Match each family of curves in List-I (A and B are arbitrary constants) with its differential equation in List-II.
| List-I (Family) | List-II (Differential equation) | ||
| (A) | \(y = Ae^{x} + Be^{-x}\) | (I) | \(\displaystyle\frac{d^2y}{dx^2} + 4y = 0\) |
| (B) | \(y = A\cos 2x + B\sin 2x\) | (II) | \(\displaystyle\frac{d^2y}{dx^2} - y = 0\) |
| (C) | \(y = Ax + B\) | (III) | \(\displaystyle\frac{d^2y}{dx^2} = 0\) |
| (D) | \(y = Ae^{2x} + B\) | (IV) | \(\displaystyle\frac{d^2y}{dx^2} - 2\frac{dy}{dx} = 0\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
Each family has two arbitrary constants, so each equation is of order 2. Differentiating twice and eliminating the constants gives the equation.
(A) \(y = Ae^{x} + Be^{-x}\)
This matches (II) \(\frac{d^2y}{dx^2} - y = 0\).
\(y' = Ae^{x} - Be^{-x}\)
\(y'' = Ae^{x} + Be^{-x} = y\)
(B) \(y = A\cos 2x + B\sin 2x\)
This matches (I) \(\frac{d^2y}{dx^2} + 4y = 0\).
\(y' = -2A\sin 2x + 2B\cos 2x\)
\(y'' = -4A\cos 2x - 4B\sin 2x = -4y\)
(C) \(y = Ax + B\)
This matches (III) \(\frac{d^2y}{dx^2} = 0\).
\(y' = A\)
\(y'' = 0\)
(D) \(y = Ae^{2x} + B\)
This matches (IV) \(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} = 0\).
\(y' = 2Ae^{2x}\)
\(y'' = 4Ae^{2x} = 2y'\)