Match each differential equation in List-I with a function in List-II that is a solution of it.
| List-I | List-II | ||
|---|---|---|---|
| Differential equation | Solution | ||
| (A) | \(x\displaystyle\frac{dy}{dx} = 2y\) | (I) | \(y = \sin x\) |
| (B) | \(\displaystyle\frac{d^2y}{dx^2} = y\) | (II) | \(y = -\displaystyle\frac{1}{x}\) |
| (C) | \(\displaystyle\frac{d^2y}{dx^2} + y = 0\) | (III) | \(y = 3x^2\) |
| (D) | \(\displaystyle\frac{dy}{dx} = y^2\) | (IV) | \(y = e^{-x}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
A function is a solution when, on substituting it and its derivatives, the two sides of the equation become equal.
(A) \(x\frac{dy}{dx} = 2y\)
This matches (III) \(y = 3x^2\). Here \(\frac{dy}{dx} = 6x\), so \(x\frac{dy}{dx} = 6x^2 = 2y\).
(B) \(\frac{d^2y}{dx^2} = y\)
This matches (IV) \(y = e^{-x}\). Differentiating twice gives \(\frac{d^2y}{dx^2} = e^{-x} = y\).
(C) \(\frac{d^2y}{dx^2} + y = 0\)
This matches (I) \(y = \sin x\). Here \(\frac{d^2y}{dx^2} = -\sin x\), so \(\frac{d^2y}{dx^2} + y = 0\).
(D) \(\frac{dy}{dx} = y^2\)
This matches (II) \(y = -\frac{1}{x}\). Here \(\frac{dy}{dx} = \frac{1}{x^2}\) and \(y^2 = \frac{1}{x^2}\).