Maths · Calculus · chapter 9 of 13

Class 12 Maths Chapter 9: Differential Equations case study question with answers

One case-based question of the kind the board asks: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

In a physics laboratory in Kolkata, a small block is attached to a light spring on a smooth table. Students pull the block 3 cm from its rest position and push it, so that it starts with a speed of 8 cm per second. If x cm is the displacement of the block from rest after t seconds, its motion is described by the differential equation d²xdt² + 16x = 0. The class wants to find the solution of this equation that fits the starting position and the starting speed.

Part 1 · 1 mark

The order and the degree of the differential equation are

(a) order 2, degree 1(b) order 1, degree 2(c) order 2, degree 2(d) order 2, degree not defined

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Answer: (a) order 2, degree 1

The highest derivative is d²x/dt², so the order is 2; the equation is a polynomial in it with power 1, so the degree is 1.

Part 2 · 1 mark

Which of the following is the general solution of the equation (A, B arbitrary constants)?

(a) x = A cos 4t + B sin 4t(b) x = A cos 16t + B sin 16t(c) x = A e^4t + B e^−4t(d) x = A cos 4t

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Answer: (a) x = A cos 4t + B sin 4t

For x = A cos 4t + B sin 4t, d²x/dt² = −16(A cos 4t + B sin 4t) = −16x, and it carries two arbitrary constants for a second order equation. (b) gives −256x; (c) gives +16x; (d) has only one constant.

Part 3 · 2 marks

Find the particular solution with x = 3 and dxdt = 8 at t = 0.

Answer and marking scheme

x = A cos 4t + B sin 4t. At t = 0, x = A = 3. dx/dt = −4A sin 4t + 4B cos 4t; at t = 0 it is 4B = 8, so B = 2. The particular solution is x = 3 cos 4t + 2 sin 4t.

A = 3 from x(0) = 3 (1)
B = 2 from x′(0) = 8; x = 3 cos 4t + 2 sin 4t (1)

Verify that x = a sin (4t + b), a and b arbitrary constants, is a solution of the equation, and state why it may be taken as the general solution.

Answer and marking scheme

dx/dt = 4a cos (4t + b), d²x/dt² = −16a sin (4t + b) = −16x, so d²x/dt² + 16x = 0. It contains two arbitrary constants, a and b, as many as the order of the equation, so it is a general solution.

finds d²x/dt² = −16x and verifies (1)
two arbitrary constants equal to the order 2 (1)

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