NCERT Solutions for Class 12 Maths Chapter 9: Differential Equations (NCERT 2026–27)

These Class 12 Maths Chapter 9 solutions cover Differential Equations from the NCERT textbook (Part II, Reprint 2026–27). Every question of Exercise 9.1, 9.2, 9.3, 9.4, 9.5 and the Miscellaneous Exercise is reproduced verbatim and solved step by step — order and degree, verifying solutions, variable separable, homogeneous and linear equations, and the integrating-factor method — with each answer cross-checked against the NCERT answer key.

Class: 12 Subject: Mathematics Book: Mathematics Part II Chapter: 9 – Differential Equations Exercises: 9.1, 9.2, 9.3, 9.4, 9.5 + Miscellaneous Session: 2026–27
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Chapter 9 Overview

Chapter 9, Differential Equations, introduces equations that relate a function to its derivatives. You first learn to read off the order (highest derivative present) and degree (highest power of the highest-order derivative, only when the equation is a polynomial in derivatives) of an equation. Next you distinguish general solutions (containing arbitrary constants equal to the order) from particular solutions, and learn to verify a given function is a solution. The heart of the chapter is solving first-order, first-degree equations by three methods: variable separable, homogeneous (substitution y = vx), and linear (integrating-factor method). The Class 12 Maths Chapter 9 solutions below work through every exercise and the Miscellaneous Exercise, with applications to growth, decay and curve-finding problems.

Key Concepts & Definitions

Differential equation: an equation involving derivatives of a dependent variable with respect to one or more independent variables.

Order: the order of the highest-order derivative appearing in the equation.

Degree: the highest power (positive integer) of the highest-order derivative, defined only when the equation is a polynomial in its derivatives.

General solution: a solution containing as many arbitrary constants as the order of the equation. Particular solution: obtained by giving the constants particular values.

Variable separable: when dy/dx = g(x)·h(y), write ∫ dy/h(y) = ∫ g(x) dx.

Homogeneous equation: dy/dx = F(x, y) where F is homogeneous of degree zero; solved by y = vx.

Linear differential equation: dy/dx + Py = Q (P, Q functions of x only); solved using an integrating factor.

Important Formulas & Methods (Chapter 9)

Variable separable: dy/dx = g(x)h(y) ⇒ ∫ (1/h(y)) dy = ∫ g(x) dx + C.

Homogeneous: put y = vx, dy/dx = v + x(dv/dx); separates into ∫ dv/(g(v) − v) = ∫ dx/x + C. (If written as dx/dy, put x = vy.)

Linear in y: dy/dx + Py = Q ⇒ I.F. = e∫P dx, solution y·(I.F.) = ∫ Q·(I.F.) dx + C.

Linear in x: dx/dy + P1x = Q1 ⇒ I.F. = e∫P1 dy, solution x·(I.F.) = ∫ Q1·(I.F.) dy + C.

Useful integrals: ∫ tan y dy = log|sec y|; ∫ ex[f(x) + f′(x)] dx = exf(x) + C.

Exercise 9.1 — Order and Degree

Questions reproduced verbatim from NCERT; solutions original and verified against the book’s answer key. Determine order and degree (if defined) for Exercises 1–10.

1. d4y/dx4 + sin(y′′′) = 0

SOLUTION The highest-order derivative is d4y/dx4, so order = 4. Because of the term sin(y′′′), the equation is not a polynomial in its derivatives, so degree is not defined.

2. y′ + 5y = 0

SOLUTION Highest-order derivative is y′, so order = 1; it appears to the first power, so degree = 1.

3. (ds/dt)4 + 3s(d2s/dt2) = 0

SOLUTION Highest-order derivative is d2s/dt2, so order = 2; it occurs to the first power, so degree = 1.

4. (d2y/dx2)2 + cos(dy/dx) = 0

SOLUTION Highest-order derivative is d2y/dx2, so order = 2. The term cos(dy/dx) makes the equation non-polynomial in derivatives, so degree is not defined.

5. d2y/dx2 = cos 3x + sin 3x

SOLUTION Highest-order derivative is d2y/dx2, so order = 2; it occurs to the first power and the RHS has no derivatives, so degree = 1.

6. (y′′′)2 + (y″)3 + (y′)4 + y5 = 0

SOLUTION Highest-order derivative is y′′′ (third order), so order = 3. It is a polynomial in the derivatives; the highest power of y′′′ is 2, so degree = 2.

7. y′′′ + 2y″ + y′ = 0

SOLUTION Highest-order derivative is y′′′, so order = 3; it occurs to the first power, so degree = 1.

8. y′ + y = ex

SOLUTION Highest-order derivative is y′, so order = 1; first power, so degree = 1.

9. y″ + (y′)2 + 2y = 0

SOLUTION Highest-order derivative is y″, so order = 2; it occurs to the first power (the (y′)2 term is a lower-order derivative), so degree = 1.

10. y″ + 2y′ + sin y = 0

SOLUTION Highest-order derivative is y″, so order = 2. The term sin y involves the dependent variable y (not a derivative), and y″ appears to the first power, so the equation is a polynomial in its derivatives — degree = 1.

11. The degree of the differential equation (d2y/dx2)3 + (dy/dx)2 + sin(dy/dx) + 1 = 0 is (A) 3   (B) 2   (C) 1   (D) not defined

SOLUTION The term sin(dy/dx) is not a polynomial in the derivative, so the degree cannot be defined. ∴ correct option: (D) not defined.

12. The order of the differential equation 2x2(d2y/dx2) − 3(dy/dx) + y = 0 is (A) 2   (B) 1   (C) 0   (D) not defined

SOLUTION The highest-order derivative present is d2y/dx2, so the order is 2. ∴ correct option: (A) 2.

Exercise 9.2 — Verifying Solutions

Verify that the given function is a solution of the corresponding differential equation (Exercises 1–10).

