When the substitution \(y = vx\) is made in the differential equation \(\displaystyle\frac{dy}{dx} = \frac{x^{2} + y^{2}}{2xy}\), the equation becomes
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(x\displaystyle\frac{dv}{dx} = \frac{1 - v^{2}}{2v}\)
The right side is a ratio of two expressions of the same degree 2, so the equation is homogeneous. With y = vx, the product rule gives
\( \displaystyle \frac{dy}{dx} = v + x\frac{dv}{dx} \)
The right side becomes
\( \displaystyle \frac{x^{2} + v^{2}x^{2}}{2x \cdot vx} = \frac{1 + v^{2}}{2v} \)
Equating the two sides:
\( \displaystyle v + x\frac{dv}{dx} = \frac{1 + v^{2}}{2v} \)
\( \displaystyle x\frac{dv}{dx} = \frac{1 + v^{2}}{2v} - v \)
\( \displaystyle = \frac{1 + v^{2} - 2v^{2}}{2v} \)
\( \displaystyle = \frac{1 - v^{2}}{2v} \)
Why the other options are incorrect
Option (1) \(x\frac{dv}{dx} = \frac{1 + v^{2}}{2v}\) — Incorrect. It leaves out the term v of \(\frac{dy}{dx} = v + x\frac{dv}{dx}\); taking v to the right side gives \(1 - v^{2}\) in the numerator.
Option (3) \(\frac{dv}{dx} = \frac{1 - v^{2}}{2v}\) — Incorrect. The factor x of \(x\frac{dv}{dx}\), which comes from the product rule, is missing.
Option (4) \(x\frac{dv}{dx} = \frac{1 + 3v^{2}}{2v}\) — Incorrect. This adds v instead of subtracting it: \(\frac{1 + v^{2}}{2v} + v = \frac{1 + 3v^{2}}{2v}\).