Differential Equations — Class 12 MCQs with Answers

Differential Equations — important questions & MCQs with answers (Class 12 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Order of d²y/dx² + dy/dx + y = 0:

   A3B1C2D0
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   Answer: 2

   Explanation: Order is the highest derivative present, not the number of terms. Here d²y/dx² appears, so order = 2 — don't stop at dy/dx and call it order 1.

2. Q2 of 6Easy

   dy/dx = y type of equation:

   ALinear onlyBVariable separableCBernoulliDExact
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   Answer: Variable separable

   Explanation: dy/dx = y separates cleanly into dy/y = dx, with only y-terms on one side and only dx on the other — that split is the defining test for variable separable.

3. Q3 of 6Medium

   DE for exponential growth: dy/dt = ky (k > 0). Solution:

   Ay = 1/y₀By = ktCy = e^tDy = y₀ e^(kt)
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   Answer: y = y₀ e^(kt)

   Explanation: dy/dt = ky separates to dy/y = k dt; integrating from the initial value y₀ gives y = y₀e^(kt) — the constant k sits in the exponent, driving growth when k > 0.

4. Q4 of 6Medium

   Assertion (A): The general solution of dy/dx = e^(x − y) is e^y = eˣ + C. Reason (R): The equation dy/dx = e^(x − y) can be written as e^y dy = eˣ dx.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: R is true: e^(x − y) = eˣ·e^(−y), so multiplying by e^y separates the variables as e^y dy = eˣ dx. Integrating both sides gives e^y = eˣ + C, so A is true, and R is the very step that leads to A.

5. Q5 of 6Hard

   Solve dy/dx = xy / (1 + x²):

   Ay = ln(1 + x²)By = 1 + xCy = √(1 + x²)Dy² = C(1 + x²)
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   Answer: y² = C(1 + x²)

   Explanation: Separating gives dy/y = x/(1+x²) dx; integrating gives ln y = (1/2)ln(1+x²) + C, so squaring clears the half-power to give y² = C(1+x²).

6. Q6 of 6Hard

   Assertion (A): The general solution of the differential equation y dx + (x − y²) dy = 0 (y > 0) is x = y²/3 + C/y. Reason (R): The function F(x, y) = y² + 2xy is a homogeneous function of degree 2.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true but R is not the correct explanation of A

   Explanation: A is true: the equation is dx/dy + x/y = y, linear in x with integrating factor e^(∫dy/y) = y, so xy = ∫y² dy = y³/3 + C, i.e. x = y²/3 + C/y. R is true: F(λx, λy) = λ²(y² + 2xy). But R is about a different function and plays no part in solving A.

Differential Equations notes: key concepts (Class 12 Mathematics)

A differential equation relates a function to its derivatives. The chapter defines order and degree, distinguishes general from particular solutions, and solves the three first-order types — variable separable, homogeneous (via y = vx) and linear (via an integrating factor) — with growth, decay and cooling as applications.

Differential equation

An equation involving an unknown function and its derivatives, e.g. dy/dx + y = eˣ; solving it recovers the function.

Order of a DE

The order of the highest derivative present; d²y/dx² + dy/dx + y = 0 is order 2.

Degree of a DE

The power of the highest-order derivative once the equation is made polynomial in its derivatives (free of radicals).

General vs particular solution

A general solution of an order-n equation carries n arbitrary constants; initial conditions fix them to give a particular solution.

Forming a DE

Differentiate a family of curves and eliminate its arbitrary constants; y = A cos x + B sin x gives d²y/dx² + y = 0.

Variable separable

If dy/dx = g(x)h(y), rearrange to dy/h(y) = g(x)dx and integrate both sides separately.

Homogeneous DE

Of the form dy/dx = F(y/x); substitute y = vx so dy/dx = v + x dv/dx, turning it separable in v and x.

Linear first-order DE

Standard form dy/dx + P(x)y = Q(x); solved with an integrating factor rather than separation.

Integrating factor

IF = e^(∫P dx); multiplying the linear DE by it makes the left side d/dx(y·IF).

Solution of a linear DE

y·(IF) = ∫Q·(IF) dx + C; e.g. for dy/dx + (1/x)y = x² the IF is x.

Growth and decay

dy/dt = ky solves to y = y₀e^(kt); k > 0 models growth, k < 0 decay such as radioactive decline.

Applications

Newton's law of cooling dT/dt = −k(T − Tₛ) and first-order reaction kinetics d[A]/dt = −k[A] are standard modelled examples.