If \(y = y(x)\) is the solution of the differential equation \(x^2\displaystyle\frac{dy}{dx} = x^2 + xy + y^2\), \(x \gt 0\), with \(y(1) = 1\), then the value of y at \(x = e^{\frac{\pi}{12}}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\sqrt{3}\,e^{\frac{\pi}{12}}\)
Dividing by \(x^2\):
\( \displaystyle \frac{dy}{dx} = 1 + \frac{y}{x} + \frac{y^2}{x^2} \)
This is a homogeneous equation. With \(y = vx\), \(\frac{dy}{dx} = v + x\frac{dv}{dx}\):
\( \displaystyle v + x\frac{dv}{dx} = 1 + v + v^2 \)
\( \displaystyle \frac{dv}{1 + v^2} = \frac{dx}{x} \)
\( \tan^{-1} v = \log x + C \)
\( \displaystyle \tan^{-1}\left(\frac{y}{x}\right) = \log x + C \)
Using \(y(1) = 1\):
\( \displaystyle \tan^{-1} 1 = 0 + C \Rightarrow C = \frac{\pi}{4} \)
At \(x = e^{\frac{\pi}{12}}\), \(\log x = \frac{\pi}{12}\):
\( \displaystyle \tan^{-1}\left(\frac{y}{x}\right) = \frac{\pi}{12} + \frac{\pi}{4} = \frac{\pi}{3} \)
\( \displaystyle \frac{y}{x} = \tan\frac{\pi}{3} = \sqrt{3} \)
\( y = \sqrt{3}\,e^{\frac{\pi}{12}} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{1}{\sqrt{3}}\,e^{\frac{\pi}{12}}\) — Incorrect. This uses \(\tan\frac{\pi}{6}\), from \(\frac{\pi}{4} - \frac{\pi}{12}\); but \(\log x\) is added to the constant \(\frac{\pi}{4}\), not subtracted.
Option (3) \(e^{\frac{\pi}{12}}\) — Incorrect. This keeps \(\frac{y}{x} = 1\), its value at \(x = 1\); the ratio changes with x.
Option (4) \(2e^{\frac{\pi}{12}}\) — Incorrect. The angle is \(\frac{\pi}{3}\), whose tangent is \(\sqrt{3}\), not 2.