NCERT Class 12 Mathematics Chapter 9 Differential Equations

NCERT Class 12 Mathematics Chapter 9 Differential Equations Solutions, NCERT Solutions For Class 12 Maths, NCERT Class 12 Mathematics Chapter 9 Differential Equations Notes to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 12 Mathematics Chapter 9 Differential Equations Question Answer and select needs one.

NCERT Class 12 Mathematics Chapter 9 Differential Equations

Join Telegram channel

Follow us:

Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 12 Mathematics Chapter 9 Differential Equations Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 12 Mathematics Chapter 9 Differential Equations Solutions for All Subject, You can practice these here.

Differential Equations

Chapter – 9

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

Ans: 

Highest order derivative in the differential equation is . Its order is four.

Differential equation is not a polynomial equation in its derivatives. Its degree is not defined.

2. y’ + 5y = 0

Ans: y’ + 5y = 0 

Highest order derivative in the differential equation is y’. Its order in one.

It is a polynomial equation in y’ Highest power y’ is 1. Its degree is one.

Ans: 

Highest order derivative in the given differential equation is d²s/dt² Its order is two. 

It is a polynomial equation in d²s/dt² and ds/dt.

The power d²s/dt² is 1. Degree is one.

Ans: 

Highest order derivative in the given differential equation is d² y/dx²  Order is 2.

Given differential equation is not a polynomial equation in its derivatives. Degree is not defined.

Ans: 

Highest order derivative in the given differential equation is d² y/dx² Its order is two. 

It is a polynomial equation in d² y/dx²  and the power is 1. Its degree is 1.

Ans: 

Ans:

Ans: 

Highest order derivative present in the differential equation is y’. Its order is one. 

Given differential equation is a polynomial equation in y’ and the highest power is one. Its degree is one.

Ans: 

Ans: 

11. The degree of the differential equation

(A) 3 

(B) 2

(C) 1 

(D) not defined.

Ans: 

Differential equation is not a polynomial equation in its derivatives. Its degree is not defined. Thus, the Correct option is D.

12. The order of the differential equation

(A) 2

(B) 1

(C) 0 

(D) not defined. 

Ans: 

Highest order derivative present in the given differential equation is d² y/dx² Its order is two. Thus, the correct option is A.

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

1. y = eˣ + 1

Ans: 

From (1) and (2)

= 0

Thus, the given function is the solution of corresponding differential equation.

2. y = x² + 2x + C

Ans: y = x² + 2x + c

y’ = 2x + 2

Therefore,

y’ – 2x – 2 

= 2x + 2 – 2x – 2 

= 0

Thus, the given function is the solution of the differential equation.

3. y = cos x + C :

Ans: y = cos x + C

Therefore, 

Thus, the given function is the solution of the differential equation.

Ans: 

Thus, the given function is the solution of the differential equation.

5. y = Ax

Ans: y = Ax

= Ax

= y

Thus, the given function is the solution of the differential equation.

6. y = x sin x

Ans: y = x sin x

= sin x + x cos x

Therefore,

Thus, the given function is the solution of the differential equation.

7. xy = log y + C

Ans: 

Thus, the given function is the solution of the differential equation.

8. y – cos y = x

Ans: y – cos y = x

Therefore, 

Thus, the given function is the solution of the differential equation.

9. x + y = tan⁻¹ y

Ans: x + y = tan⁻¹ y

Therefore, 

= – 1 – y² + y² + 1 

= 0

Thus, the given function is the solution of the differential equation.

10.

Ans: 

Therefore,

= x – x 

= 0

Thus, the given function is the solution of the differential equation.

11. 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 

Ans: We know that the number of constants in the general solution of a differential equation of order n is equal to its order.

The number of constants in general equation of fourth order differential equation is 4. 

Thus, the correct option is D.

12. 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

Ans: In a particular solution of a differential equation, there are no arbitrary constants. 

Thus, the correct option is D.

