Question

If the solution curve $y=y\left(x\right)$ of the differential equation $y^{2}dx+(x^2–xy+y^2)dy=0$, which passes through the point $(1,1)$ and intersects the line $y=\sqrt{3}x$ at the point $(\alpha ,\sqrt{3}\alpha )$ , then value of ${\log }_e\left(\sqrt{3}\alpha \right)$ is equal to:

Answer 1

$\frac{dy}{dx}=\frac{y^2}{xy-x^2-y^2}$
Put $y=vx$ we get
$v+x\frac{dv}{dx}=\frac{v^2}{v-1-v^2}$
$\Rightarrow x\frac{dv}{dx}=\frac{v^2-v^2+v+v^3}{v-1-v^2}$
$\Rightarrow \int \frac{v-1-v^2}{v(1+v^2)}dv=\int \frac{dx}{x}$
$\Rightarrow {\tan }^{-1}v-\ln v=\ln x+C$
putting $v=\frac{y}{x}$, we get
${\tan }^{-1}\left(\frac{y}{x}\right)-\ln \left(\frac{y}{x}\right)=\ln x+C$
As it passes through $(1,1)$
$C=\frac{\pi }{4}$
$\Rightarrow {\tan }^{-1}\left(\frac{y}{x}\right)-\ln \left(\frac{y}{x}\right)=\ln x+\frac{\pi }{4}$
Put $y=\sqrt{3}x$, we get
$\Rightarrow \frac{\pi }{3}-\ln \sqrt{3}=\ln x+\frac{\pi }{4}$
$\Rightarrow \ln x=\frac{\pi }{12}-\ln \sqrt{3}=\ln \alpha$
Now, $\ln \left(\sqrt{3}\alpha \right)=\ln \sqrt{3}+\ln \alpha$
$=\ln \sqrt{3}+\frac{\pi }{12}-\ln \sqrt{3}=\frac{\pi }{12}$