Differential Equations Practice

Take timed practice tests on Differential Equations for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.

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Differential Equations Practice Tests

Six 20-question timed sessions plus one 60-question module test. Each question is original and calibrated from the uploaded material pattern without copying PDF wording.

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JEE Main Session 1

2Questions

60s/QTimer

+4 / -1Scoring

Preview all 12 questions in Differential Equations Practice Tests (no login required)

Order and degree
1. The order of the differential equation d²y/dx² + (dy/dx)³ = x is:
- A. 2 (Correct)
- B. 3
- C. 1
- D. 4
Explanation: Order = order of the highest derivative present. Highest derivative is d²y/dx² (second order) → order = 2. Degree = power of the highest derivative after rationalising → degree = 1.
Separable DE
2. The differential equation dy/dx = y is solved by:
- A. y = Ceˣ (Correct)
- B. y = Cx
- C. y = x + C
- D. y = Ce^(−x)
Explanation: Separate variables: dy/y = dx. Integrate: ln|y| = x + C₁ → y = Ce^x.
Variable separable
3. For dy/dx = (1+y²)/(1+x²), the solution is:
- A. tan⁻¹y = tan⁻¹x + C (Correct)
- B. y = x + C
- C. y² = x² + C
- D. y/(1+y²) = x/(1+x²)
Explanation: Separate: dy/(1+y²) = dx/(1+x²). Integrate both sides: tan⁻¹y = tan⁻¹x + C.
Homogeneous DE
4. The substitution y = vx is used to solve:
- A. Homogeneous differential equations where f(x,y) = f(λx,λy) (Correct)
- B. Linear DEs
- C. Bernoulli DEs
- D. Exact DEs
Explanation: A DE is homogeneous if dy/dx = f(y/x). Substituting y = vx → v + x(dv/dx) = f(v), which is separable in v and x.
Linear DE
5. The integrating factor (IF) for the linear DE dy/dx + Py = Q (P, Q functions of x) is:
- A. e^(∫P dx) (Correct)
- B. e^(−∫P dx)
- C. ∫P dx
- D. P itself
Explanation: For dy/dx + P(x)y = Q(x): multiply through by IF = e^(∫P dx). The left side becomes d/dx[y × IF] = Q × IF. Integrate to find y.
Particular solutions
6. If dy/dx = x and y(0) = 1, then y is:
- A. y = x²/2 + 1 (Correct)
- B. y = x²/2
- C. y = x + 1
- D. y = eˣ
Explanation: Integrate: y = x²/2 + C. Apply y(0) = 1: 1 = 0 + C → C = 1. Solution: y = x²/2 + 1.
JEE Main
7. The general solution of dy/dx = e^(x+y) is:
- A. e^(−y) + eˣ = C (Correct)
- B. eˣ + e^(−y) = C
- C. e^(−y) = eˣ + C
- D. e^y = eˣ + C
Explanation: Separate: e^(−y) dy = eˣ dx. Integrate: −e^(−y) = eˣ + C₁ → e^(−y) = −eˣ + C → e^(−y) + eˣ = C.
Geometric interpretation
8. A curve passes through (1, 2) with slope equal to 2y/x at every point. The equation of the curve is:
- A. y = 2x² (Correct)
- B. y = x²
- C. y = 2x
- D. y = 2/x
Explanation: dy/dx = 2y/x → dy/y = 2dx/x → ln|y| = 2ln|x| + C → y = Ax². At (1,2): 2 = A → A = 2 → y = 2x².
JEE Advanced
9. The orthogonal trajectories of the family of circles x² + y² = r² are:
- A. Straight lines through the origin: y = kx (Correct)
- B. Parabolas
- C. Ellipses
- D. Hyperbolas
Explanation: Family of circles: x²+y²=r². DE: differentiate → 2x+2y(dy/dx)=0 → dy/dx=−x/y. For orthogonal trajectories, replace dy/dx with −y'/dx = −1/(−x/y) = y/x. New DE: dy/dx = y/x → ln y = ln x + C → y = kx (straight lines through origin).
Mixed
10. A differential equation of order n has:
- A. n arbitrary constants in the general solution (Correct)
- B. n solutions
- C. n variables
- D. n² terms
Explanation: The general solution of an nth order ODE contains exactly n arbitrary constants. Applying n initial/boundary conditions determines these constants uniquely.
Mixed
11. Solve: (1+x²) dy + 2xy dx = d(x² sin y)?
- A. This requires checking if exact: ∂M/∂y = ∂N/∂x (Correct)
- B. Direct integration
- C. Variable separable directly
- D. Linear ODE method
Explanation: Write as M dx + N dy = 0 where M = 2xy, N = 1+x². Check exactness: ∂M/∂y = 2x, ∂N/∂x = 2x → exact DE. Solution: ∫M dx = x²y + g(y). Differentiate w.r.t. y: x² + g'(y) = N = 1+x² → g'(y) = 1 → g(y) = y. Solution: x²y + y = C → y(x²+1) = C.
Mixed
12. Newton's law of cooling: dT/dt = −k(T−T₀). After solving, T(t) is:
- A. T₀ + (T_initial − T₀)e^(−kt) (Correct)
- B. T₀ × e^(−kt)
- C. T_initial × e^(−kt)
- D. T₀ + T_initial e^(kt)
Explanation: dT/dt = −k(T−T₀). Let u = T−T₀: du/dt = −ku → u = Ae^(−kt) → T−T₀ = (T_i−T₀)e^(−kt) → T = T₀ + (T_i−T₀)e^(−kt). As t→∞, T→T₀.

