Each homogeneous differential equation in List-I is changed by the substitution \(y = vx\). Match List-I with the equation in v and x it reduces to, given in List-II.
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\displaystyle\frac{dy}{dx} = \frac{2x + y}{x}\) | (I) | \(x\displaystyle\frac{dv}{dx} = v^2\) |
| (B) | \(\displaystyle\frac{dy}{dx} = \frac{x^2 + y^2}{xy}\) | (II) | \(x\displaystyle\frac{dv}{dx} = 2\) |
| (C) | \(\displaystyle\frac{dy}{dx} = \frac{xy + y^2}{x^2}\) | (III) | \(x\displaystyle\frac{dv}{dx} = -\frac{1 + v^2}{1 + v}\) |
| (D) | \(\displaystyle\frac{dy}{dx} = \frac{y - x}{y + x}\) | (IV) | \(x\displaystyle\frac{dv}{dx} = \frac{1}{v}\) |
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)-(I), (D)-(III)
With \(y = vx\),
\( \displaystyle \frac{dy}{dx} = v + x\frac{dv}{dx} \)
So each equation becomes \(x\frac{dv}{dx} = (\text{right-hand side in } v) - v\).
(A) \(\frac{dy}{dx} = \frac{2x + y}{x}\)
This matches (II). The right side is \(2 + v\), so \(x\frac{dv}{dx} = 2\).
(B) \(\frac{dy}{dx} = \frac{x^2 + y^2}{xy}\)
This matches (IV). The right side is \(\frac{1}{v} + v\), so \(x\frac{dv}{dx} = \frac{1}{v}\).
(C) \(\frac{dy}{dx} = \frac{xy + y^2}{x^2}\)
This matches (I). The right side is \(v + v^2\), so \(x\frac{dv}{dx} = v^2\).
(D) \(\frac{dy}{dx} = \frac{y - x}{y + x}\)
This matches (III). The right side is \(\frac{v - 1}{v + 1}\), so
\( \displaystyle x\frac{dv}{dx} = \frac{v - 1}{v + 1} - v \)
\( \displaystyle x\frac{dv}{dx} = \frac{v - 1 - v^2 - v}{v + 1} = -\frac{1 + v^2}{1 + v} \)