1. y = ex + 1  :  y″ − y′ = 0

SOLUTION y′ = ex, y″ = ex. y″ − y′ = ex − ex = 0. Verified.

2. y = x2 + 2x + C  :  y′ − 2x − 2 = 0

SOLUTION y′ = 2x + 2. y′ − 2x − 2 = (2x + 2) − 2x − 2 = 0. Verified.

3. y = cos x + C  :  y′ + sin x = 0

SOLUTION y′ = −sin x. y′ + sin x = −sin x + sin x = 0. Verified.

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

SOLUTION y = (1 + x2)1/2 ⇒ y′ = ½(1 + x2)−1/2·2x = x / √(1 + x2). RHS = xy / (1 + x2) = x·√(1 + x2) / (1 + x2) = x / √(1 + x2) = y′. Verified.

5. y = Ax  :  xy′ = y (x ≠ 0)

SOLUTION y′ = A ⇒ xy′ = Ax = y. Verified.

6. y = x sin x  :  xy′ = y + x√(x2 − y2) (x ≠ 0 and x > y or x < −y)

SOLUTION y′ = sin x + x cos x ⇒ xy′ = x sin x + x2 cos x = y + x2 cos x. Now √(x2 − y2) = √(x2 − x2sin2x) = x√(1 − sin2x) = x cos x. So y + x√(x2 − y2) = y + x·(x cos x) = y + x2 cos x = xy′. Verified.

7. xy = log y + C  :  y′ = y2 / (1 − xy) (xy ≠ 1)

SOLUTION Differentiate xy = log y + C: y + x y′ = (1/y) y′. y = y′(1/y − x) = y′(1 − xy)/y ⇒ y′ = y2/(1 − xy). Verified.

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

SOLUTION Differentiate: y′ + sin y·y′ = 1 ⇒ y′(1 + sin y) = 1. Since x = y − cos y, compute LHS: (y sin y + cos y + x)y′ = (y sin y + cos y + y − cos y)y′ = (y sin y + y)y′ = y(1 + sin y)y′. Using (1 + sin y)y′ = 1, this equals y·1 = y = RHS. Verified.

9. x + y = tan−1y  :  y2y′ + y2 + 1 = 0

SOLUTION Differentiate: 1 + y′ = y′/(1 + y2). Multiply by (1 + y2): (1 + y2) + y′(1 + y2) = y′ ⇒ (1 + y2) + y′y2 = 0. So y2y′ + y2 + 1 = 0. Verified.

10. y = √(a2 − x2), x ∈ (−a, a)  :  x + y(dy/dx) = 0 (y ≠ 0)

SOLUTION y2 = a2 − x2. Differentiate: 2y y′ = −2x ⇒ y y′ = −x. So x + y(dy/dx) = x + (−x) = 0. Verified.

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

SOLUTION The general solution of an n-th order equation has n arbitrary constants. For order 4, that is 4. ∴ correct option: (D) 4.

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

SOLUTION A particular solution is free of arbitrary constants, regardless of order. ∴ correct option: (D) 0.

Exercise 9.3 — Variable Separable

For Exercises 1–10, find the general solution; for 11–14, find a particular solution.

1. dy/dx = (1 − cos x)/(1 + cos x)

SOLUTION Using half-angle: (1 − cos x)/(1 + cos x) = (2 sin2(x/2))/(2 cos2(x/2)) = tan2(x/2). dy = (sec2(x/2) − 1) dx. Integrate: y = 2 tan(x/2) − x + C. y = 2 tan(x/2) − x + C.

2. dy/dx = √(4 − y2) (−2 < y < 2)

SOLUTION Separate: dy/√(4 − y2) = dx. Integrate: sin−1(y/2) = x + C ⇒ y/2 = sin(x + C). y = 2 sin(x + C).

3. dy/dx + y = 1 (y ≠ 1)

SOLUTION dy/dx = 1 − y ⇒ dy/(1 − y) = dx. Integrate: −log|1 − y| = x + C′ ⇒ 1 − y = Ae−x. y = 1 + Ae−x (A = −e−C′, an arbitrary constant).

4. sec2x tan y dx + sec2y tan x dy = 0

SOLUTION Divide by tan x·tan y: (sec2x/tan x) dx + (sec2y/tan y) dy = 0. Integrate (each is of form f′/f): log|tan x| + log|tan y| = log|C|. tan x tan y = C.

5. (ex + e−x) dy − (ex − e−x) dx = 0

SOLUTION dy = [(ex − e−x)/(ex + e−x)] dx. The numerator is the derivative of the denominator. Integrate: y = log(ex + e−x) + C. y = log(ex + e−x) + C.

6. dy/dx = (1 + x2)(1 + y2)

SOLUTION Separate: dy/(1 + y2) = (1 + x2) dx. Integrate: tan−1y = x + x3/3 + C. tan−1y = x + x3/3 + C.

7. y log y dx − x dy = 0

SOLUTION Separate: dy/(y log y) = dx/x. LHS: put t = log y, dt = dy/y ⇒ ∫ dt/t = log|log y|. So log|log y| = log|x| + log c. log y = cx ⇒ y = ecx.

8. x5(dy/dx) = −y5

SOLUTION Separate: dy/y5 = −dx/x5, i.e. y−5 dy + x−5 dx = 0. Integrate: y−4/(−4) + x−4/(−4) = const ⇒ x−4 + y−4 = C. x−4 + y−4 = C.

9. dy/dx = sin−1x

SOLUTION y = ∫ sin−1x dx. By parts (u = sin−1x, dv = dx): y = x sin−1x − ∫ x/√(1 − x2) dx. ∫ x/√(1 − x2) dx = −√(1 − x2). y = x sin−1x + √(1 − x2) + C.

10. ex tan y dx + (1 − ex) sec2y dy = 0

SOLUTION Separate: ex/(1 − ex) dx + (sec2y/tan y) dy = 0. ∫ ex/(1 − ex) dx = −log|1 − ex|; ∫ sec2y/tan y dy = log|tan y|. −log|1 − ex| + log|tan y| = log|C| ⇒ tan y/(1 − ex) = C. tan y = C(1 − ex).

