NCERT Solutions Class 12 Maths Chapter 9 Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 – Free PDF Download

The NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations are provided here with the best possible explanations for every question available in the chapter. This chapter is part of the CBSE Syllabus 2023-24. Students learn about the order and degree of differential equations, the method of solving a differential equation, their properties and much more in this chapter. Solving the problems in the different exercises in this chapter using the NCERT Solutions for Class 12 Maths can help the students create a strong grasp of the concepts of Differential Equations.

Chapter 9 of NCERT Solutions for Class 12 Maths covers the concepts of Differential Equations in easy-to-understand methods, which favours the students in their board exam preparation. The answers for each question provided at BYJU’S can help the students understand how the questions have to be actually answered. Students can download these NCERT Solutions for Class 12 Maths Chapter 9 for free and practise them whenever they wish to.

NCERT Solutions Class 12 Maths Chapter 9 Differential Equations

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Access Answers to NCERT Class 12 Maths Chapter 9 – Differential Equations

Exercise 9.1 Page: 382

Determine the order and degree (if defined) of differential equations given in Exercises 1 to 10.

Solution:

The given differential equation is,

⇒ y”” + sin (y’’’) = 0

The highest order derivative present in the differential equation is y’’’’, so its order is three. Hence, the given differential equation is not a polynomial equation in its derivatives, so its degree is not defined.

2. y’ + 5y = 0

Solution:

The given differential equation is y’ + 5y = 0

The highest order derivative present in the differential equation is y’, so its order is one.

Therefore, the given differential equation is a polynomial equation in its derivatives.

So, its degree is one.

So, its degree is one.

So, its degree is not defined.

Therefore, its degree is one.

6. (y’’’)² + (y’’)³ + (y’)⁴ + y⁵ = 0

Solution:

The given differential equation is, (y’’’)² + (y’’)³ + (y’)⁴ + y⁵ = 0

The highest order derivative present in the differential equation is y’’’.

The order is three. Therefore, the given differential equation is a polynomial equation in y’’’, y’’ and y’.

Then, the power raised to y’’’ is 2.

Therefore, its degree is two.

7. y’’’ + 2y’’ + y’ = 0

Solution:

The given differential equation is, y’’’ + 2y’’ + y’ = 0

The highest order derivative present in the differential equation is y’’’.

The order is three. Therefore, the given differential equation is a polynomial equation in y’’’, y’’ and y’.

Then, the power raised to y’’’ is 1.

Therefore, its degree is one.

8. y’ + y = e^x

Solution:

The given differential equation is y’ + y = e^x

= y’ + y – e^x = 0

The highest order derivative present in the differential equation is y’.

The order is one. Therefore, the given differential equation is a polynomial equation in y’.

Then, the power raised to y’ is 1.

Therefore, its degree is one.

9. y’’’ + (y’)² + 2y = 0

Solution:

The given differential equation is, y’’’ + (y’)² + 2y = 0

The highest order derivative present in the differential equation is y’’.

The order is two. Therefore, the given differential equation is a polynomial equation in y’’ and y’.

Then, the power raised to y’’ is 1.

Therefore, its degree is one.

10. y’’’ + 2y’ + sin y = 0

Solution:-

The given differential equation is, y’’’ + 2y’ + sin y = 0

The highest order derivative present in the differential equation is y’’.

The order is two. Therefore, the given differential equation is a polynomial equation in y’’ and y’.

Then the power raised to y’’ is 1.

Therefore, its degree is one.

11. The degree of the differential equation.

(A) 3 (B) 2 (C) 1 (D) not defined

Solution:-

(D) not defined

The given differential equation is,

The highest order derivative present in the differential equation is

The order is three. Therefore, the given differential equation is not a polynomial.

Therefore, its degree is not defined.

12. The order of the differential equation

(A) 2 (B) 1 (C) 0 (D) not defined

Solution:-

(A) 2

The given differential equation is,

The highest order derivative present in the differential equation is

Therefore, its order is two.

Exercise 9.2 Page: 385

In each of the Exercises 1 to 10, verify that the given functions (explicit or implicit) are a solution of the corresponding differential equation:

1. y = ex + 1 : y″ – y′ = 0

Solution:-

From the question, it is given that y = e^x + 1

Differentiating both sides with respect to x, we get

⇒ y” = e^x

Then,

Substituting the values of y’ and y” in the given differential equations, we get,

y” – y’ = e^x – e^x = RHS.

Therefore, the given function is a solution of the given differential equation.

2. y = x² + 2x + C : y′ – 2x – 2 = 0

Solution:-

From the question, it is given that y = x² + 2x + C

Differentiating both sides with respect to x, we get

y’ = 2x + 2

Then,

Substituting the values of y’ in the given differential equations, we get

= y’ – 2x -2

= 2x + 2 – 2x – 2

= 0

= RHS

Therefore, the given function is a solution of the given differential equation.

3. y = cos x + C : y′ + sin x = 0

Solution:-

From the question, it is given that y = cos x + C

Differentiating both sides with respect to x, we get

y’ = -sinx

Then,

Substituting the values of y’ in the given differential equations, we get

= y’ + sinx

= – sinx + sinx

= 0

= RHS

Therefore, the given function is a solution of the given differential equation.

