The solution of the differential equation d y d x = - x 2 + 3 y 2 3 x 2 + y 2 , y 1 = 0 is

Given,

Differential equation,

$\frac{dy}{dx}=-\left(\frac{x^2+3y^2}{3x^2+y^2}\right) \& y\left(1\right)=0$

Now let $y=vx$

$v+x\frac{dv}{dx}=\frac{dy}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=-\left(\frac{1+3v^2}{3+v^2}\right)$

$\Rightarrow x\frac{dv}{dx}=-\frac{{\left(v+1\right)}^3}{3+v^2}$

$\Rightarrow \frac{\left(3+v^2\right)dv}{{\left(v+1\right)}^3}+\frac{dx}{x}=0$

$\Rightarrow \int \frac{4dv}{{\left(v+1\right)}^3}+\int \frac{dv}{v+1}-\int \frac{2dv}{{\left(v+1\right)}^2}+\int \frac{dx}{x}=0$

$\Rightarrow \frac{-2}{{\left(v+1\right)}^2}+\ln \left|v+1\right|+\frac{2}{v+1}+\ln \left|x\right|=c$

$\Rightarrow \frac{-2x^2}{{\left(x+y\right)}^2}+\ln \left|\frac{x+y}{x}\right|+\frac{2x}{x+y}+\ln \left|x\right|=c$

$\Rightarrow \frac{2xy}{{\left(x+y\right)}^2}+\ln \left|x+y\right|=c$

$\because y\left(1\right)=0$

$\therefore c=0$, as $x=1,y=0$

$\Rightarrow \frac{2xy}{{\left(x+y\right)}^2}+\ln \left|x+y\right|=0$