Answer
\[(b)\;\; y^2-x^2=Kx^3\]
Work Step by Step
$(a)\;$ $x^2+3y^2=2Cy$ ___(1)
Differentiate (1) with respect to $x$
$2x+6y\frac{dy}{dx}=2C\frac{dy}{dx}$ ___(2)
Put $2C=\frac{x^2+3y^2}{y}$ from (1) in (2)
$2x+6y\frac{dy}{dx}=\left(\frac{x^2+3y^2}{y}\right)\frac{dy}{dx}$
$2x=\frac{dy}{dx}\left(\frac{x^2+3y^2-6y^2}{y}\right)$
$\frac{dy}{dx}=\frac{2xy}{x^2-3y^2}$ ____(3)
Hence differential equation of (1) is $\frac{dy}{dx}=\frac{2xy}{x^2-3y^2}$.
$(b)\;$ Replace $\frac{dy}{dx}$ by $-\frac{dx}{dy}$ in (3) [ for orthogonal trajectories]
$-\frac{dx}{dy}=\frac{2xy}{x^2-3y^2}$
$\frac{dy}{dx}=\frac{3y^2-x^2}{2xy}$
$\frac{dy}{dx}=\Large\frac{3\left(\frac{y}{x}\right)^2-1}{\frac{2y}{x}}$ ___(4)
Put $v=\frac{y}{x}$ ____(5)
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
(4) becomes
$v+x\frac{dv}{dx}=\frac{3v^2-1}{2v}$
$x\frac{dv}{dx}=\frac{3v^2-1}{2v}-v=\frac{v^2-1}{2v}$
$\frac{2v}{v^2-1}dv=\frac{1}{x}dx$
Integrating,
$\int\frac{2v}{v^2-1}dv=\int\frac{1}{x}dx$
$\ln|v^2-1|=\ln x+\ln K$
$K$ is constant of integration
$v^2-1=Kx$
$\frac{y^2}{x^2}-1=Kx$ $\;\;\;$ [From (5)]
$y^2-x^2=Kx^3$
Hence orthogonal trajectories of (1) is $y^2-x^2=Kx^3$.