Ans: 

Integrating both sides, we get:

Ans: 

Integrating both sides, we get:

Ans: 

Integrating both sides, we get:

⇒ – log(y – 1) = x + log C 

⇒ – log C – log (y – 1) = x

⇒ – [log C + log (y – 1)] = x 

⇒ log C (y – 1) = – x

⇒ C(y – 1)= e⁻ˣ

⇒ y = 1 + 1/C e⁻ˣ

⇒ y = 1 + Ae⁻ˣ  (where A = 1/C)

4. sec² x tan y dx + sec² y tan x dy = 0

Ans: sec² x tan y dx + sec² y tan x dy = 0

Dividing both sides by tan x tan y

Integrating both sides, we get:

Now,

= log t 

= log(tan x) ………..(2)

Similarly, 

Using (1), (2) and (3) 

⇒ log(tan x) = – log(tan y) + log C 

⇒ log(tan x) = log (C/tan y)

⇒ tan x = C/tan y

⇒ tan x tan y = C

5. (eˣ + e⁻ˣ) dy – (eˣ – e⁻ˣ) dx = 0

Ans: (eˣ + e⁻ˣ)dy – (e⁻ˣ – e⁻ˣ)dx = 0

⇒ (eˣ + e⁻ˣ) dy = (eˣ – e⁻ˣ)dx

Integrating both sides, we get:

Let (eˣ + e⁻ˣ) = t 

⇒ d/dx(eˣ + e⁻ˣ) = dt/dx 

⇒ (eˣ – e⁻ˣ)dx = dt 

Putting these values in equation (1), we get:

Ans: 

Integrating both sides, we get:

7. y log y dx – x dy = 0

Ans: y log y dx – x dy = 0

⇒ y log y dx = x dy

Integrating both sides, we get:

Let log y = t

Putting these values in equation (1), we get:

⇒ log t = log x + log C 

⇒ log (log y) = log Cx 

⇒ log y = eᶜˣ

Ans: 

Integrating both sides, we get:

Ans: 

⇒ dy = sin⁻¹ xdx

 Integrating both sides, we get:

 ∫ dy =  ∫(sin⁻¹ xdx 

⇒ y = ( sin⁻¹x.1)dx

Let 1 – x² – t

Putting these values in equation (1), we get:

10. eˣ tan y dx + (1 – eˣ) sec² y dy = 0

Ans: eˣ tan y dx + (1 – eˣ) sec² y dy = 0

Integrating both sides, we get:

Integrating both sides, we get:

= log u

= log(tan y)            ……(2) 

Now, let (1- eˣ) = t

Integrating both sides, we get:

= log t

= log(1 – eˣ) ……(3) 

Using (1), (2) and (3) 

⇒ log(tan y) = log(1 – eˣ) + log C 

⇒ log(tan y) = log [C(1 – ex)] 

⇒ tan y = C(1 – eˣ)

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

Ans: 

Integrating both sides, we get:

= 2x² + x = Ax² + A + Bx² + Bx + Cx + C

= 2x² + x = (A + B)x² + (B + C)x + (A + C) 

Comparing the coefficients of x² and x, we get: 

A + B = 2

B + C = 2 

A + C = 0 

Therefore,

A = 1/2, B = 3/2 and C= -1/2 

Substituting these values in (2), we get: 

Hence, equation (1) becomes:

Now, y = 1 when x = 0

Ans:

Integrating both sides, we get:

Comparing the coefficients of x² and x, we get:

A = – 1

B – C = 0 

A + B + C = 0 

Therefore, 

A = – 1, B = 1/2 and C = 1/2

Substituting these values in (2), we get:

Hence, equation (1) becomes:

Now, y = 0, when x = 2

Thus,

Ans: 

Integrating both sides, we get: 

∫ dy = cos⁻¹ a∫ dx 

⇒ y = cos⁻¹ a.x + C 

⇒ y = x cos⁻¹ a + C 

Now, y = 1 when x = 0

⇒ 1 = 0.cos⁻¹ a + C 

= C = 1 

Thus,

14. dy/dx = y tan x; y = 1 when x = 0

Ans:

Integrating both sides, we get:

⇒ log y = log(sec x) + log C 

⇒ log y = log(sec x.C) 

⇒ y = C sec x 

Now, y = 1 when x = 0 

⇒ 1 = C × sec 0 

⇒ 1 = C × 1 

⇒ C = 1 

Thus, y = sec x

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

Ans: 

Integrating both sides, we get: 

∫ dy = ∫ eˣ sin xdx

y = ∫ eˣ sin xdx ……..(1) 

Let I = ∫ eˣ sin xdx

Substituting this value in equation (1), we get:

Since, the curve passes through (0,0), we have:

Thus, 

Hence, the required equation of the curve is 2y – 1 = eˣ (sin x – cos x)

16. For the differential equation xy dy/dx = (x + 2)(y + 2). fine the solution curve passing through the point (1,- 1). 

Ans: 

Integrating both sides, we get:

Since, the curve passes through (1,- 1), we have: 

1 – 1 – C = log [(1) (-1+2)²] 

⇒ – 2 – C = log 1 

⇒ – 2 – C = 0

⇒ C = – 2

Thus, y – x + 2 = log [x² (y + 2)²] is the required solution of the curve.

