Solve: $(3xy+y^2)dx+(x^2+xy)dy=0 \qquad y(1)=1$

Question

$(a)\;x^2(2x+y)=3 \\ (b)\;y^2(2x+y)=3 \\ (c)\;x^2y(2x+y)=3 \\ (d)\;x^2y(2x+3)=9 $

Answer 1

$\large\frac{dy}{dx}=-\large\frac{3xy+y^2}{x^2+xy}$

$y=vx$

$v+xv'= -\large\frac{3v+v^2}{1+v}$

$xv'= -[-v+\large\frac{3v+v^2}{1+v}]$

$\qquad =\large\frac{-2v (2+v)}{1+v}$

Separating variables ,

$\large\frac{(1+v)dv}{v(2+v)}=\large\frac{-2}{x} $$dx$

$\large\frac{1}{2} \bigg[\large\frac{1}{v}+\frac{1}{2+v}\bigg]dv=-\large\frac{-2}{x}$$dx$

Integrate $\log (v+2+v))+4 \log x =2 \log c$

$v (2+v)x^4=c^2$

$v=\large\frac{y}{x}$

$x^2y (2x+y)=c^2=k$

=> $y(1)=1$

$x^2y (2x+y)=3$

Hence c is the correct answer.