NCERT Solutions for Class 12 Maths Chapter 9: Differential Equations
The requirement of Class 12 Maths Chapter 9: Differential Equations deals with all the aspects related to the solution of differential equations. This Chapter, class 12 maths chapter 9, enunciates an idea over the basic concepts and ways to form a differential equation along with the methods of solving it. The solutions provided in this chapter will be of great help in understanding the nature and characteristics of differential equations together with its applications. So, comprehension of Class 12 Differential Equations bears immense significance for the students since most of the questions relating to the same appear in almost all types of entrance tests apart from their great importance in studying mathematics and engineering further. Chapter 9 Maths Class 12 Pdf provides a kind of base or conceptual clarity in solving various kinds of differential equations, such as first-order and first-degree differential equations, linear differential equations, and homogeneous differential equations. Solutions with step-by-step reasoning help to build up a very strong conceptual framework in which the student is able to develop a greater capacity for problem-solving. Such solutions for class 12 strictly cover the CBSE syllabus requirements and are comprehensive in terms of the different concepts explained in this chapter. The chapter also includes multiple examples and exercises, making the learner practice thoroughly.
Download PDF For NCERT Solutions for Maths Differential Equations
The NCERT Solutions for Class 12 Maths Chapter 9: Differential Equations are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Solutions for Class 12 Maths Chapter 9: Differential Equations
Students can access the NCERT Solutions for Class 12 Maths Chapter 9: Differential Equations. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Differential Equations - Exercise 9.1
Determine order and degree (if defined) of differential equations given in Questions 1 to 10:
Given:
The highest order derivative present in the differential equation is y”” and its order is 4.
The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.
y’ + 5y = 0
The given differential equation is:
y’ + 5y = 0
The highest order derivative present in the differential equation is y’. Therefore, its order is one.
It is a polynomial equation in y’. The highest power raised to y’is 1. Hence, its degree is one.
Given:
The highest order derivative present in the given differential equation is d2s/dt2 Therefore, its order is two.
It is a polynomial equation in d2s/dt2 and ds/dt. The power raised to d2s/dt2 is 1.
Hence, its degree is one.
Given:
The highest order derivative present in the given differential equation is d2y/dx2 Therefore, its order is 2.
The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.
Given:
The highest order derivative present in the differential equation is d2y/dx2 and its order is 2.
The given differential equation is a polynomial equation in derivatives and the highest power raised to highest order d2y/dx2 is one, so its degree is 1.
Hence, order is 2 and degree is 1.
Given:
The highest order derivative present in the differential equation is y”’. Therefore, its order is three.
The given differential equation is a polynomial equation in y”’, y”, and y’.
The highest power raised to y”’ is 2. Hence, its degree is 2.
y”’ + 2y” + y’ = 0
Given: y”’ + 2y” + y’ = 0
The highest order derivative present in the differential equation is y”’. Therefore, its order is three.
It is a polynomial equation in y”’, y” and y’. The highest power raised to y”’is 1. Hence, its degree is 1.
y′ + y = ex
Given: y′ + y = ex
⇒ y′ + y – ex = 0
The highest order derivative present in the differential equation is y’. Therefore, its order is one.
The given differential equation is a polynomial equation in y’ and the highest power raised to y’ is one. Hence, its degree is one.
y’′ + (y’)2 + 2y = 0
Given: y’′ + (y’)2 + 2y = 0
The highest order derivative present in the differential equation is y”. Therefore, its order is two.
The given differential equation is a polynomial equation in y”and y’ and the highest power raised to y” is one.
Hence, its degree is one.
y’′ + 2y’ + sin y = 0
Given: y’′ + 2y’ + sin y = 0
The highest order derivative present in the differential equation is y”. Therefore, its order is two.
This is a polynomial equation in y”. and y’.and the highest power raised to y”. is one. Hence, its degree is one.
The degree of the differential equation is:
(A) 3
(B) 2
(C) 1
(D) Not defined
Given: 
The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined.
