Question

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

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

Answer 1

Answer: (D)

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