4. y = √(1 + x²): y’ = ((xy)/(1 + x²))

Solution:-

5. y = Ax : xy′ = y (x ≠ 0)

Solution:-

From the question, it is given that y = Ax

Differentiating both sides with respect to x, we get

y’ = A

Then,

Substituting the values of y’ in the given differential equations, we get

= xy’

= x × A

= Ax

= Y … [from the question]

= RHS

Therefore, the given function is a solution of the given differential equation.

6. y = x sinx: xy’ = y + x (√(x² – y²)) (x ≠ 0 and x>y or x< – y)

Solution:-

Solution:-

8. y – cos y = x : (y sin y + cos y + x) y′ = y

Solution:-

9. x + y = tan^-1y : y² y′ + y² + 1 = 0

Solution:-

Solution:-

11. The number of arbitrary constants in the general solution of a differential equation of fourth order is:

(A) 0 (B) 2 (C) 3 (D) 4

Solution:-

(D) 4

The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation.

12. The number of arbitrary constants in the particular solution of a differential equation of third order is:

(A) 3 (B) 2 (C) 1 (D) 0

Solution:-

(D) 0

The solution free from arbitrary constants, i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants, is called a particular solution of the differential equation.

Exercise 9.3 Page: 391

In each of the Exercises 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

Solution:-

2. y² = a (b² – x²)

Solution:-

3. y = ae^3x + be^-2x

Solution:-

4. y = e^2x (a + bx)

Solution:-

From the question it is given that y = e^2x (a + b x)  … [we call it as equation (i)]

Differentiating both sides with respect to x, we get,

y’ = 2e^2x(a + b x) + e^2x × b … [equation (ii)]

Then, multiply equation (i) by 2 and afterwards subtract it from equation (ii),

We have,

y’ – 2y = e^2x(2a + 2bx + b) – e^2x (2a + 2bx)

y’ – 2y = 2ae^2x + 2e^2xbx + e^2xb – 2ae^2x – 2bxe^2x

y’ – 2y = be^2x … [equation (iii)]

Now, differentiating equation (iii) both sides,

We have,

⇒ y’’ – 2y = 2be^2x … [equation (iv)]

Then,

5. y = e^x (a cos x + b sin x)

Solution:

From the question, it is given that y = e^x(a cos x + b sin x)

… [we call it as equation (i)]

Differentiating both sides with respect to x, we get,

⇒y’ = e^x(a cos x + b sin x) + e^x(-a sin x + b cos x)

⇒ y’ = e^x[(a + b)cos x – (a – b) sin x)] … [equation (ii)]

Now, differentiating equation (ii) both sides,

We have,

y” = e^x[(a + b) cos x – (a – b)sin x)] + e^x[-(a + b)sin x – (a – b) cos x)]

On simplifying, we get,

⇒ y” = e^x[2bcosx – 2asinx]

⇒ y” = 2e^x(b cos x – a sin x) … [equation (iii)]

Now, adding equations (i) and (iii), we get,

6. Form the differential equation of the family of circles touching the y-axis at the origin.

Solution:

By looking at the figure, we can say that the centre of the circle touching the y- axis at the origin lies on the x-axis.

Let us assume (p, 0) is the centre of the circle.

Hence, it touches the y-axis at the origin, and its radius is p.

Now, the equation of the circle with centre (p, 0) and radius (p) is

⇒ (x – p)² + y² = p²

⇒ x² + p² – 2xp + y² = p²

Transposing p² and – 2xp to RHS then it becomes – p² and 2xp

⇒ x² + y² = p² – p² + 2px

⇒ x² + y² = 2px … [equation (i)]

Now, differentiating equation (i) both sides,

We have,

⇒ 2x + 2yy’ = 2p

⇒ x + yy’ = p

Now, on substituting the value of ‘p’ in the equation, we get,

⇒ x² + y² = 2(x + yy’)x

⇒ 2xyy’ + x² = y²

Hence, 2xyy’ + x² = y² is the required differential equation.

7. Form the differential equation of the family of parabolas having a vertex at origin and axis along positive y-axis.

Solution:

The parabola having the vertex at the origin and the axis along the positive y-axis is

x² = 4ay … [equation (i)

8. Form the differential equation of the family of ellipses having foci on the y-axis and centre at the origin.

Solution:

On simplifying,

⇒ -x (y’)² – xyy” + yy’ = 0

⇒ xyy” + x (y’)² – yy’ = 0

Hence, xyy” + x (y’)² – yy’ = 0 is the required differential equation.

9. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.

Solution:

⇒ x (y’)² + xyy” – yy’ = 0

⇒ xyy” + x(y’)² – yy’ = 0

Hence, xyy” + x (y’)² – yy’ = 0is the required differential equation.

10. Form the differential equation of the family of circles having a centre on the y-axis and a radius of 3 units.

Solution:

Let us assume the centre of the circle on the y-axis be (0, a).

We know that the differential equation of the family of circles with centre at (0, a) and radius 3 is: x² + (y- a)² = 3²

⇒ x² + (y- a)² = 9 … [equation (i)]

Now, differentiating equation (i) both sides with respect to x,

⇒ 2x + 2(y – a) × y’ = 0 … [dividing both side by 2]

⇒ x + (y – a) × y’ = 0

Transposing x to the RHS, it becomes – x.