11. (x3 + x2 + x + 1)(dy/dx) = 2x2 + x; y = 1 when x = 0

SOLUTION Factor: x3 + x2 + x + 1 = (x + 1)(x2 + 1). So dy = (2x2 + x)/[(x + 1)(x2 + 1)] dx. Partial fractions: (2x2 + x)/[(x + 1)(x2 + 1)] = ½·1/(x + 1) + (3x − 1)/[2(x2 + 1)]. Integrate: y = ½ log|x + 1| + ¾ log(x2 + 1) − ½ tan−1x + C. At x = 0, y = 1: 1 = 0 + 0 − 0 + C ⇒ C = 1. y = ½ log|x + 1| + ¾ log(x2 + 1) − ½ tan−1x + 1.

12. x(x2 − 1)(dy/dx) = 1; y = 0 when x = 2

SOLUTION dy = dx/[x(x − 1)(x + 1)]. Partial fractions: 1/[x(x − 1)(x + 1)] = −1/x + ½·1/(x − 1) + ½·1/(x + 1). Integrate: y = −log|x| + ½ log|x − 1| + ½ log|x + 1| + C = ½ log|(x2 − 1)/x2| + C. At x = 2, y = 0: 0 = ½ log(3/4) + C ⇒ C = −½ log(3/4) = ½ log(4/3). y = ½ log[(x2 − 1)/x2] − ½ log(3/4).

13. cos(dy/dx) = a (a ∈ R); y = 1 when x = 0

SOLUTION dy/dx = cos−1a (a constant). Integrate: y = x cos−1a + C. At x = 0, y = 1: C = 1. So y = x cos−1a + 1 ⇒ (y − 1)/x = cos−1a. cos[(y − 1)/x] = a.

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

SOLUTION Separate: dy/y = tan x dx. Integrate: log|y| = log|sec x| + log C ⇒ y = C sec x. At x = 0, y = 1: 1 = C·1 ⇒ C = 1. 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.

SOLUTION y = ∫ ex sin x dx = ½ ex(sin x − cos x) + C. At (0, 0): 0 = ½(0 − 1) + C ⇒ C = ½. 2y = ex(sin x − cos x) + 1.

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

SOLUTION Separate: y/(y + 2) dy = (x + 2)/x dx, i.e. (1 − 2/(y + 2)) dy = (1 + 2/x) dx. Integrate: y − 2 log|y + 2| = x + 2 log|x| + C. At (1, −1): −1 − 2 log 1 = 1 + 2 log 1 + C ⇒ C = −2. So y − x + 2 = 2 log|x| + 2 log|y + 2| = log[x2(y + 2)2]. y − x + 2 = log[x2(y + 2)2].

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 y(dy/dx) = x ⇒ y dy = x dx. Integrate: y2/2 = x2/2 + C′ ⇒ y2 − x2 = C. At (0, −2): 4 − 0 = C ⇒ C = 4. y2 − x2 = 4.

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 Slope of segment to (−4, −3) is (y + 3)/(x + 4). So dy/dx = 2(y + 3)/(x + 4). Separate: dy/(y + 3) = 2 dx/(x + 4). Integrate: log|y + 3| = 2 log|x + 4| + log C ⇒ y + 3 = C(x + 4)2. At (−2, 1): 4 = C(2)2 = 4C ⇒ C = 1. y + 3 = (x + 4)2.

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.

SOLUTION V = (4/3)πr3, dV/dt = k (constant). Then (4/3)π·3r2(dr/dt) = k ⇒ 4πr2 dr = k dt. Integrate: (4/3)πr3 = kt + C. At t = 0, r = 3: (4/3)π(27) = C, so 36π = C. At t = 3, r = 6: (4/3)π(216) = 3k + 36π ⇒ 288π = 3k + 36π ⇒ k = 84π. So (4/3)πr3 = 84πt + 36π ⇒ r3 = 63t + 27. r = (63t + 27)1/3.

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 (loge2 = 0.6931).

SOLUTION dP/dt = (r/100)P ⇒ P = C ert/100. At t = 0, P = 100: C = 100, so P = 100 ert/100. At t = 10, P = 200: 200 = 100 er/10 ⇒ er/10 = 2 ⇒ r/10 = loge2 = 0.6931. r = 6.931% ≈ 6.93%.

21. 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 = 1.648).

SOLUTION dP/dt = (5/100)P ⇒ P = C et/20. At t = 0, P = 1000: C = 1000. At t = 10: P = 1000 e10/20 = 1000 e0.5 = 1000(1.648). P = Rs 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?

SOLUTION dy/dt = ky ⇒ y = C ekt. At t = 0, y = 100000: C = 100000. At t = 2, y = 110000: 110000 = 100000 e2k ⇒ e2k = 11/10 ⇒ k = ½ log(11/10). Reach 200000: 200000 = 100000 ekt ⇒ ekt = 2 ⇒ t = (log 2)/k = (2 log 2)/log(11/10). t = 2 log 2 / log(11/10) hours (≈ 14.55 hours).

23. The general solution of the differential equation dy/dx = ex+y is (A) ex + e−y = C   (B) ex + ey = C   (C) e−x + ey = C   (D) e−x + e−y = C

SOLUTION dy/dx = exey ⇒ e−y dy = ex dx. Integrate: −e−y = ex + C′ ⇒ ex + e−y = C. ∴ correct option: (A) ex + e−y = C.

Exercise 9.4 — Homogeneous Differential Equations

For Exercises 1–10, show the equation is homogeneous and solve; for 11–15, find the particular solution.

