Class 12 Maths Chapter 9 Differential Equations Worksheet with Answers, free PDF (CBSE and JEE Main; based on NCERT)

A 20-mark worksheet on Differential Equations for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

If y = x^m, x > 0, is a solution of x² y″ − 2x y′ − 4y = 0 and m > 0, then m equals

(a) 1(b) 4(c) 2(d) 3

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Answer: (b) 4

y′ = m x^m−1, y″ = m(m − 1) x^m−2. Substituting: [m(m − 1) − 2m − 4] x^m = 0, so m² − 3m − 4 = 0, (m − 4)(m + 1) = 0. The positive root is m = 4.

Question 2

The general solution of dydx = x e^−y is

(a) y = x² e^y + C(b) e^−y = x²2 + C(c) e^y = x² + C(d) e^y = x²2 + C

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Answer: (d) e^y = x²2 + C

Separating, e^y dy = x dx. Integrating both sides: e^y = x²/2 + C. Option (b) integrates e^y wrongly as e^−y; (c) drops the factor 1/2.

Question 3

The sum of the order and the degree of (d³ydx³)² − 4(d²ydx²)⁵ + y = 0 is

(a) 7(b) 8(c) 5(d) 4

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Answer: (c) 5

The highest order derivative is d³y/dx³, so the order is 3; its power is 2, so the degree is 2. The sum is 5. Option (b) uses the power 5 of the second derivative as the degree.

Question 4

For which of the following differential equations is the degree not defined?

(a) y″ + cos y = 0(b) y′ + e^y′ = x(c) (y′)² = sin x(d) y‴ + y y′ = 0

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Answer: (b) y′ + e^y′ = x

e^y′ is not a polynomial in y′, so (b) is not a polynomial equation in its derivatives and its degree is not defined. In (a) cos y involves y, not a derivative, so (a) is a polynomial in y″ of degree 1; (c) has degree 2 and (d) degree 1.

Question 5

The degree of the differential equation (y″)² + (y′)³ − 7y = 0 is

(a) 1(b) 3(c) 2(d) 5

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Answer: (c) 2

The equation is a polynomial in y′ and y″. The highest order derivative is y″, and its highest power is 2, so the degree is 2. Option (b) takes the highest power of any derivative; (d) adds the two powers.

Question 6 · 2 marks

Find the order and the degree of the differential equation [1 + (dydx)²]^3/2 = 5 d²ydx².

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The highest order derivative is d²y/dx², so the order is 2. The equation is not yet a polynomial in the derivatives because of the power 3/2; squaring both sides gives [1 + (dy/dx)²]³ = 25 (d²y/dx²)², a polynomial equation in y′ and y″. The highest power of y″ is 2, so the degree is 2.

order 2 (1)
squares to a polynomial in the derivatives and states degree 2 (1)

Question 7 · 2 marks

Find the integrating factor of cos x dydx + y sin x = 1, 0 < x < π2, and hence its general solution.

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Dividing by cos x: dy/dx + (tan x) y = sec x, so P = tan x, Q = sec x. I.F. = e^{∫ tan x dx} = e^{log sec x} = sec x. Then y sec x = ∫ sec² x dx = tan x + C, so y = sin x + C cos x.

standard form and I.F. = sec x (1)
y = sin x + C cos x (1)

Question 8 · 2 marks

A colony of bacteria grows at a rate proportional to the number N present, and it trebles in 4 hours. Write the differential equation for N and show that N(t + 4) = 3N(t) for every t, so the trebling time does not depend on the number at the start.

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dN/dt = kN. Separating, dN/N = k dt, so N = N₀ e^kt, N₀ the starting number; trebling in 4 hours gives e^4k = 3. Then N(t + 4) = N₀ e^kt e^4k = 3 N₀ e^kt = 3N(t) for every t, whatever N₀ is.

dN/dt = kN, N = N₀ e^kt and e^4k = 3 (1)
N(t + 4) = N₀ e^kt e^4k = 3N(t), independent of N₀ (1)

Question 9 · 3 marks

Find the general solution of dydx + 4y = 8x.

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P = 4, Q = 8x. I.F. = e^4x. Then y e^4x = ∫ 8x e^4x dx = 8[x e^4x/4 − e^4x/16] + C = 2x e^4x − e^4x/2 + C (integration by parts). Dividing by e^4x: y = 2x − 1/2 + C e^−4x.

I.F. = e^4x (1)
∫ 8x e^4x dx = 2x e^4x − 12 e^4x by parts (1)
y = 2x − 12 + C e^−4x (1)

Question 10 · 3 marks

Find the particular solution of (1 + x²) dydx = xy, y > 0, given that y = 2 when x = 0.

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Separating, dy/y = x dx/(1 + x²). Integrating, log y = (1/2) log (1 + x²) + log C, so y = C√(1 + x²). At x = 0, y = 2: C = 2. The particular solution is y = 2√(1 + x²).

separates the variables (1)
integrates: y = C√(1 + x²) (1)
C = 2 and y = 2√(1 + x²) (1)

Question 11 · 3 marks

Solve x dydx − 3y = x⁴ cos x, x > 0.

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Dividing by x: dy/dx − (3/x) y = x³ cos x, so P = −3/x. I.F. = e^{−3 log x} = e^{log x⁻³} = 1/x³. Then y/x³ = ∫ x³ cos x · (1/x³) dx = ∫ cos x dx = sin x + C. Hence y = x³ sin x + C x³.

standard form with P = −3x (1)
I.F. = 1x³ (1)
y = x³ sin x + C x³ (1)

Class 12 Maths Chapter 9: Differential Equations worksheet with answer key