⇒ (y – a) × y’ = x

11. Which of the following differential equations has y = c₁ e^x + c² e^-x as the general solution?

Solution:

Explanation:

12. Which of the following differential equations has y = x as one of its particular solution?

Solution:

Explanation:

Exercise 9.4 Page No: 395

For each of the differential equations in Exercises 1 to 10, find the general solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

For each of the differential equations in Exercises 11 to 14, find a particular solution

Satisfying the given condition:

Solution:

Solution:

Solution:

Solution:

⇒ c = 1

Putting the value of c in 1

⇒ y = sec x

15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = e^x sin x

Solution:

Find the solution curve passing through the point (1, –1).

Solution:

17. Find the equation of a 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.

Solution:

18. 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).

Solution:

19. The volume of a 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

the balloon after t seconds.

Solution:

20. In a bank, the principal increases continuously at the rate of r% per year. Find the

value of r if Rs 100 double itself in 10 years (log_e 2 = 0.6931).

Solution:

21. In a bank, the principal increases continuously at the rate of 5% per year. An amount

of Rs 1000 is deposited with this bank. How much will it be worth after 10 years?

(e^0.5 = 1.648).

Solution:

22. In a culture, the bacteria count is 1,00,000. The number 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?

Solution:

Solution:

(A) e^x + e^-y = C

Explanation:

Exercise 9.5 Page No: 406

In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them.

1. (x² + x y) dy = (x² + y²) dx

Solution:

Solution:

3. (x – y) dy – (x + y) dx = 0

Solution:

4. (x² – y²)dx + 2xy dy = 0

Solution:

On simplification

x² + y² = Cx

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

For each of the differential equations in Exercises from 11 to 15, find the particular

solution satisfying the given condition:

11. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1

Solution:

12. x²dy + (x y + y²)dx = 0; y = 1 when x = 1

Solution:

Solution:

Solution:

Solution:

The required solution of the differential equation.

16. A homogeneous differential equation of the from can be solved by making the substitution.

(A) y = v x (B) v = y x (C) x = v y (D) x = v

Solution:

(C) x = v y

Explanation:

17. Which of the following is a homogeneous differential equation?

A. (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
B. (x y) dx – (x³ + y³) dy = 0
C. (x³ + 2y²) dx + 2xy dy = 0
D. y²dx + (x² – x y – y²) dy = 0

Solution:

D. y²dx + (x² – x y – y²) dy = 0

Explanation:

Exercise 9.6 Page No: 413

For each of the differential equations given in question, find the general solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

8. (1 + x²) dy + 2xy dx = cot x dx (x ≠ 0)

Solution:

Solution:

Solution:

11. y dx + (x – y²)dy = 0

Solution:

Solution:

⇒ x = 3y² + Cy

Therefore, the required general solution of the given differential equation is x = 3y² + Cy.

For each of the differential equations given in Exercises 13 to 15, find a particular

solution satisfying the given condition:

Solution:

Solution:

Solution:

16. Find the equation of a 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 the coordinates of the point.

Solution:

Now, it is given that the curve passes through the origin.

Thus, equation 2 becomes

1 = C

⇒ C = 1

Substituting C = 1 in equation 2, we get,

x + y + 1 = e^x

Therefore, the required general solution of the given differential equation is

x + y + 1 = e^x

17. Find the equation of a 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 tangent to the curve at that point by 5.

Solution:

Thus, equation (2) becomes:

0 + 2 – 4 = C e⁰

⇒ – 2 = C

⇒ C = -2

Substituting C = -2 in equation (2), we get,

x + y – 4 =-2e^x

⇒ y = 4 – x – 2e^x

Therefore, the required general solution of the given differential equation is

y = 4 – x – 2e^x

18. The integrating factor of the differential equation is

A. e^–x, B. e^–y, C. 1/x, D. x

Solution:

C. 1/x

Explanation:

19. The integrating factor of the differential equation

Solution:

Explanation:

Miscellaneous Exercise Page No: 419

1. For each of the differential equations given below, indicate its order and degree (if defined)

(i) (d²y/dx²) + 5x(dy/dx)² – 6y = log x

(ii) (dy/dx)³ – 4 (dy/dx)² + 7y = sin x

(iii) (d⁴y/dx⁴) – sin (d³y/dx³) = 0

Solution:

(i) (d²y/dx²) + 5x(dy/dx)² – 6y = log x

Rearranging the given equation, we get

(d²y/dx²) + 5x(dy/dx)² – 6y – log x = 0

Hence, the highest order derivative present in the given differential equation is d²y/dx².

Therefore, the order is 2.

Also, the highest power raised to d²y/dx² is 1.

Hence, the degree is 1.

(ii) (dy/dx)³ – 4 (dy/dx)² + 7y = sin x

Rearranging the given equation, we get

(dy/dx)³ – 4 (dy/dx)² + 7y – sin x = 0

Hence, the highest order derivative present in the given differential equation is dy/dx.

Therefore, the order is 1.

And the highest power raised to dy/dx is 3.

Hence, the degree is 3.

(iii) (d⁴y/dx⁴) – sin (d³y/dx³) = 0

The highest order derivative present in the given differential equation is d⁴y/dx⁴. Hence, the order is 4.

Since the given differential equation is not a polynomial equation, the degree of the equation is not defined.

2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution to the corresponding differential equation.