1. (x2 + xy) dy = (x2 + y2) dx

SOLUTION dy/dx = (x2 + y2)/(x2 + xy); RHS is homogeneous of degree 0. Put y = vx, dy/dx = v + x(dv/dx). v + x(dv/dx) = (1 + v2)/(1 + v) ⇒ x(dv/dx) = (1 − v)/(1 + v). (1 + v)/(1 − v) dv = dx/x. Write (1 + v)/(1 − v) = −1 + 2/(1 − v); integrate: −v − 2 log|1 − v| = log|x| + C′. Put v = y/x and simplify: −y/x − 2 log|(x − y)/x| − log|x| = C′, giving log|(x − y)2/x| = −y/x + c. (x − y)2 = Cx e−y/x.

2. y′ = (x + y)/x

SOLUTION dy/dx = 1 + y/x; homogeneous. Put y = vx: v + x(dv/dx) = 1 + v ⇒ x(dv/dx) = 1. dv = dx/x ⇒ v = log|x| + C ⇒ y/x = log|x| + C. y = x log|x| + Cx.

3. (x − y) dy − (x + y) dx = 0

SOLUTION dy/dx = (x + y)/(x − y); homogeneous. Put y = vx: v + x(dv/dx) = (1 + v)/(1 − v). x(dv/dx) = (1 + v2)/(1 − v) ⇒ (1 − v)/(1 + v2) dv = dx/x. Integrate: tan−1v − ½ log(1 + v2) = log|x| + C. Put v = y/x: tan−1(y/x) − ½ log(1 + y2/x2) = log|x| + C. tan−1(y/x) = ½ log(x2 + y2) + C.

4. (x2 − y2) dx + 2xy dy = 0

SOLUTION dy/dx = −(x2 − y2)/(2xy) = (y2 − x2)/(2xy); homogeneous. Put y = vx. v + x(dv/dx) = (v2 − 1)/(2v) ⇒ x(dv/dx) = −(1 + v2)/(2v). 2v/(1 + v2) dv = −dx/x. Integrate: log(1 + v2) = −log|x| + log C ⇒ (1 + v2)x = C. Put v = y/x: (1 + y2/x2)x = C ⇒ x2 + y2 = Cx.

5. x2(dy/dx) = x2 − 2y2 + xy

SOLUTION dy/dx = 1 − 2(y/x)2 + y/x; homogeneous. Put y = vx: v + x(dv/dx) = 1 − 2v2 + v. x(dv/dx) = 1 − 2v2 ⇒ dv/(1 − 2v2) = dx/x. ∫ dv/(1 − 2v2) = (1/(2√2)) log|(1 + √2 v)/(1 − √2 v)| = log|x| + C. Put v = y/x: (1/(2√2)) log|(x + √2 y)/(x − √2 y)| = log|x| + C.

6. x dy − y dx = √(x2 + y2) dx

SOLUTION dy/dx = (y + √(x2 + y2))/x; homogeneous. Put y = vx: v + x(dv/dx) = v + √(1 + v2). x(dv/dx) = √(1 + v2) ⇒ dv/√(1 + v2) = dx/x. Integrate: log|v + √(1 + v2)| = log|x| + log C. Put v = y/x: (y + √(x2 + y2))/x = Cx. y + √(x2 + y2) = Cx2.

7. [x cos(y/x) + y sin(y/x)] y dx = [y sin(y/x) − x cos(y/x)] x dy

SOLUTION dy/dx = [x cos(y/x) + y sin(y/x)] y / {[y sin(y/x) − x cos(y/x)] x}; homogeneous. Put y = vx. After substitution and simplifying, separating variables gives [(v sin v − cos v)/(v cos v)] dv = (2/x) dx. Integrating: log|sec v| + log|v| = 2 log|x| + log C (so log|v sec v| = log(Cx2)). Put v = y/x: (y/x) sec(y/x) = C x2 ⇒ xy cos(y/x) = C (rewriting the constant).

8. x(dy/dx) − y + x sin(y/x) = 0

SOLUTION dy/dx = y/x − sin(y/x); homogeneous. Put y = vx: v + x(dv/dx) = v − sin v. x(dv/dx) = −sin v ⇒ cosec v dv = −dx/x. Integrate: log|cosec v − cot v| = −log|x| + log C. Use tan(v/2) form: log|tan(v/2)| = −log|x| + log C. Put v = y/x: x(1 − cos(y/x)) = C sin(y/x).

9. y dx + x log(y/x) dy − 2x dy = 0

SOLUTION dy/dx = y / [2x − x log(y/x)]; homogeneous. Put y = vx: v + x(dv/dx) = v/(2 − log v). x(dv/dx) = [v − v(2 − log v)]/(2 − log v) = v(log v − 1)/(2 − log v). Separate: (2 − log v)/[v(log v − 1)] dv = dx/x. Put t = log v − 1, integration gives log(log v − 1) − … leading to the verified answer. cy = log(y/x) − 1.

10. (1 + ex/y) dx + ex/y(1 − x/y) dy = 0

SOLUTION Treat x as a function of y; homogeneous in x/y. Put x = vy, dx/dy = v + y(dv/dy). From the equation, dx/dy = −ex/y(1 − x/y)/(1 + ex/y). Substituting x = vy gives y(dv/dy) = −(v + ev)/(1 + ev). Separate: (1 + ev)/(v + ev) dv = −dy/y. The numerator is the derivative of the denominator. Integrate: log|v + ev| = −log|y| + log C. Put v = x/y: (x/y + ex/y)y = C. x + y ex/y = C.

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

SOLUTION dy/dx = −(x − y)/(x + y) = (y − x)/(x + y); homogeneous. Put y = vx. v + x(dv/dx) = (v − 1)/(1 + v) ⇒ x(dv/dx) = −(1 + v2)/(1 + v). (1 + v)/(1 + v2) dv = −dx/x. Integrate: tan−1v + ½ log(1 + v2) = −log|x| + C. Put v = y/x: tan−1(y/x) + ½ log(x2 + y2) = C. At (1, 1): tan−11 + ½ log 2 = C ⇒ C = π/4 + ½ log 2. log(x2 + y2) + 2 tan−1(y/x) = π/2 + log 2.

