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)

$\log _e|x+y|+\frac{2 x y}{(x+y)^2}=0$

Answer 2

Put y=vx
v+xdxdv​=−(3+v21+3v2​)
xdxdv​=−3+v2(v+1)3​
(v+1)3(3+v2)dv​+xdx​=0
∫(v+1)34dv​+∫v+1dv​−∫(v+1)22dv​+∫xdx​=0
(v+1)2−2​+ln(v+1)+v+12​+lnx=c
(x+y)2−2x2​+ln(xx+y​)+x+y2x​+lnx=c
(x+y)22xy​+ln(x+y)=c
∴c=0, as x=1,y=0
∴(x+y)22xy​+ln(x+y)=0
So, the correct answer is (D) : $\log _e|x+y|+\frac{2 x y}{(x+y)^2}=0$