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=√3x…

Given ${y}^{2}dx+({x}^{2}-xy+{y}^{2})dy=0$y2dx+(x2−xy+y2)dy=0

$$\Rightarrow \frac{dx}{dy}=-\frac{({x}^{2}-xy+{y}^{2})}{{y}^{2}}...(i)$$⇒dydx​=−y2(x2−xy+y2)​...(i)

Now let $x=vy$x=vy $\Rightarrow v+y\frac{dv}{dy}=\frac{dx}{dy}$⇒v+ydydv​=dydx​

Now putting in equation $(i)$(i) we get

$$\Rightarrow v+y\frac{dv}{dy}=-({v}^{2}-v+1)$$⇒v+ydydv​=−(v2−v+1)

$$\Rightarrow y\frac{dv}{dy}=-{v}^{2}-1$$⇒ydydv​=−v2−1

$$\Rightarrow \frac{-dv}{1+{v}^{2}}=\frac{dy}{y}$$⇒1+v2−dv​=ydy​

Now Integrating Both side

$$\Rightarrow -\int \frac{dy}{1+{v}^{2}}=\int \frac{dy}{y}$$⇒−∫1+v2dy​=∫ydy​

$\Rightarrow -{\mathrm{tan}}^{-1}v=\mathrm{log}|y|+c$⇒−tan−1v=log∣y∣+c $\Rightarrow -{\mathrm{tan}}^{-1}(\frac{x}{y})=\mathrm{log}|y|+c$⇒−tan−1(yx​)=log∣y∣+c

Also this curve passes through $(1,1)$(1,1)

$\Rightarrow {\mathrm{tan}}^{-1}1=\mathrm{log}|1|+c$⇒tan−11=log∣1∣+c $\Rightarrow c=-\frac{\pi }{4}$⇒c=−4π​

$$\Rightarrow -{\mathrm{tan}}^{-1}\frac{x}{y}=\mathrm{log}|y|-\frac{\pi }{4}$$⇒−tan−1yx​=log∣y∣−4π​

Now putting $y=\sqrt{3}x$y=3​x and $x=\alpha$x=α we get

$=-{\mathrm{tan}}^{-1}\frac{x}{\sqrt{3}x}=\mathrm{log}|\sqrt{3}\alpha |-\frac{\pi }{4}$=−tan−13​xx​=log∣3​α∣−4π​ $=-{\mathrm{tan}}^{-1}\frac{1}{\sqrt{3}}=\mathrm{log}(\sqrt{3}\alpha )-\frac{\pi }{4}$=−tan−13​1​=log(3​α)−4π​

$$=-\frac{\pi }{6}+\frac{\pi }{4}=\mathrm{log}|\sqrt{3}\alpha |$$=−6π​+4π​=log∣3​α∣

$$\Rightarrow \mathrm{log}|\sqrt{3}\alpha |=\frac{\pi }{12}$$⇒log∣3​α∣=12π​