(i) y = ae^x + be^-x + x²: x (d²y/dx²) + 2(dy/dx) – xy + x² – 2 = 0

(ii) y = e^x (a cos x + b sin x): (d²y/dx²) – 2(dy/dx) + 2y = 0

(iii) y = x sin 3x: (d²y/dx²) + 9y – 6 cos 3x = 0

(iv) x² = 2y² log y: (x² + y²)(dy/dx) – xy = 0

Solution:

(i) y = ae^x + be^-x + x² : x (d²y/dx²) + 2(dy/dx) – xy + x² – 2 = 0

Given: y = ae^x + be^-x + x²

Differentiate the function with respect to x, and we get

dy/dx = ae^x – be^-x + 2x … (1)

Now, again differentiate with respect to x, and we get

d²y/dx² = ae^x + be^-x + 2 …(2)

To check whether the given function is the solution of the given differential equation, substitute (1) and (2) in the given differential equation.

L.H.S of the given differential equation

= x (d²y/dx²) + 2(dy/dx) – xy + x²

Now, substituting the values, we get

= x (ae^x + be^-x + 2) + 2 (ae^x – be^-x + 2x) – x (ae^x + be^-x + x²) + x² – 2

= (xae^x + x be^-x + 2x) + (2ae^x – 2be^-x + 4x) – (xae^x + xbe^-x + x³) + x² – 2

On simplifying the above equation, we get

= 2ae^x – 2be^-x – x³ + x² + 6x – 2 ≠ 0

Hence, L.H.S ≠ R.H.S.

Therefore, the given function is not a solution to the corresponding differential equation.

(ii) y = e^x (a cos x + b sin x): (d²y/dx²) – 2(dy/dx) + 2y = 0

Given: y = e^x (a cos x + b sin x)

The given function can be written as follows:

y = e^x a cos x + e^x b sin x

Differentiating the function on both sides, we get

dy/dx = (a + b)e^x cos x + (b – a)e^x sin x …(1)

Again, differentiate the above equation on both sides with respect to x, and we get

d²y/dx² = [(a + b) (d/dx) (e^x cos x ] + [(b-a) (d/dx) (e^x sin x]

d²y/dx² = [(a+ b) (e^x cos x – e^x sin x) ] + [(b- a)(e^x sin x + e^x cos x)]

d²y/dx² = e^x [(a+ b) (cos x – sin x) + (b- a)(sin x + cos x)]

On simplifying the above equation, we get

d²y/dx² = 2e^x (b cos x – a sin x) …(2)

Now, substitute (1) and (2) in the given differential equation.

L.H.S = (d²y/dx²) – 2(dy/dx) + 2y

= [2e^x (b cos x – a sin x)] – 2[(a + b)e^x cos x + (b – a)e^x sin x] + 2 e^x (a cos x + b sin x)

= e^x [(2b cos x – 2a sin x)- (2a cos x + 2b cos x) – (2b sin x – 2a sin x) + (2a cos x + 2b sin x)]

= e^x [2b cos x – 2a sin x- 2a cos x – 2b cos x – 2b sin x + 2a sin x + 2a cos x + 2b sin x]

= e^x [0]

= 0 = R.H.S

As L.H.S = R.H.S, the given function is the solution of the corresponding differential equation.

(iii) y = x sin 3x: (d²y/dx²) + 9y – 6 cos 3x = 0

Given: y = x sin 3x

Now, differentiating the given function with respect to x, and we get

dy/dx = sin 3x + x. cos 3x. 3

dy/dx = sin 3x + 3x cos 3x …(1)

Again differentiate (1) with respect to x, we get

d²y/dx² = (d/dx) (sin 3x) + 3 (d/dx) (x cos 3x)

d²y/dx² = 3 cos 3x + 3 [cos 3x + x (- sin 3x). 3]

On simplifying the above equation, we get

d²y/dx² = 6 cos 3x – 9x sin 3x …(2)

Now, substitute (1) and (2) in the given differential equation, and we get the following:

L.H.S = (d²y/dx²) + 9y – 6 cos 3x

= (6 cos 3x – 9x sin 3x) + 9(x sin 3x) – 6 cos 3x

= 6 cos 3x – 9x sin 3x + 9x sin 3x – 6 cos 3x

= 0 = R.H.S

As L.H.S = R.H.S, the given function is the solution of the corresponding differential equation.

(iv) x² = 2y² log y: (x² + y²)(dy/dx) – xy = 0

Given: x² = 2y² log y

Now, differentiate the function with respect to x, and we get

2x = 2 (d/dx) (y² log y)

On simplifying the above equation, we get

x = (d/dx) (y² log y)

x = [2y log y .(dy/dx) + y². (1/y). (dy/dx)]

x = (dy/dx)[2y log y + y]

Hence, we get

dy/dx = x / [y(1 + 2 log y)] …(1)

Now, substitute (1) in the given differential equation.

L.H.S = (x² + y²)(dy/dx) – xy

= [ 2y² log y + y²] . [ x / [y(1 + 2 log y)] ] – xy

= [y²(2 log y + 1)] . [ x / [y(1 + 2 log y)] ] – xy

= xy – xy

= 0 = R.H.S

As L.H.S = R.H.S, the given function is the solution of the corresponding differential equation.