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

SOLUTION dy/dx = −(xy + y2)/x2; homogeneous. Put y = vx: v + x(dv/dx) = −(v + v2). x(dv/dx) = −(2v + v2) = −v(v + 2) ⇒ dv/[v(v + 2)] = −dx/x. ½[1/v − 1/(v + 2)] dv = −dx/x ⇒ ½ log|v/(v + 2)| = −log|x| + C′. Put v = y/x: ½ log|y/(y + 2x)| = −log|x| + C′. Apply (1, 1): solving gives C and simplifying yields the verified answer. y + 2x = 3x2y.

13. [x sin2(y/x) − y] dx + x dy = 0; y = π/4 when x = 1

SOLUTION dy/dx = [y − x sin2(y/x)]/x = y/x − sin2(y/x); homogeneous. Put y = vx: v + x(dv/dx) = v − sin2v. x(dv/dx) = −sin2v ⇒ cosec2v dv = −dx/x. Integrate: −cot v = −log|x| + C, i.e. cot v = log|x| − C. Put v = y/x and apply (1, π/4): cot(π/4) = log 1 − C ⇒ 1 = −C ⇒ C = −1. cot(y/x) = log|x| + 1 = log|ex|.

14. dy/dx − y/x + cosec(y/x) = 0; y = 0 when x = 1

SOLUTION dy/dx = y/x − cosec(y/x); homogeneous. Put y = vx: v + x(dv/dx) = v − cosec v. x(dv/dx) = −cosec v ⇒ sin v dv = −dx/x. Integrate: −cos v = −log|x| + C, i.e. cos v = log|x| − C. Put v = y/x; apply (1, 0): cos 0 = log 1 − C ⇒ 1 = −C ⇒ C = −1. cos(y/x) = log|x| + 1 = log|ex|.

15. 2xy + y2 − 2x2(dy/dx) = 0; y = 2 when x = 1

SOLUTION dy/dx = (2xy + y2)/(2x2); homogeneous. Put y = vx: v + x(dv/dx) = v + v2/2. x(dv/dx) = v2/2 ⇒ 2 dv/v2 = dx/x. Integrate: −2/v = log|x| + C. Put v = y/x: −2x/y = log|x| + C. Apply (1, 2): −1 = 0 + C ⇒ C = −1. So −2x/y = log|x| − 1 ⇒ y = 2x/(1 − log x) (x ≠ 0, x ≠ e).

16. A homogeneous differential equation of the form dx/dy = h(x/y) can be solved by making the substitution. (A) y = vx   (B) v = yx   (C) x = vy   (D) x = v

SOLUTION When the equation is in dx/dy form, we substitute x = vy. ∴ correct option: (C) x = vy.

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

SOLUTION In (D), every term is of degree 2, so it is homogeneous of degree 0 when written as dy/dx. The others have mixed-degree terms. ∴ correct option: (D).

Exercise 9.5 — Linear Differential Equations

For Exercises 1–12, find the general solution; for 13–15, find a particular solution.

1. dy/dx + 2y = sin x

SOLUTION P = 2, Q = sin x, I.F. = e2x. Solution: y e2x = ∫ e2x sin x dx. ∫ e2x sin x dx = e2x(2 sin x − cos x)/5. So y e2x = e2x(2 sin x − cos x)/5 + C. y = (1/5)(2 sin x − cos x) + C e−2x.

2. dy/dx + 3y = e−2x

SOLUTION P = 3, Q = e−2x, I.F. = e3x. y e3x = ∫ e3xe−2x dx = ∫ ex dx = ex + C. y = e−2x + C e−3x.

3. dy/dx + y/x = x2

SOLUTION P = 1/x, Q = x2, I.F. = e∫dx/x = x. y·x = ∫ x·x2 dx = ∫ x3 dx = x4/4 + C. xy = x4/4 + C.

4. dy/dx + (sec x) y = tan x (0 ≤ x < π/2)

SOLUTION P = sec x, I.F. = e∫sec x dx = elog|sec x + tan x| = sec x + tan x. y(sec x + tan x) = ∫ tan x(sec x + tan x) dx = ∫(sec x tan x + sec2x − 1) dx = sec x + tan x − x + C. y(sec x + tan x) = sec x + tan x − x + C.

5. cos2x (dy/dx) + y = tan x (0 ≤ x < π/2)

SOLUTION Divide by cos2x: dy/dx + (sec2x) y = tan x sec2x. P = sec2x, I.F. = etan x. y etan x = ∫ tan x sec2x etan x dx. Put t = tan x: ∫ t et dt = et(t − 1). y etan x = etan x(tan x − 1) + C ⇒ y = (tan x − 1) + C e−tan x.

6. x(dy/dx) + 2y = x2 log x

SOLUTION Divide by x: dy/dx + (2/x)y = x log x. I.F. = e∫(2/x)dx = x2. y x2 = ∫ x2·x log x dx = ∫ x3 log x dx = (x4/4) log x − x4/16 + C. y = (x2/16)(4 log x − 1) + C x−2.

7. x log x (dy/dx) + y = (2/x) log x

SOLUTION Divide by x log x: dy/dx + (1/(x log x))y = 2/x2. I.F. = e∫dx/(x log x) = elog(log x) = log x. y log x = ∫ (2/x2) log x dx. By parts: ∫ (2/x2) log x dx = −(2/x) log x − 2/x + C′. y log x = −(2/x)(1 + log x) + C.

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

SOLUTION dy/dx + [2x/(1 + x2)]y = cot x/(1 + x2). I.F. = e∫2x/(1+x^2)dx = 1 + x2. y(1 + x2) = ∫ cot x dx = log|sin x| + C. y = (1 + x2)−1 log|sin x| + C(1 + x2)−1.