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.

Ans: Let x and y be 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 

Therefore,

Integrating both sides, we get:

Since, the curve passes through (0,- 2), we have: 

⇒ (- 2)² – 0² = 2C

= 2C = 4 

Thus, y² – x² = 4 is the required equation of the curve.

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).

Ans: (xy) is point of contact of curve and tangent.

Slope (m₁) of segment joining (x,y) and (- 4,- 3) is y + 3/x + 4 

We know that the slope of a tangent to the curve in the coordinate axis is given by the relation, dy/dx.

Therefore, slope (m₂) of tangent is dy/dx. 

Since, m₂ = 2m₁

Integrating both sides, we get:

⇒ log(y + 3) = 2 log(x + 4) + log C

⇒ log(y + 3) = log C(x + 4)² 

⇒ y + 3 = C(x + 4)²

Since, the curve passes through (- 2,1), we have: 

⇒ 1 + 3 = C(- 2 + 4)²

⇒ 4 = 4C 

⇒ C = 1

Thus, y + 3 = (x + 4)² is the required equation of the curve.

19. The volume of 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.

Ans: Let the rate of change of volume of the balloon be k.

Integrating both sides, we get: 

 ∫ 4πr²dr =  ∫ kdt 

⇒ 4π r³/3 = kt + C 

⇒ 4πr³ = 3(kt + C) 

At t = 0, r = 3

⇒ 4π × 27 =3(k × 0 + C) 

⇒ 108π = 3C 

⇒ C = 36π 

Now, at t = 3, r = 6 

⇒ 4π × 6³ = 3(k × 3 + C) 

⇒ 864π = 3(3k + 36π) 

⇒ 3k = 288π – 36π = 252π 

⇒ k = 84π 

Hence, 

4πι³ = 3[84πt + 36π] 

4πrι³ = 4π (63t + 27)

r³ = 63t + 27 

Thus, the radius  of the balloon after t seconds is

20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (log, 2 = 0.6931).

Ans: Let p, t and r represent the principle, time and rate of interest respectively. 

The principle increases continuously at the rate of r% per year. 

Integrating both sides, we get:

It is given that p = 100 when t = 0 

Therefore, 

⇒ 100 = eᵏ

Now, if t = 10 then p = 2 × 100 = 200 

Hence,

Thus, the rate of interest, r = 6.931%.

21. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is worth after 10 years ( = 1.648).

Ans: Let p and t be the principle and time, respectively. 

The principle increases continuously at the rate of 5% per year.

Integrating both sides, we get:

Now, p =1000 when t = 0 

Therefore, 

⇒ 1000 = eᶜ

Now, at t = 10 and eᶜ = 1000

Thus, after 10 years the amount will worth ₹1648.

22. 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?

Ans: Let y be the number of bacteria at any instant t. 

Rate of growth of the bacteria is proportional to the number present.

Integrating both sides, we get:

Let y₀ be the number of bacteria at t = 0. 

⇒ log y₀  = C 

⇒ log y = kt + log y₀

⇒ log y – log y₀ = kt

Since, the number of bacteria increases by 10% in 2 hours.

Taking log on both the sides

Therefore,

Now, let the time when the number of bacteria increases from 1,00,000 to 2,00,000 be t₁

Therefore, y =  y₀  at t = t₁

 Hence,

Thus, in hours, the number of bacteria increases from 1,00,000 to 2,00,000.

23. The general solution of the differential equation is

Ans: 

Integrating both sides, we get:

(where, C = – k) 

Thus, the correct option is (A).

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

1. (x² + xy) dy = (x² + y²) dx

Ans: can be written as:

Equation is a homogeneous equation. Let y = vx

Ans: 

Equation is a homogeneous equation.