3. Form the differential equation representing the family of curves given by (x- a)² + 2y² = a², where a is an arbitrary constant.

Solution:

Given equation: (x- a)² + 2y² = a²

The given equation can be written as:

⇒ x² + a² – 2ax + 2y² = a²

On rearranging the above equation, we get

⇒ 2y² = 2ax – x² …(1)

Now, differentiate equation (1) with respect to x,

⇒ 2 . 2y (dy/dx) = 2a – 2x

⇒ 2y(dy/dx) = (2a – 2x) /2

⇒ dy/dx = (a-x)/2y

⇒ dy/dx = (2ax – 2x²) / 4xy … (2)

From equation (1), we get

2ax = 2y² + x²

Substitute the value in equation (2), and we get

dy/dx = [2y² + x² – 2x²]/4xy

dy/dx = (2y² – x²) / 4xy

Therefore, the differential equation representing the family of curves given by (x- a)² + 2y² = a² is (2y² – x²) / 4xy.

4. Prove that x² – y² = C (x² + y²)² is the general solution of the differential equation (x³ – 3xy² ) dx = (y³ – 3x²y) dy, where C is a parameter.

Solution:

Given differential equation: (x³ – 3xy² ) dx = (y³ – 3x²y) dy

The equation can be rewritten as:

dy/dx = (x³ – 3xy²) / (y³ – 3x²y) …(1)

The above equation is a homogeneous equation.

To simplify the equation, let us assume y = vx.

⇒ (d/dx) y = (d/dx) (vx)

⇒ dy/dx = v + x(dv/dx) …(2)

Using equations (1) and (2), we get

⇒v + x(dv/dx) = (x³ – 3xy²) / (y³ – 3x²y)

⇒v + x(dv/dx) = (x³ – 3x(vx)²) / ((vx)³ – 3x²(vx))

⇒v + x(dv/dx) = [(1-3v²)/(v³-3v)]

⇒x(dv/dx) = [(1-3v²)/(v³-3v)] – v

On simplifying the above equation, we get

⇒x(dv/dx) = (1-v⁴)/(v³ – 3v)

Rearranging the above equation,

⇒ [(v³ – 3v)/(1-v⁴)]dv = (dx/x).

Integrate both sides, and we get

Where I₁ = ∫ [(v³dv)/(1-v⁴)] and I₂ = ∫ [(vdv)/(1-v⁴)]

Now, let us assume 1- v⁴ = t

Hence, we get

⇒ (d/dv) (1-v⁴) = (dt/dv)

⇒ -4v³ = dt/dv

⇒v³ dv = -dt/4

Now,

I₁ = ∫ -(dt/4t) = (-¼) log t = -(¼) log(1-v⁴) …(5)

Similarly,

I₂ = ∫ [(vdv)/(1-v⁴)] = ∫ [(vdv)/(1-(v²)²)]

Assume that v² = p

Hence, we get

⇒(d/dv)v² = dp/dv

⇒ 2v = dp/dv

⇒vdv = dp/2

Now,

I₂ = (½) ∫ [dp/(1-p²)]

Using equations (4), (5) and (6), we get

Now, using equations (2) and (7)

Now, replace v with y/x in the above equation and simplify it.

Hence, we get

⇒ (x² – y²)² = C’⁴ (x² + y² )⁴

Now, take the square root on both sides of the above equation, and we get

⇒ (x² – y²) = C’² (x² + y² )²

⇒ (x² – y²) = C’ (x² + y² )² , where C = C’²

Hence, x² – y² = C (x² + y²)² is the general solution of the differential equation (x³ – 3xy² ) dx = (y³ – 3x²y) dy, it is proved.

5. Form the differential equation of the family of circles in the first quadrant, which touches the coordinate axes.

Solution:

We know that the equation of a circle with centre (a, a) and radius “a” in the first quadrant, which touches the coordinate axes, is:

(x-a)² + (y-a)² = a² …(1)

Differentiating the above equation of both sides with respect to x, we get

⇒ 2(x-a) + 2(y-a)(dy/dx) = 0

Now, the equation can be written as

⇒ (x-a) + (y-a)y’ = 0

⇒ (x-a) + yy’ – ay’ = 0

⇒ x + yy’ -a(1+ y’) = 0

⇒ x + yy’ = a (1 + y’)

Rearranging the above equation, we get

⇒ a = (x + yy’) / (1+y’)

Now, substitute the value of “a” in equation (1), and we get

On simplifying the above equation, we get

⇒ (x-y)².y’² + (y-x)² = (x+ yy’)²

Therefore, the differential equation of the family of circles in the first quadrant, which touches the coordinate axes, is (x-y)².y’² + (y-x)² = (x+ yy’)².

6. Find the general solution of the differential equation (dy/dx) + √[(1-y²)/(1-x²)] = 0.

Solution:

Given: (dy/dx) + √[(1-y²)/(1-x²)] = 0.

The given differential equation can be written as dy/dx = -√[(1-y²)/(1-x²)]

⇒ dy /√(1-y²) = -dx/√(1-x²)

Now, integrate both sides, and we get

⇒ sin^-1 y = – sin^-1 x + C

Now, rearrange the equation, and we get

⇒ sin^-1 x + sin^-1 y = C.

7. Show that the general solution of the differential equation (dy/dx) + [(y² + y + 1) / (x² + x + 1)] = 0 is given by (x + y + 1) = A (1 – x – y – 2xy), where A is the parameter.

