The particular solution of the differential equation \(\displaystyle\frac{dy}{dx} = \frac{2x}{y}\), \(y \gt 0\), satisfying \(y(1) = 2\), is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(y^2 = 2x^2 + 2\)
Separating the variables:
\(y\,dy = 2x\,dx\)
Integrating both sides:
\( \displaystyle \frac{y^2}{2} = x^2 + C \)
\(y^2 = 2x^2 + 2C\)
Putting x = 1, y = 2:
\(4 = 2 + 2C\)
\(2C = 2\)
So the particular solution is
\(y^2 = 2x^2 + 2\)
Why the other options are incorrect
Option (1) \(y^2 = x^2 + 3\) — Incorrect. It passes through (1, 2), but differentiating gives \(2y\frac{dy}{dx} = 2x\), that is \(\frac{dy}{dx} = \frac{x}{y}\), not \(\frac{2x}{y}\).
Option (2) \(y = x + 1\) — Incorrect. Here \(\frac{dy}{dx} = 1\), while \(\frac{2x}{y} = \frac{2x}{x + 1}\), which is not 1 for every x (at x = 0 it is 0).
Option (3) \(y^2 = 2x^2 + 4\) — Incorrect. It solves the equation but does not pass through (1, 2): at x = 1 it gives \(y^2 = 6\).