Hence, the correct answer is D.
Hence, option (D) is correct.
The order of the differential equation is:
(A) 2
(B) 1
(C) 0
(D) Not defined
Given:
The highest order derivative present in the differential equation is d2y/dx2 and its order is 2.
Therefore, option (A) is correct.
Differential Equations - Exercise 9.2
Given: y = √1 + x2
Hence, the given function is the solution of the corresponding differential equation.
In each of the Questions, 1 to 6 verify that the given functions (explicit) is a solution of the corresponding differential equation:
y = ex + 1 : y″ – y′ = 0
Given: y = ex + 1
Thus, the given function is the solution of the corresponding differential equation.
y = x2 + 2x + C : y′ – 2x – 2 = 0
Given: y = x2 + 2x + C
Differentiating both sides of this equation with respect to x, we get:
Hence, the given function is the solution of the corresponding differential equation.
y = cos x + C : y′ + sin x = 0
Given: y = cos x + C
Differentiating both sides of this equation with respect to x, we get:
Hence, the given function is the solution of the corresponding differential equation.
Given: y = √1 + x2
Hence, the given function is the solution of the corresponding differential equation.
y = Ax : xy′ = y (x ≠ 0)
Given: y = Ax
Differentiating both sides with respect to x, we get:
Hence, the given function is the solution of the corresponding differential equation.
Given: y = x sin x
Differentiating both sides of this equation with respect to x, we get:
Hence, the given function is the solution of the corresponding differential equation.
xy = log y + C :
Given: xy = log y + C
Differentiating both sides of this equation with respect to x, we get:
Hence, the given function is the solution of the corresponding differential equation.
y – cos y = x : (y sin y + cos y + x) y′ = y
Given: y – cos y = x
Differentiating both sides of the equation with respect to x, we get:
Hence, the given function is the solution of the corresponding differential equation.
x + y = tan-1 y : y2y’ + y2 + 1 = 0
Given: x + y = tan-1 y
Differentiating both sides of this equation with respect to x, we get:
Hence, the given function is the solution of the corresponding differential equation.
Given: y = √a2 – x2
Differentiating both sides of this equation with respect to x, we get:
Hence, the given function is the solution of the corresponding differential equation.
Choose the correct answer:
The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4
Option (D) is correct.
We know that the number of constants in the general solution of a differential equation of order n is equal to its order.
Therefore, the number of constants in the general equation of fourth order differential equation is four.
The number of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3
(B) 2
(C) 1
(D) 0
The number of arbitrary constants in a particular solution of a differential equation of any order is zero (0) as a particular solution is a solution which contains no arbitrary constant.
Therefore, option (D) is correct.
Differential Equations - Exercise 9.3
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
The equation of the family of hyperbolas with the centre at origin and foci along the x-axis is:
….(i)
This is the required differential equation.
Given: Equation of the family of curves 
Differentiating both sides of the given equation with respect to x, we get:
Hence, the required differential equation of the given curve is y” = 0.
Given: Equation of the family of curves
Differentiating both sides with respect to x, we get:
This is the required differential equation of the given curve.
Given: Equation of the family of curves ….(i)
Differentiating both sides with respect to x, we get:
This is the required differential equation of the given curve.
Given: Equation of the family of curves
Differentiating both sides with respect to x, we get:
This is the required differential equation of the given curve.
Given: Equation of the family of curves….(i)
Differentiating both sides with respect to x, we get:
This is the required differential equation of the given curve.
Form the differential equation of the family of circles touching the y-axis at the origin.
The centre of the circle touching the y-axis at origin lies on the x-axis.
Let (a, 0) be the centre of the circle.
Since it touches the y-axis at origin, its radius is a.
Now, the equation of the circle with centre (a, 0) and radius (a) is
This is the required differential equation.