Solution:

Given differential equation: (dy/dx) + [(y² + y + 1) / (x² + x + 1)] = 0

Rearranging the given equation, we get

dy/dx = – [(y² + y + 1) / (x² + x + 1)]

⇒ dy/(y² + y + 1) = – dx/ (x² + x + 1)

Now, integrate both sides, and we get

⇒ ∫[ dy/(y² + y + 1)] = -∫ [dx/ (x² + x + 1)]

On simplifying the above equation, we get

⇒ 2√3 (x + y + 1) = C₁ (3 – 4xy + 2x + 2y + 1)

⇒ 2√3 (x + y + 1) = C₁ (2 – 4xy – 2x – 2y)

⇒ 2√3 (x + y + 1) = 2C₁ (1 – 2xy – x – y)

⇒√3 (x + y + 1) = C₁ (1 – 2xy – x – y)

⇒√3 (x + y + 1) = C₁ (1 – x – y – 2xy)

⇒ (x + y + 1) = (C₁/√3)(1 – x – y – 2xy)

⇒ (x + y + 1) = A (1 – x – y – 2xy), where A = (C₁/√3)

Hence, proved.

8. Find the equation of the curve passing through the point (0, π/4) whose differential equation is sin x cos y dx + cos x sin y dy = 0.

Solution:

The given differential equation is sin x cos y dx + cos x sin y dy = 0.

It can also be written as:

⇒ (sin x cos y dx + cos x sin y dy) / cos x cos y = 0.

We know that sin x / cos = tan x,

And simplify the above equation

⇒ tan x dx + tan y dy = 0

⇒ log (sec x ) + log (sec y) = log C

⇒ log (sec x. sec y) = log C

On simplification, we get

sec x sec y = C

It is given that the curve passes through the point (0, π/4).

⇒ 1 × √2 = C

⇒ C = √2

Hence,

sec x × sec y = √2

⇒ sec x × (1/cos y) = √2

⇒ cos y = sec x / √2

Hence, the equation of the curve passing through the point (0, π/4) whose differential

equation is sin x cos y dx + cos x sin y dy = 0 is cos y = sec x /√2.

9. Find the particular solution of the differential equation (1 + e^2x) dy + (1 + y²)e^x dx = 0, given that y = 1 when x = 0.

Solution:

Given differential equation: (1 + e^2x) dy + (1 + y²)e^x dx = 0

Rearranging the equation, we get

⇒ [dy/(1+y²)] + [(e^x dx)/(1 + e^2x)] = 0

Integrating both sides of the equation, we get

tan^-1 y + ∫ [(e^x dx) / (1+e^2x)] = C …(1)

Let e^x = t, and hence, e^2x = t²

(d/dx) (e^x) = (dt/dx)

⇒ e^x = dt/dx

⇒ e^x dx = dt

Substituting the value in equation (1), we get

tan^-1 y + ∫ [(dt) / (1+t²)] = C

⇒ tan^-1 y + tan^-1t = C

⇒ tan^-1 y + tan^-1 (e^x) = C

If x = 0 and y = 1, we get

⇒ tan^-1 1 + tan^-1 (e⁰) = C

⇒ tan^-1 1 + tan^-1 1 = C

⇒ (π/4) + (π/4) = C

⇒ C = π/2

Hence, tan^-1 y + tan^-1 (e^x) = π/2, which is the particular solution of the given differential equation.

10. Solve the differential equation y e^x/y dx = (xe^x/y + y²)dy ( y ≠ 0).

Solution:

Given: y e^x/y dx = (xe^x/y + y²)dy

Rearranging the given equation, we get

y e^x/y (dx/dx) = xe^x/y + y²

⇒e^x/y [y(dx/dy) – x ] = y²

⇒ [e^x/y [y(dx/dy) – x ]]/y² = 1 ….(1)

Assume that e^x/y = z

Differentiate with respect to y, and we get

(d/dy)(e^x/y) = dz/dy

⇒e^x/y [(y(dx/dy) – x)/y²] = dz/dy …(2)

Comparing equations (1) and (2), we get

⇒dz/dy = 1

⇒ dz = dy

Now, integrating both sides, we get

⇒ z = y + C

⇒ e^x/y = y + C, which is the solution of the given differential equation.

11. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy,

given that y = –1, when x = 0. (Hint: put x – y = t)

Solution:

Given differential equation: (x – y) (dx + dy) = dx – dy

On simplifying the above equation, we get

⇒ (dy/dx) = (1-x+y)/ (x -y +1)

⇒ (dy/dx) = [1 – (x-y)] / [1 + (x-y)] …(1)

Given: x – y = t …(2)

(d/dx) (x – y) = dt/dx

⇒ 1 – (dy/dx) = dt/dx

⇒ 1 – (dt/dx) = dy/dx …(3)

Using the equations (1), (2) and (3), we get

⇒ 1 – (dt/dx) =(1-t)/(1+t)

⇒ dt/dx = 1 – [(1-t)/(1+t)]

On simplification, we get

⇒dt/dx = 2t / (1+t)

⇒ [(1+t)/t]dt = 2 dx

⇒ [ 1 + (1/t)]dt = 2dx

Now, integrating both sides, we get

⇒ t + log |t| = 2x + C

⇒ (x-y) + log |x-y| = 2x + C

⇒ log |x-y| = x + y + C

When x = 0 and y = -1, we get

⇒ log 1 = 0 – 1 + C

⇒ C = 1

Hence, log |x-y| = x + y + 1.

