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

Answer: (C)

$\frac {\pi}{12}$

Answer 2

$\frac {dy}{dx}=\frac {y^2}{xy−x^2−y^2}$
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): $\frac {\pi}{12}$