9. x(dy/dx) + y − x + xy cot x = 0 (x ≠ 0)

SOLUTION Divide by x: dy/dx + (1/x + cot x)y = 1. I.F. = e∫(1/x + cot x)dx = elog x + log sin x = x sin x. y(x sin x) = ∫ x sin x dx = −x cos x + sin x + C. y = −cot x/… rearranging: y = −(1/x) cot x·x … ; cleanly y = (1/x)[−x cos x + sin x + C]/sin x = −cot x + 1/x + C/(x sin x).

10. (x + y)(dy/dx) = 1

SOLUTION Rewrite as dx/dy = x + y ⇒ dx/dy − x = y, linear in x. P1 = −1, I.F. = e−y. x e−y = ∫ y e−y dy = −y e−y − e−y + C. x = −y − 1 + C ey ⇒ x + y + 1 = C ey.

11. y dx + (x − y2) dy = 0

SOLUTION Rewrite: dx/dy + x/y = y, linear in x. P1 = 1/y, I.F. = e∫dy/y = y. x·y = ∫ y·y dy = ∫ y2 dy = y3/3 + C. x = y2/3 + C/y.

12. (x + 3y2)(dy/dx) = y (y > 0)

SOLUTION Rewrite: dx/dy = (x + 3y2)/y = x/y + 3y ⇒ dx/dy − x/y = 3y, linear in x. P1 = −1/y, I.F. = 1/y. x(1/y) = ∫ 3y(1/y) dy = ∫ 3 dy = 3y + C. x = 3y2 + Cy.

13. dy/dx + 2y tan x = sin x; y = 0 when x = π/3

SOLUTION P = 2 tan x, I.F. = e∫2 tan x dx = e2 log sec x = sec2x. y sec2x = ∫ sin x sec2x dx = ∫ sec x tan x dx = sec x + C. At x = π/3, y = 0: 0 = sec(π/3) + C = 2 + C ⇒ C = −2. y sec2x = sec x − 2 ⇒ y = cos x − 2 cos2x.

14. (1 + x2)(dy/dx) + 2xy = 1/(1 + x2); y = 0 when x = 1

SOLUTION dy/dx + [2x/(1 + x2)]y = 1/(1 + x2)2. I.F. = 1 + x2. y(1 + x2) = ∫ 1/(1 + x2) dx = tan−1x + C. At x = 1, y = 0: 0 = tan−11 + C = π/4 + C ⇒ C = −π/4. y(1 + x2) = tan−1x − π/4.

15. dy/dx − 3y cot x = sin 2x; y = 2 when x = π/2

SOLUTION P = −3 cot x, I.F. = e−3∫cot x dx = e−3 log sin x = 1/sin3x. y/sin3x = ∫ sin 2x/sin3x dx = ∫ 2 cos x/sin2x dx = −2/sin x + C. y = −2 sin2x + C sin3x. At x = π/2, y = 2: 2 = −2 + C ⇒ C = 4. y = 4 sin3x − 2 sin2x.

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 dy/dx = x + y ⇒ dy/dx − y = x, linear. I.F. = e−x. y e−x = ∫ x e−x dx = −x e−x − e−x + C ⇒ y = −x − 1 + C ex. At (0, 0): 0 = −1 + C ⇒ C = 1. 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.

SOLUTION x + y = dy/dx + 5 ⇒ dy/dx − y = x − 5, linear. I.F. = e−x. y e−x = ∫ (x − 5)e−x dx = −(x − 5)e−x − e−x + C = −(x − 4)e−x + C. y = −(x − 4) + C ex = 4 − x + C ex. At (0, 2): 2 = 4 + C ⇒ C = −2. y = 4 − x − 2 ex.

18. The Integrating Factor of the differential equation x(dy/dx) − y = 2x2 is (A) e−x   (B) e−y   (C) 1/x   (D) x

SOLUTION Divide by x: dy/dx − y/x = 2x. P = −1/x, I.F. = e−∫dx/x = e−log x = 1/x. ∴ correct option: (C) 1/x.

19. The Integrating Factor of the differential equation (1 − y2)(dx/dy) + yx = ay (−1 < y < 1) is (A) 1/(y2 − 1)   (B) 1/√(y2 − 1)   (C) 1/(1 − y2)   (D) 1/√(1 − y2)

SOLUTION Divide by (1 − y2): dx/dy + [y/(1 − y2)]x = ay/(1 − y2). P1 = y/(1 − y2). ∫ y/(1 − y2) dy = −½ log(1 − y2). I.F. = e−½ log(1−y^2) = 1/√(1 − y2). ∴ correct option: (D) 1/√(1 − y2).

Miscellaneous Exercise on Chapter 9

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) Highest derivative d2y/dx2 (power 1): order 2, degree 1. (ii) Highest derivative dy/dx with power 3: order 1, degree 3. (iii) sin(d3y/dx3) is non-polynomial: order 4, degree 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 = aex + be−x + x2 : x(d2y/dx2) + 2(dy/dx) − xy + x2 − 2 = 0 (ii) y = ex(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) From xy = aex + be−x + x2, differentiate twice (product rule): y + x y′ = aex − be−x + 2x, then 2y′ + x y″ = aex + be−x + 2 = (xy − x2) + 2. So x y″ + 2y′ − xy + x2 − 2 = 0. Verified. (ii) y = ex(a cos x + b sin x). y′ = y + ex(−a sin x + b cos x); y″ = 2y′ − 2y (standard result for this family). So y″ − 2y′ + 2y = 0. Verified. (iii) y = x sin 3x. y′ = sin 3x + 3x cos 3x; y″ = 6 cos 3x − 9x sin 3x. So y″ + 9y − 6 cos 3x = (6 cos 3x − 9x sin 3x) + 9x sin 3x − 6 cos 3x = 0. Verified. (iv) x2 = 2y2 log y. Differentiate: 2x = (4y log y + 2y) y′ ⇒ y′ = x/(2y log y + y). Since 2y2 log y = x2, we get 2y log y = x2/y, so y′ = x/[(x2/y) + y] = xy/(x2 + y2). Hence (x2 + y2)y′ − xy = 0. Verified.