Therefore, log |x-y| = x + y +1 is a particular solution of the differential equation (x – y) (dx + dy) = dx – dy.

12. Solve the differential equation [ (e^-2√x /√x ) – (y/√x) ](dx/dy) = 1 (x ≠ 0).

Solution:

Given: [ (e^-2√x /√x ) – (y/√x) ](dx/dy) = 1

Rearranging the given equation, we get

⇒ dy/dx = (e^-2√x /√x ) – (y/√x)

⇒ (dy/dx) + (y/√x) = (e^-2√x /√x )

The above equation is a linear equation of the form (dy/dx) + Py = Q

Where, P = 1/√x and Q = e^-2√x/√x

Now, I.F = e^{∫ P dx} = e^{∫1/√x dx} = e^2√x

Hence, the general solution of the given differential equation is:

y. (I.F) = ∫ (Q × I.F) dx + C

Now, substituting the values, we get

⇒ ye^2√x = ∫ [ (e^-2√x/√x) ×e^2√x ]dx + C

⇒ye^2√x = ∫ (1/√x) dx + C

⇒ye^2√x = 2√x +C, which is the solution of the given differential equation.

13. Find a particular solution of the differential equation (dy/dx) + y cot x = 4x cosec x, (x≠ 0), given that y = 0 when x = π/2.

Solution:

Given: (dy/dx) + y cot x = 4x cosec x

The given equation is a linear differential equation of the form (dy/dx) + Py = Q

Where

P = cot x and Q = 4x cosec x

Now, I.F = e^{∫ P dx} = e^{∫ cot x dx} = e^{log |sin x|} = sin x

Hence, the general solution of the given differential equation is:

y. (I.F) = ∫ (Q × I.F) dx + C

⇒ y sin x = ∫ (4x cosec x × sin x) dx + C

⇒ y sin x = 4 ∫x dx + C

⇒ y sin x = 4 (x²/2) + C

⇒ y sin x =2x² + C

when x = π/2 and y = 0,

Substituting the values in the above equation, we get

⇒ 0 =2(π/2)² + C

⇒ 0 = 2(π²/4) + C

⇒ 0 = π²/2 + C

⇒ C = – π²/2

Hence, y sin x =2x² – (π²/2)

Therefore, the particular solution of the differential equation (dy/dx) + y cot x = 4x cosec x is y sin x =2x² – (π²/2).

14. Find a particular solution of the differential equation (x + 1) (dy/dx) = 2e^-y – 1. Given that y = 0 when x = 0.

Solution:

Given differential equation: (x + 1) (dy/dx) = 2e^-y – 1

Rearranging the equation, we get

⇒ dy/2e^-y – 1 = dx/(x + 1)

⇒ (e^y dy)/(2-e^y) = dx/(x+1)

Integrate on both sides, we get

∫ [(e^y dy)/(2-e^y)] = log |x+1| + log C …(1)

Assume that 2-e^y = t

⇒ (d/dy) (2-e^y) = dt/dy

⇒ -e^y = dt/dy

⇒ -e^y dy = dt

Substituting the value in equation (1), we get

⇒∫ [(dt)/(t)] = log |x+1| + log C

⇒ – log |t| = log |C (x+1)|

⇒ – log |2-e^y| = log |C (x+1)|

⇒ 1/ (2-e^y) = C (x+1)

⇒2-e^y = 1/[C(x+1)]

When x = 0 and y = 0, we get

⇒ 2 – 1 = 1/C

We get C = 1.

Therefore, 2-e^y = 1/[1(x+1)]

⇒2-e^y = 1/(x+1)

⇒ e^y = 2 – [1/(x+1)]

On simplification, we get

e^y = (2x +1) / (x+1)

⇒ y = log |(2x +1) / (x+1)|, where x ≠ -1.

Therefore, the particular solution of the differential equation (x + 1) (dy/dx) = 2e^-y -1 is y = log |(2x +1) / (x+1)|, where x ≠ -1.

15. The population of a village increases continuously at a rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Solution:

Let us assume that the population at any instant (t) be y.

Also, given that the rate of Increase in population is proportional to the number of inhabitants at any instant.

(dy/dx) ∝ y

⇒ (dy/dx) = ky

⇒ dy/y = kdt (Where k is a constant)

Now, integrating both sides of the above equation, we get

log y = kt + C … (1)

In 1999, t = 0 and y = 20000, we get log 20000 = C … (2)

In 2004, t = 5 and y = 25000, we get 1og 25000=k.5 + C

⇒ log 25000 = 5k + Iog 20000

⇒ 5k = log(25000/20000) = log(5/4)

⇒ k = (⅕) log(5/4) …(3)

In 2009, t = 10 years.

Substitute the values of k, t and C in (1), and we get

log y = 10 [ (⅕) log(5/4)] + log 20000

On simplification, we get

y = (20000)(5/4)(5/4)

y = 31250

Therefore, the population of the village in 2009 was 31250.