Let y = vx

3. (x – y) dy – (x + y) dx = 0

Ans: 

Equation is a homogeneous equation.

4. (x² – y²) dx + 2xy dy = 0

Ans: 

Given differential equation is a homogeneous equation.

Let y = vx

Ans: 

Given differential equation is a homogeneous equation.

Let y = vx

Ans: 

Given differential equation is a homogeneous equation.

Let y = vx

Ans: 

Given differential equation is a homogeneous equation. 

Let y = vx

Ans: 

Given differential equation is a homogeneous equation.

Let y = vx 

9.

Ans: 

Given differential equation is a homogeneous equation.

Let y = vx

Required solution of the given differential equation.

Ans: 

Given differential equation is a homogeneous equation.

Let x = vy

For each of the differential equations in Exercises from 11 to 15 find the particular solution satisfying in the given condition: 

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

Ans: 

Given differential equation is a homogeneous equation.

Let y = vx

12. x² dy + (xy + y²) dx = 0; y = 1 when x = 1

Ans: 

Given differential equation is a homogeneous equation.

y = vx

13.

Ans: 

Given differential equation is a homogeneous equation.

Let y = vx

14.

Ans: 

Given differential equation is a homogeneous equation. 

Let y = vx

Required solution of the given differential equation.

15.

Ans: 

Given differential equation is a homogeneous equation.

Let y = vx

16. A homogeneous differential equation of the from dx/dy = h (x/y) can be solved by making the substitution.

(A) y = vx 

(B) v = yx 

(C) x = vy 

(D) x = v

Ans: For solving homogeneous equation of form we need to make substitution as x = vy 

Thus, the correct option is C.

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

(A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 

(B) (xy) dx – (x³ + y³) dy = 0 

(C) (x³ + 2y²) dx + 2xy dy = 0

(D) y² dx + (x² – xy – y²) dy = 0

Ans: F(x,y) is homogeneous function of degree n, if F( λx,λv) = λ’ F(x,y) for non-zero constant (λ). 

Consider equation given in D: 

y² dx + (x² – xy² – y²)dy = 0

Differential equation given in D is a homogeneous equation.

For each of the differential equations given in Exercises 1 to 12, find the general solution:

Ans: Differential equation is dy,/ dx + 2y ,sin x 

This is in the form dy/dx + py = Q where p = 2 and  Q =  sin x 

2.

Ans: 

Differential equation is dy/dx + py = Q

Ans: 

Ans: 

Ans: 

(where, p = sec² x and Q = sec² x. tan x)

Ans: 

Ans: 

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

Ans: 

Ans: 

Ans: 

11. y dx + (x – y²) dy = 0

Ans: 

Ans: 

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

Ans: 

⇒ 0 = 2 + C 

⇒ C = – 2

 y sec² x = sec x – 2 

⇒ y = cos x – 2 cos² x

Ans: 

Ans: 

y = 2 at x = π/2 

2 = – 2 + C 

⇒ C = 4

y = – 2 sin² x + 4 sin³ x 

⇒ y = 4sin³ x – 2 sin² x

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.

Ans: Let F(x, y) be the curve passing through origin.

At (x,y), slope of curve will be dy/dx

⇒ ye⁻ˣ = – xe⁻ˣ + [- e⁻ˣ) + C 

⇒ ye⁻ˣ = – e⁻ˣ + (x + 1)+C 

⇒ y = – (x + 1) + Ceˣ 

⇒ x + y + 1 = Ceˣ

Curve passes through origin.

1 = C 

⇒ x + y + 1 = eˣ

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.

Ans: F(x,y) be curve and let (x,y) be a point on the curve. Slope of the tangent to curve at (x,y)

(where, p = – 1 and Q = x – 5)

Curve passes through (0,2)

0 + 2 – 4 = C.0 

⇒ – 2 = C 

⇒ C = – 2 

x + y – 4 = – 2eˣ 

⇒ y = 4 – x – 2eˣ

18. The Integrating Factor of the differential equation is

(A) e⁻ˣ

(B) e⁻ʸ

(C) 1/x

(D) x

Ans: 

Thus, the correct option is C.

19. The Integrating Factor of the differential equation is

Ans: 

Thus, the correct option is D.

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

Ans: 

Highest order derivative present in differential equation is d² y/dx² . Its order is two. 

Highest power raised to d² y/dx²  is one. Its degree is one.

