Question

If the solution curve \(y = y(x)\) of the differential equation\( y^2dx + (x^2 – xy + y^2)dy = 0\), which passes through the point \((1,1)\) and intersects the line \(y=\sqrt 3x\) at the point \((α,\sqrt 3α)\), then value of \(log_e(\sqrt 3α)\) is equal to :

• A. \(\frac {\pi}{3}\)
• B. \(\frac {\pi}{2}\)
• C. \(\frac {\pi}{12}\)
• D. \(\frac {\pi}{6}\)

Answer 1

The Correct Option is C

\(\frac {dy}{dx}=\frac {y^2}{xy−x^2−y^2}\)
Put \(y = vx\) we get
\(v+x\frac {dv}{dx}=\frac {v^2}{v−1−v^2}\)

\(⇒ x\frac {dv}{dx}=\frac {v^2−v^2+v+v^3}{v−1−v^2}\)

\(⇒∫\frac {v−1−v^2}{v(1+v^2)}dv=∫\frac {dx}{x}\)

\(tan^{−1}⁡(\frac yx)−ln(\frac yx)=ln\ x+c\)
As it passes through \((1, 1)\)
\(c=\frac {\pi}{4}\)
\(⇒ tan^{−1⁡}(\frac yx)−ln(\frac yx)=ln\ x+\frac {\pi}{4}\)
Put \(y=\sqrt 3x\) we get.
\(⇒ \frac {\pi}{3}−ln\sqrt 3=ln\ x+\frac {\pi}{4}\)
\(⇒ ln\ x=\frac {\pi}{12}−ln\sqrt 3=ln α\)
\(∴ln(\sqrt 3α)=ln\sqrt 3+ln α\)
\(=ln\sqrt 3+\frac {π}{12}−ln\sqrt 3\)
\(=\frac {\pi}{12}\)

So, the correct option is (C): \(\frac {\pi}{12}\)