Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Differential equation | Type | ||
| (A) | \(\displaystyle\frac{dy}{dx} = e^{x}\sec y\) | (I) | Linear, of the form \(\displaystyle\frac{dx}{dy} + P_1x = Q_1\) |
| (B) | \(\displaystyle\frac{dy}{dx} = \frac{x^2 + y^2}{2x^2}\) | (II) | Variables separable |
| (C) | \(\displaystyle\frac{dy}{dx} + y = x^2\) | (III) | Linear, of the form \(\displaystyle\frac{dy}{dx} + Py = Q\) |
| (D) | \(y\displaystyle\frac{dx}{dy} + x = y^2\) | (IV) | Homogeneous |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
(A) \(\frac{dy}{dx} = e^{x}\sec y\)
This matches (II) Variables separable: it can be written as \(\cos y\,dy = e^{x}\,dx\).
(B) \(\frac{dy}{dx} = \frac{x^2 + y^2}{2x^2}\)
This matches (IV) Homogeneous: the right side equals \(\frac{1}{2}\left(1 + \left(\frac{y}{x}\right)^2\right)\), a function of \(\frac{y}{x}\) only (homogeneous of degree zero).
(C) \(\frac{dy}{dx} + y = x^2\)
This matches (III) Linear in y, with \(P = 1\) and \(Q = x^2\).
(D) \(y\frac{dx}{dy} + x = y^2\)
This matches (I) Linear in x: dividing by y gives \(\frac{dx}{dy} + \frac{1}{y}x = y\), with \(P_1 = \frac{1}{y}\) and \(Q_1 = y\).