Find the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
The equation of the parabola having the vertex at origin and the axis along the positive y-axis is:
x2 = 4ay
Differentiating equation (1) with respect to x, we get:
2x = 4ay’
Dividing equation (2) by equation (1), we get:
This is the required differential equation.
Form the differential equation of family of ellipse having foci on y-axis and centre at the origin.
The equation of the family of ellipses having foci on the y-axis and the centre at origin is as follows:
This is the required differential equation.
Form the differential equation of the family of circles having centres on y-axis and radius 3 units.
Let the centre of the circle on y-axis be (0, b).
The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:
This is the required differential equation.
Which of the following differential equation has as the general solution:
Given: 
Differentiating with respect to x, we get:
This is the required differential equation of the given equation of curve.
Hence, the correct answer is B.
Which of the following differential equations has y = x as one of its particular solutions:
The given equation of curve is y = x.
Differentiating with respect to x, we get:
dy/dx = 1 …(1)
Again, differentiating with respect to x, we get:
d2y/dx2 = 0 ….(2)
Now, on substituting the values of y, d2y/dx2 and dy/dx from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct.
Therefore, option (C) is correct.
Differential Equations - Exercise 9.4
In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years
(e0.5 = 1.645).
Let p and t be the principal and time respectively.
It is given that the principal increases continuously at the rate of 5% per year.
Hence, after 10 years the amount will be worth Rs 1648.
For each of the differential equations in Questions 1 to 4, find the general solution:
Given: Differential equation
This is the required general solution of the given differential equation.
The given differential equation is:
This is the required general solution of the given differential equation.
Given: Differential equation
This is the required general solution of the given differential equation.
Given: Differential equation
This is the required general solution of the given differential equation.
Given: Differential equation
This is the required general solution of the given differential equation.
Given: Differential equation y log y dx – x dy = 0
This is the required general solution of the given differential equation.
Given: Differential equation
Given: Differential equation
This is the required general solution of the given differential equation.
Given: Differential equation
Comparing the coefficients of x2 and x, we get:
A + B = 2
B + C = 1
A + C = 0
Solving these equations, we get:
Substituting C = 1 in equation (1), we get:
y = sec x
Find the equation of the curve passing through the point (0, 0) and whose differential equation is y’ = ex sin x
The differential equation of the curve is:
For the differential equation 
Find the equation of the curve passing through the point (0,-2) given that at any point (x,y) on the curve the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point.
Let x and y be the x-coordinate and y-coordinate of the curve respectively.
We know that the slope of a tangent to the curve in the coordinate axis is given by the relation,
dy/dx
According to the given information, we get:
Now, the curve passes through point (0, –2).
∴ (–2)2 – 02 = 2C
⇒ 2C = 4
Substituting 2C = 4 in equation (1), we get:
y2 – x2 = 4
This is the required equation of the curve.
At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (–4, –3). Find the equation of the curve given that it passes through (–2, 1).
It is given that (x, y) is the point of contact of the curve and its tangent.
The slope (m1) of the line segment joining (x, y) and (–4, –3) is y+3/x+4.
We know that the slope of the tangent to the curve is given by the relation,
dy/dx
Substituting C = 1 in equation (1), we get:
y + 3 = (x + 4)2
This is the required equation of the curve.
The volume of the spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Let the rate of change of the volume of the balloon be k (where k is a constant).
⇒ 4π × 33 = 3 (k × 0 + C)
⇒ 108π = 3C
⇒ C = 36π
At t = 3, r = 6:
⇒ 4π × 63 = 3 (k × 3 + C)
⇒ 864π = 3 (3k + 36π)
⇒ 3k = –288π – 36π = 252π
⇒ k = 84π
Substituting the values of k and C in equation (1), we get:
Thus, the radius of the balloon after t seconds is
In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (loge 2 = 0.6931).
Let p, t, and r represent the principal, time, and rate of interest respectively.
It is given that the principal increases continuously at the rate of r% per year.