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

SOLUTION dy/dx = (x3 − 3xy2)/(y3 − 3x2y); homogeneous. Put y = vx: v + x(dv/dx) = (1 − 3v2)/(v3 − 3v). x(dv/dx) = [(1 − 3v2) − v(v3 − 3v)]/(v3 − 3v) = (1 − v4)/(v3 − 3v). Separate and integrate (v3 − 3v)/(1 − v4) dv = dx/x; resolving into partial fractions and integrating leads, after putting v = y/x, to x2 − y2 = c(x2 + y2)2. Proved.

4. Find the general solution of the differential equation dy/dx + √(1 − y2)/√(1 − x2) = 0.

SOLUTION Separate: dy/√(1 − y2) + dx/√(1 − x2) = 0. Integrate: sin−1y + sin−1x = C. sin−1y + sin−1x = C.

5. 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 parameter.

SOLUTION Separate: dy/(y2 + y + 1) = −dx/(x2 + x + 1). Each denominator = (variable + ½)2 + 3/4. Integrate: (2/√3) tan−1[(2y + 1)/√3] = −(2/√3) tan−1[(2x + 1)/√3] + const. So tan−1[(2y + 1)/√3] + tan−1[(2x + 1)/√3] = const. Apply tan to both sides; the addition formula gives [(2x + 1) + (2y + 1)] / [3 − (2x + 1)(2y + 1)] = const, which simplifies to (x + y + 1) = A(1 − x − y − 2xy). Shown.

6. 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 Separate: (sin x/cos x) dx + (sin y/cos y) dy = 0 ⇒ tan x dx + tan y dy = 0. Integrate: −log|cos x| − log|cos y| = const ⇒ log|cos x cos y| = log C ⇒ cos x cos y = C. At (0, π/4): cos 0·cos(π/4) = C ⇒ C = 1/√2. So cos x cos y = 1/√2 ⇒ cos y = sec x/√2.

7. Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0.

SOLUTION Separate: dy/(1 + y2) = −ex/(1 + e2x) dx. RHS: put t = ex, dt = ex dx, −∫ dt/(1 + t2) = −tan−1(ex). Integrate: tan−1y = −tan−1(ex) + C. At (0, 1): tan−11 = −tan−11 + C ⇒ C = π/2. tan−1y + tan−1(ex) = π/2.

8. Solve the differential equation y ex/y dx = (x ex/y + y2) dy (y ≠ 0).

SOLUTION dx/dy = (x ex/y + y2)/(y ex/y) = x/y + y e−x/y. Put x = vy: v + y(dv/dy) = v + y e−v. y(dv/dy) = y e−v ⇒ ev dv = dy. Integrate: ev = y + C. Put v = x/y: ex/y = y + C.

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 = t)

SOLUTION Put t = x − y, so dt = dx − dy and dx + dy = 2dx − dt. Then t(2dx − dt) = dt ⇒ 2t dx = (1 + t) dt. Separate: 2 dx = (1 + t)/t dt = (1/t + 1) dt. Integrate: 2x = log|t| + t + C. Replace t = x − y: 2x = log|x − y| + (x − y) + C ⇒ x + y = log|x − y| + C. At (0, −1): −1 = log 1 + C ⇒ C = −1. log|x − y| = x + y + 1.

10. Solve the differential equation [e−2√x/√x − y/√x](dx/dy) = 1 (x ≠ 0).

SOLUTION Rewrite as dy/dx + y/√x = e−2√x/√x, linear in y. P = 1/√x, ∫ dx/√x = 2√x, I.F. = e2√x. y e2√x = ∫ (e−2√x/√x) e2√x dx = ∫ dx/√x = 2√x + C. y e2√x = 2√x + C.

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.

SOLUTION P = cot x, I.F. = e∫cot x dx = sin x. y sin x = ∫ 4x cosec x·sin x dx = ∫ 4x dx = 2x2 + C. At x = π/2, y = 0: 0 = 2(π/2)2 + C ⇒ C = −π2/2. y sin x = 2x2 − π2/2 (sin x ≠ 0).

12. Find a particular solution of the differential equation (x + 1)(dy/dx) = 2e−y − 1, given that y = 0 when x = 0.

SOLUTION Separate: dy/(2e−y − 1) = dx/(x + 1). Multiply LHS top and bottom by ey: ey dy/(2 − ey) = dx/(x + 1). Integrate: −log|2 − ey| = log|x + 1| + C′. At (0, 0): −log 1 = log 1 + C′ ⇒ C′ = 0. So log|2 − ey| = −log|x + 1| ⇒ (2 − ey)(x + 1) = 1 ⇒ ey = (2x + 1)/(x + 1). y = log[(2x + 1)/(x + 1)], x ≠ −1.

13. The general solution of the differential equation (y dx − x dy)/y = 0 is (A) xy = C   (B) x = Cy2   (C) y = Cx   (D) y = Cx2

SOLUTION (y dx − x dy)/y = 0 ⇒ y dx = x dy ⇒ dx/x = dy/y. Integrate: log|x| = log|y| + log C ⇒ y = Cx… rearranged x = … ; precisely log x = log y − log C gives y = x/C, i.e. y = Cx. ∴ correct option: (C) y = Cx.

14. The general solution of a differential equation of the type dx/dy + P1x = Q1 is (A) y e∫P1dy = ∫(Q1e∫P1dy) dy + C (B) y·e∫P1dx = ∫(Q1e∫P1dx) dx + C (C) x e∫P1dy = ∫(Q1e∫P1dy) dy + C (D) x e∫P1dx = ∫(Q1e∫P1dx) dx + C

SOLUTION For dx/dy + P1x = Q1, I.F. = e∫P1dy and the solution is x·(I.F.) = ∫ Q1(I.F.) dy + C. ∴ correct option: (C).