Ans: 

Highest order derivative in differential equation is dy/dx. its order is one. 

Highest power raised to dy/dx is three. Its degree is three.

Ans: 

Highest order derivative in differential equation is . Order is four. 

The given differential equation is not a polynomial equation. Degree is not defined.

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

(i) xy = a eˣ + b e⁻ˣ + x²

Ans: xy = a eˣ + b e⁻ˣ + x²

xy = aeˣ + be⁻ˣ + x² ……….(1)

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

= aeˣ + be⁻ˣ + 2-(aeˣ + be⁻ˣ + x²) + x² – 2        [using (1) and (2)]

= aeˣ + be⁻ˣ + 2 – aeˣ – be⁻ˣ – x² + x² – 2 

= 0 

= RHS

Thus, the given function is a solution of the corresponding differential equation.

(ii) y = eˣ (a cos x + b sin x)

Ans: y = eˣ (a cos x + b sin x)

y = eˣ (a cos x + b sin x) ………..(1) 

Differentiating both sides with respect to I, we get:

 y = eˣ (a cos x + b sin x) = aeˣ cos x + beˣ sin x

Again, differentiating both sides with respect to x, we get:

Now, we have 

= 0

= RHS

Thus, the given function is a solution of the corresponding differential equation.

(iii) y = x sin 3x

Ans: y = x sin 3x

y = x sin 3x ………(1)

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Thus, the given function is a solution of the corresponding differential equation.

(iv) x² = 2y² log y

Ans: x² = 2y² log y

x² = 2y² log y      ……(1) 

Differentiating both sides with respect to x, we get:

= xy – xy

= 0 

= RHS

Thus, the given function is a solution of the corresponding differential equation.

3. Prove that x² – y² = c (x² + y²) is the general solution of differential equation (x³ – 3x y²) dx = (y³ – 3x² y) dy, where c is a parameter.

Ans: 

This is a homogeneous equation, to simplify it, let y = vx

Using (1) and (2)

Integrating both sides, we get:

Let 1 – v⁴ = t

Therefore,

Now,

And

Let v² = p

Therefore,

Now, 

Using (4), (5) and (6)

Using (2) and (7)

Taking square root on both sides

4. Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Ans: Equation of a circle in first quadrant with centre (a, a) and radius (a) which touches coordinate axes is: 

(x – a)² + (y – a)²  = a²  ………(1)

Differentiating both sides with respect to x, we get:

Substituting this value in equation (1), we get:

Hence, the differential equation of the family of circles is

5. Show that the general solution of the differential equation given by (x + y +1) = A (1 – x – y – 2xy), where A is parameter.

Ans: 

Integrating both sides, we get:

6. Find if 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.

Ans: 

The curve passes through the point (0,π/4)

Therefore,

7. Find the particular solution of the differential equation 

Ans:

Substituting this value in equation (1), we get:

Hence,

Ans: 

Differentiating it with respect to y, we get:

Integrating both sides, we get

9. 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 = 1)

Ans: 

Using (1), (2) and (3)

Integrating both sides, we get:

Ans: 

This equation is a linear differential equation of the form

The general solution of the given differential equation is given by,

11. 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

Ans: dy/dx + y cot x = 4x cosec x

This equation is a linear differential equation of the form

The general solution of the given differential equation is given by,

12. Find a particular solution of the differential equation given that y = 0 when x = 0.

Ans:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

Hence,

13. The general solution of the differential equation is

(A) xy = C

(B) x = Cy²

(C) y = Cx 

(D) y = Cx²

Ans: 

Thus, the correct option is C.

14. The general solution of a differential equation of the type is

Ans: 

Thus, the correct option is C.

15. The general solution of the differential equation eˣ dy + (y eˣ + 2x) dx = 0 is

Ans: 

This is a linear differential equation of the form

Thus, the correct option is C.

Parimol

Hi! my Name is Parimal Roy. I have completed my Bachelor’s degree in Philosophy (B.A.) from Silapathar General College. Currently, I am working as an HR Manager at Dev Library. It is a website that provides study materials for students from Class 3 to 12, including SCERT and NCERT notes. It also offers resources for BA, B.Com, B.Sc, and Computer Science, along with postgraduate notes. Besides study materials, the website has novels, eBooks, health and finance articles, biographies, quotes, and more.