $[1 + (y^{'})^{2}] \cdot y^{'''} - 3 y^{'} (y^{''})^{2} = 0$

(B) $[1 + (y^{'})^{2}] \cdot y^{'''} + (y^{''} + 2 y^{'}) (y^{''})^{2} = 0$

(C) $[1 + (y^{'})^{2}] \cdot y^{'''} + 3 y^{'} (y^{''})^{2} = 0$

(D) $[1 - (y^{'})^{2}] \cdot y^{'''} + 3 y^{'} (y^{''})^{2} = 0$

3. Solution of the differential equation $\frac{d y}{d x} = sin (x + y) + cos (x + y)$ is

(A) $tan (\frac{x + y}{2}) = a e^{x} - 1$

(B) $tan (\frac{x - y}{2}) = a e^{x} - 1$

(C) $tan (\frac{x + y}{2}) = a e^{- x} - 1$

(D) $tan (\frac{x + y}{2}) = a e^{x} + 1$

4. Solution of the differential equation $\frac{x d x + y d y}{x ^{2} + y ^{2}} = \frac{x d y - y d x}{x ^{2}}$ is

(A) $x^{2} - y^{2} = \frac{y}{x} + c$

(B) $x^{2} + y^{2} = - \frac{y}{x} + c$

(C) $x^{2} + y^{2} = \frac{x}{y} + c$

(D) $x^{2} + y^{2} = \frac{y}{x} + c$

Answer Sheet:

Question	1	2	3	4
Answer	C	C	A	D

Also Read

Divergence and Curl	Differentiability	Applications of Derivatives
Divergence Theorem	Differential Calculus	Partial Differential Equations

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What is the degree of the differential equation?

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Table of Contents

Frequently Asked Questions

What is a differential equation?

What are the types of differential equations?

What is the order of a differential equation?

What is the degree of the differential equation?

What are some different types of differential equations?

Join ALLEN!

Related Articles:-

Continuity and Differentiability

Explore The Different Methods of Differentiation

Partial Differential Equations

Differentiation

Binomial Theorem

Understanding The Conjugate Of A Complex Number

Domain and Range of a Relation

Differential Equations

1.0Order Of Differential Equations

2.0Degree of Differential Equations

3.0Solution Of Differential Equations

4.0Elementary Types of First Order and First-Degree Differential Equations

Separation of Variables

Homogeneous Differential Equations

Linear Differential Equations

5.0Solved Examples of Differential Equations

6.0Practice Question on Differential Equations

Table of Contents