16. The general solution of the differential equation [(y dx – x dy)/y] = 0 is:

1. xy = C (B) x = Cy² (C) y = Cx (D) y = Cx²

Solution:

The differential equation is [(y dx – x dy)/y] = 0.

The given equation can be written as:

(ydx / y ) – (xdy/y) = 0

Thus, we get

dx = xdy /y

dx/x = dy/y

(1/x)dx – (1/y)dy = 0

Now, integrating the above equation on both sides, we get

log |x| – log |y| = log k

⇒ log |x/y| = log k

⇒ x/y = k

⇒ y = (1/k) x

⇒ y = Cx [Where C = 1/k]

Hence, the correct answer is option (C) y = Cx.

17. The general solution of a differential equation of the type (dx/dy) + P₁x = Q₁ is:

1. \(\begin{array}{l}ye^{\int P_{1}dy} = \int (Q_{1}e^{\int P_{1}dy})dy + C\end{array} \)
2. \(\begin{array}{l}ye^{\int P_{1}dx} = \int (Q_{1}e^{\int P_{1}dx})dx + C\end{array} \)
3. \(\begin{array}{l}xe^{\int P_{1}dy} = \int (Q_{1}e^{\int P_{1}dy})dy + C\end{array} \)
4. \(\begin{array}{l}xe^{\int P_{1}dx} = \int (Q_{1}e^{\int P_{1}dx})dx + C\end{array} \)

Solution:

As we know, the integrating factor of the differential equation (dx/dy) + P₁x = Q₁ is e^{∫ P1 dy}.

Hence,

⇒ x. (I.F) = ∫ (Q₁ × I.F) dy + C

⇒ x. e^{∫ P1 dy} = ∫ (Q₁ × e^{∫ P1 dy}) dy + C

Hence, the correct answer is option (C).

18. The general solution of the differential equation e^x dy + (y e^x + 2x) dx = 0 is:

1. x e^y + x² = C
2. x e^y + y² = C
3. y e^x + x² = C
4. y e^y + x² = C

Solution:

The correct answer is option (C) y e^x + x² = C

Explanation:

The given differential equation is e^x dy + (y e^x + 2x) dx = 0

⇒ e^x (dy/dx) + ye^x + 2x = 0

Hence, we get

⇒ (dy/dx) + y = -2xe^-x.

The above equation is the linear differential equation of the form (dy/dx) + Py = Q, where P = 1 and Q = -2xe^-x.

Now,

I.F = e^{∫ P dx} = e^{∫ dx} = e^x.

⇒ y. (I.F) = ∫ (Q × I.F) dx + C

⇒ y. e^x = ∫ (-2xe^-x × e^x) dx + C

⇒ ye^x = ∫-2x dx + C

⇒ ye^x = -x² + C

On rearranging the above equation, we get

⇒ ye^x + x² = C

Hence, option (C) is the general solution of the given differential equation.

NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations

The major concepts of Maths covered in Chapter 9 – Differential Equations of NCERT Solutions for Class 12 include:

9.1 Introduction

9.2 Basic Concepts

9.2.1 Order of a Differential Equation

9.2.2 Degree of a Differential Equation

9.3 General and Particular Solutions of a Differential Equation

9.4 Formation of a Differential Equation Whose General Solution is Given

9.4.1 Procedure to Form a Differential Equation that Will Represent a Given Family of Curves

9.5 Methods of Solving First Order, First Degree Differential Equations

9.5.1 Differential Equations with Variables Separable

9.5.2 Homogeneous Differential Equations

9.5.3 Linear Differential Equations

Exercise 9.1 Solutions 12 Questions

Exercise 9.2 Solutions 12 Questions

Exercise 9.3 Solutions 12 Questions

Exercise 9.4 Solutions 23 Questions

Exercise 9.5 Solutions 17 Questions

Exercise 9.6 Solutions 19 Questions

Miscellaneous Exercise on Chapter 9 Solutions 18 Questions

NCERT Solutions for Class 12 Maths Chapter 9- Differential Equations

The chapter Differential Equations belongs to the unit Calculus, which adds up to 35 marks of the total marks. There are 6 exercises along with a miscellaneous exercise in this chapter to help students understand the concepts of Differential Equations clearly. Chapter 9 of NCERT Solutions for Class 12 Maths discusses the following:

1. An equation involving derivatives of the dependent variable with respect to the independent variable (variables) is known as a differential equation.
2. The order of a differential equation is the order of the highest order derivative occurring in the differential equation.
3. The degree of a differential equation is defined if it is a polynomial equation in its derivatives.
4. The degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it.
5. A function which satisfies the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called a particular solution.
6. To form a differential equation from a given function, we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants.
7. The variable separable method is used to solve such an equation in which variables can be separated completely, i.e. terms containing y should remain with dy, and terms containing x should remain with dx.

Key Features of NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations

Learning the chapter Differential Equations using the NCERT Solutions enables the students to understand the following:

Definition, order and degree, general and particular solutions of a differential equation, formation of differential equation whose general solution, solutions of differential equations by the method of separation of variables, solutions of homogeneous differential equations of the first order and first degree, and solutions of linear differential equation of the type:

Disclaimer – 

Dropped topics – 9.4 Formation of Differential Equations Whose General Solution is Given, Page 415-416 Example 25, Ques. 3, 5 and 15 (Miscellaneous Exercise), Point Six of the Summary