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JEE Main Session 1

2Questions

60s/QTimer

+4 / -1Scoring

Preview all 12 questions in Differential Equations Practice Tests (no login required)

Order and degree
1. The order of the differential equation d²y/dx² + (dy/dx)³ = x is:
- A. 2 (Correct)
- B. 3
- C. 1
- D. 4
Explanation: Order = order of the highest derivative present. Highest derivative is d²y/dx² (second order) → order = 2. Degree = power of the highest derivative after rationalising → degree = 1.
Separable DE
2. The differential equation dy/dx = y is solved by:
- A. y = Ceˣ (Correct)
- B. y = Cx
- C. y = x + C
- D. y = Ce^(−x)
Explanation: Separate variables: dy/y = dx. Integrate: ln|y| = x + C₁ → y = Ce^x.
Variable separable
3. For dy/dx = (1+y²)/(1+x²), the solution is:
- A. tan⁻¹y = tan⁻¹x + C (Correct)
- B. y = x + C
- C. y² = x² + C
- D. y/(1+y²) = x/(1+x²)
Explanation: Separate: dy/(1+y²) = dx/(1+x²). Integrate both sides: tan⁻¹y = tan⁻¹x + C.
Homogeneous DE
4. The substitution y = vx is used to solve:
- A. Homogeneous differential equations where f(x,y) = f(λx,λy) (Correct)
- B. Linear DEs
- C. Bernoulli DEs
- D. Exact DEs
Explanation: A DE is homogeneous if dy/dx = f(y/x). Substituting y = vx → v + x(dv/dx) = f(v), which is separable in v and x.
Linear DE
5. The integrating factor (IF) for the linear DE dy/dx + Py = Q (P, Q functions of x) is:
- A. e^(∫P dx) (Correct)
- B. e^(−∫P dx)
- C. ∫P dx
- D. P itself
Explanation: For dy/dx + P(x)y = Q(x): multiply through by IF = e^(∫P dx). The left side becomes d/dx[y × IF] = Q × IF. Integrate to find y.
Particular solutions
6. If dy/dx = x and y(0) = 1, then y is:
- A. y = x²/2 + 1 (Correct)
- B. y = x²/2
- C. y = x + 1
- D. y = eˣ
Explanation: Integrate: y = x²/2 + C. Apply y(0) = 1: 1 = 0 + C → C = 1. Solution: y = x²/2 + 1.
JEE Main
7. The general solution of dy/dx = e^(x+y) is:
- A. e^(−y) + eˣ = C (Correct)
- B. eˣ + e^(−y) = C
- C. e^(−y) = eˣ + C
- D. e^y = eˣ + C
Explanation: Separate: e^(−y) dy = eˣ dx. Integrate: −e^(−y) = eˣ + C₁ → e^(−y) = −eˣ + C → e^(−y) + eˣ = C.
Geometric interpretation
8. A curve passes through (1, 2) with slope equal to 2y/x at every point. The equation of the curve is:
- A. y = 2x² (Correct)
- B. y = x²
- C. y = 2x
- D. y = 2/x
Explanation: dy/dx = 2y/x → dy/y = 2dx/x → ln|y| = 2ln|x| + C → y = Ax². At (1,2): 2 = A → A = 2 → y = 2x².
JEE Advanced
9. The orthogonal trajectories of the family of circles x² + y² = r² are:
- A. Straight lines through the origin: y = kx (Correct)
- B. Parabolas
- C. Ellipses
- D. Hyperbolas
Explanation: Family of circles: x²+y²=r². DE: differentiate → 2x+2y(dy/dx)=0 → dy/dx=−x/y. For orthogonal trajectories, replace dy/dx with −y'/dx = −1/(−x/y) = y/x. New DE: dy/dx = y/x → ln y = ln x + C → y = kx (straight lines through origin).
Mixed
10. A differential equation of order n has:
- A. n arbitrary constants in the general solution (Correct)
- B. n solutions
- C. n variables
- D. n² terms
Explanation: The general solution of an nth order ODE contains exactly n arbitrary constants. Applying n initial/boundary conditions determines these constants uniquely.
Mixed
11. Solve: (1+x²) dy + 2xy dx = d(x² sin y)?
- A. This requires checking if exact: ∂M/∂y = ∂N/∂x (Correct)
- B. Direct integration
- C. Variable separable directly
- D. Linear ODE method
Explanation: Write as M dx + N dy = 0 where M = 2xy, N = 1+x². Check exactness: ∂M/∂y = 2x, ∂N/∂x = 2x → exact DE. Solution: ∫M dx = x²y + g(y). Differentiate w.r.t. y: x² + g'(y) = N = 1+x² → g'(y) = 1 → g(y) = y. Solution: x²y + y = C → y(x²+1) = C.
Mixed
12. Newton's law of cooling: dT/dt = −k(T−T₀). After solving, T(t) is:
- A. T₀ + (T_initial − T₀)e^(−kt) (Correct)
- B. T₀ × e^(−kt)
- C. T_initial × e^(−kt)
- D. T₀ + T_initial e^(kt)
Explanation: dT/dt = −k(T−T₀). Let u = T−T₀: du/dt = −ku → u = Ae^(−kt) → T−T₀ = (T_i−T₀)e^(−kt) → T = T₀ + (T_i−T₀)e^(−kt). As t→∞, T→T₀.