Hence, the value of r is 6.93%.
In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present.
Let y be the number of bacteria at any instant t.
It is given that the rate of growth of the bacteria is proportional to the number present.
The general solution of the differential equation 
Therefore, option (A) is correct.
Differential Equations - Exercise 9.5
The given differential equation i.e., (x2 + xy) dy = (x2 + y2) dx can be written as:
This shows that equation (1) is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Differentiating both sides with respect to x, we get:
dy/dx = v + x dv/dx
Substituting the values of v and dy/dx in equation (1), we get:
This is the required solution of the given differential equation.
The given differential equation is:
Thus, the given equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Differentiating both sides with respect to x, we get:
dy/dx = v + x dv/dx
Substituting the values of y and dy/dx in equation (1), we get:
This is the required solution of the given differential equation.
The given differential equation is
The given differential equation is:
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
This is the required solution of the given differential equation.
The given differential equation is:
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
This is the required solution for the given differential equation.
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
This is the required solution of the given differential equation.
The given differential equation is:
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
This is the required solution of the given differential equation.
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
This is the required solution of the given differential equation.
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as: y = vx
Integrating both sides, we get:
This is the required solution of the given differential equation.
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as: x = vy
This is the required solution of the given differential equation.
(x + y) dy + (x – y) dx = 0; y = 1 when x = 1
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the value of 2k in equation (2), we get:
This is the required solution of the given differential equation.
x2 dy + (xy + y2 ) dx = 0; y = 1 when x = 1
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
This is the required solution of the given differential equation.
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
This is the required solution of the given differential equation.
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
This is the required solution of the given differential equation.
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
A homogeneous differential equation of the form
(A) y = vx
(B) v = yx
(C) x = vy
(D) x = v
We know that a homogeneous differential equation of the form
Therefore, option (C) is correct.
Which of the following is a homogeneous differential equation:
Out of the given four options, option (D) is the only option in which all coefficients of x and y are of same degree i.e., 2. It may be noted that y2 is a term of second degree.
Hence differential equation in option (D) is a Homogeneous differential equation.
Differential Equations - Exercise 9.6
Given: Differential equation
Given: Differential equation
This is the required general solution of the given differential equation.
Given: Differential equation
The given differential equation is:
Given: Differential equation
Given: Differential equation
Given: Differential equation
Given: Differential equation
Given: Differential equation
Given: Differential equation
Given: Differential equation
Given: Differential equation
This is a linear equation of the form:
Hence, the required solution of the given differential equation is y = cos x – 2 cos2 x.
Given: Differential equation
This is the required general solution of the given differential equation.
Given: Differential equation
This is the required particular solution of the given differential equation.
Find the equation of the curve passing through the origin, given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of coordinates of that point.
Let F (x, y) be the curve passing through the origin.
At point (x, y), the slope of the curve will be dy/dx.
According to the given information:
The curve passes through the origin.
Therefore, equation (2) becomes:
1 = C
⇒ C = 1
Substituting C = 1 in equation (2), we get:
x + y + 1 = ex
Hence, the required equation of curve passing through the origin is x + y + 1 = ex.
Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangents to the curve at that point by 5.
Let F (x, y) be the curve and let (x, y) be a point on the curve. The slope of the tangent to the curve at (x, y) is dy/dx.
According to the given information:
The curve passes through point (0, 2).
Therefore, equation (2) becomes:
0 + 2 – 4 = Ce0
⇒ – 2 = C
⇒ C = – 2
Substituting C = –2 in equation (2), we get:
x + y -4 = – 2ex
⇒y = 4 – x – 2ex
This is the required equation of the curve.
Choose the correct answer:
The integrating factor of the differential equation is:
(A) e–x
(B) e–y
(C) 1/x
(D) x
Given: Differential equation
Therefore, option (C) is correct.
Choose the correct answer:
The integrating factor of the differential equation
Given: Differential equation
Therefore, option (D) is correct.
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