Question

The solution of the differential equation $\frac{d y}{d x}=-\left(\frac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0$ is

• A. $\log _e|x+y|-\frac{x y}{(x+y)^2}=0$
• B. $\log _e|x+y|-\frac{2 x y}{(x+y)^2}=0$
• C. $\log _e|x+y|+\frac{x y}{(x+y)^2}=0$
• D. $\log _e|x+y|+\frac{2 x y}{(x+y)^2}=0$

Answer 1

Answer: (D)

Put $y=v x$
$ v+x \frac{d v}{d x}=-\left(\frac{1+3 v^2}{3+v^2}\right)$
$ x \frac{d v}{d x}=-\frac{(v+1)^3}{3+v^2} $
$ \frac{\left(3+v^2\right) d v}{(v+1)^3}+\frac{d x}{x}=0 $
$ \int \frac{4 d v}{(v+1)^3}+\int \frac{d v}{v+1}-\int \frac{2 d v}{(v+1)^2}+\int \frac{d x}{x}=0$
$ \frac{-2}{(v+1)^2}+\ln (v+1)+\frac{2}{v+1}+\ln x=c $
$ \frac{-2 x^2}{(x+y)^2}+\ln \left(\frac{x+y}{x}\right)+\frac{2 x}{x+y}+\ln x=c $
$\frac{2 x y}{(x+y)^2}+\ln (x+y)=c$
$ \therefore c=0, \text { as } x=1, y=0$
$\therefore \frac{2 x y}{(x+y)^2}+\ln (x+y)=0$