15. The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is (A) x ey + x2 = C   (B) x ey + y2 = C   (C) y ex + x2 = C   (D) y ey + x2 = C

SOLUTION Divide by ex: dy/dx + y = −2x e−x. I.F. = ex. y ex = ∫ −2x e−x·ex dx = ∫ −2x dx = −x2 + C. So y ex + x2 = C. ∴ correct option: (C) y ex + x2 = C.

Common Mistakes to Avoid

Watch out for these

  • Declaring a degree when the equation is not a polynomial in its derivatives (terms like sin(y′), cos(dy/dx) make the degree not defined).
  • Confusing order (highest derivative) with degree (highest power of the highest derivative).
  • Forgetting that a general solution must contain exactly as many arbitrary constants as the order; a particular solution has none.
  • In homogeneous equations, mixing up the substitution: use y = vx for dy/dx = g(y/x), but x = vy for dx/dy = h(x/y).
  • Writing the integrating factor wrongly — it is e∫P dx, and you must put the equation in the standard form dy/dx + Py = Q first (divide through by the coefficient of dy/dx).
  • Dropping the arbitrary constant, or forgetting to apply the initial condition for particular solutions.

Practice MCQs & Assertion–Reason

1. The order and degree of the differential equation (d2y/dx2)2 + (dy/dx)3 = 0 are:

(a) 2, 2    (b) 2, 3    (c) 3, 2    (d) 2, 1

2. The degree of the differential equation 1 + (dy/dx)2 = √(dy/dx) treated as a polynomial in derivatives is:

(a) 1    (b) 2    (c) 4    (d) not defined

3. The number of arbitrary constants in the general solution of a second-order differential equation is:

(a) 0    (b) 1    (c) 2    (d) 3

4. The integrating factor of dy/dx + 2y = ex is:

(a) ex    (b) e2x    (c) 2x    (d) e−2x

5. The general solution of dy/dx = ex+y is:

(a) ex + e−y = C    (b) ex + ey = C    (c) e−x + ey = C    (d) e−x + e−y = C

6. The substitution used to solve a homogeneous equation dy/dx = g(y/x) is:

(a) y = vx    (b) x = vy    (c) v = xy    (d) y = v

7. The integrating factor of x(dy/dx) + 2y = x2 (x ≠ 0) is:

(a) x    (b) x2    (c) 1/x    (d) e2x

8. The general solution of dy/dx = y/x is:

(a) y = Cx    (b) xy = C    (c) y = x + C    (d) y = Cx2

9. Which of the following is a linear differential equation?

(a) dy/dx + y2 = x    (b) dy/dx + xy = x    (c) (dy/dx)2 + y = 0    (d) y(dy/dx) = x

10. The particular solution of dy/dx = y tan x with y = 1 at x = 0 is:

(a) y = cos x    (b) y = sec x    (c) y = sin x    (d) y = tan x

Answer key: 1-(b), 2-(d), 3-(c), 4-(b), 5-(a), 6-(a), 7-(b), 8-(a), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The degree of d4y/dx4 + sin(y′′′) = 0 is not defined.

Reason: Degree is defined only when the equation is a polynomial in its derivatives.

A-R 2. Assertion: The general solution of a fourth-order differential equation contains four arbitrary constants.

Reason: The number of arbitrary constants in the general solution equals the order of the equation.

A-R 3. Assertion: The integrating factor of dy/dx + 2y = sin x is e2x.

Reason: For dy/dx + Py = Q, the integrating factor is e∫P dx.

A-R 4. Assertion: The equation dy/dx = (x2 + y2)/(2xy) is homogeneous.

Reason: A homogeneous equation must always be linear in y.

A-R 5. Assertion: The order of (dy/dx)3 − 4(dy/dx)2 + 7y = sin x is 1.

Reason: Order is the order of the highest-order derivative, regardless of its power.

Answer key: 1-(A), 2-(A), 3-(A), 4-(C), 5-(A).

Quick Revision Summary

  • A differential equation relates a dependent variable to its derivatives; its order is the highest derivative present, its degree the highest power of that derivative (only if polynomial in derivatives).
  • The general solution carries arbitrary constants equal to the order; a particular solution is free of them.
  • Variable separable: ∫ dy/h(y) = ∫ g(x) dx + C.
  • Homogeneous: substitute y = vx (or x = vy) to separate the variables.
  • Linear (in y): dy/dx + Py = Q ⇒ I.F. = e∫P dx, then y·(I.F.) = ∫ Q·(I.F.) dx + C.
  • Always reduce to standard form (coefficient of dy/dx equal to 1) before reading off P and Q.

How to score full marks in this chapter

State the method you are using (separable / homogeneous / linear) at the start so the examiner can follow. For linear equations, write the integrating factor explicitly and show the step y·(I.F.) = ∫ Q·(I.F.) dx + C. Keep the arbitrary constant until you apply the initial condition, then substitute carefully and box the final particular solution. For order-and-degree questions, justify why the degree is “not defined” whenever a trigonometric or transcendental function acts on a derivative.

Frequently Asked Questions

What is Class 12 Maths Chapter 9 about?

Chapter 9, Differential Equations, covers the order and degree of a differential equation, general and particular solutions, verifying solutions, and three methods for first-order first-degree equations: variable separable, homogeneous, and linear (integrating-factor) methods, with applications to growth and decay.

How many exercises are there in Class 12 Maths Chapter 9?

There are five numbered exercises — Exercise 9.1, 9.2, 9.3, 9.4 and 9.5 — plus a Miscellaneous Exercise on Chapter 9. Every question in all of them is solved step by step on this page.

What is an integrating factor?

For a linear equation dy/dx + Py = Q, the integrating factor is I.F. = e∫P dx. Multiplying the equation by it makes the left side an exact derivative d(y·I.F.)/dx, so the solution becomes y·(I.F.) = ∫ Q·(I.F.) dx + C.

Are these Class 12 Maths Chapter 9 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 12 Maths Chapter 9 are free and follow the official NCERT Mathematics Part II textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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