Match the following differential equation in List–I with the general solutions in      

List–I		List–II
(P)	$(xy+2x^2{y}^2)ydx+(xy-x^2{y}^2)xdy=0$	(1)	$xy+\mathcal{l}n\left|\frac{x}{y}\right|=\frac{1}{xy}+c$
(Q)	$(x^2{y}^2+xy+1)ydx+(x^2{y}^2-xy+1)xdy=0$	(2)	$\mathcal{l}n\left|\frac{x}{y}\right|=xy+c$
(R)	$(x^2+y^2+2x)dx+2ydy=0$	(3)	$x+\mathcal{l}n(x^2+y^2)=c$
(S)	$y(1-xy)dx=x(1+xy)dy$	(4)	$2\mathcal{l}n\left|x\right|=\mathcal{l}n\left|y\right|+\frac{1}{xy}+c$

(where ‘c’ is arbitrary constant)

(P)   $xy(ydx+xdy)+x^2{y}^2(2ydx-xdy)=0$
 $$\begin{gathered} \frac{1}{x^2{y}^2}d\left(xy\right)+\frac{2dx}{x}-\frac{dy}{y}=0 \\ 2\mathcal{l}n\left|x\right|=\mathcal{l}n\left|y\right|+\frac{1}{xy}+C \end{gathered}$$
 (Q)   $ydx+xdy+xy(ydx-xdy)+x^2{y}^2(ydx+xdy)=0$
 $$\begin{gathered} \Rightarrow (1+x^2{y}^2)d\left(xy\right)+xy(ydx-xdy)=0\Rightarrow (\frac{1}{x^2{y}^2}+1)d\left(xy\right)+\frac{dx}{x}-\frac{dy}{y}=0 \\ \Rightarrow \frac{-1}{xy}+xy+\ln \left|x\right|-\ln \left|y\right|+c \end{gathered}$$
(R) $(x^2+y^2+2x)dx+2ydy=0$

$$\begin{gathered} (x^2+y^2)dx+d(x^2+y^2)=0 \\ dx+d\left(\ln (x^2+y^2)\right)=0 \\ x+\ln (x^2+y^2)=C \\ \mathbf{\left(}\mathbf{S}\mathbf{\right)} ydx-xdy=xy(xdy+ydx) \\ \frac{dx}{x}-\frac{dy}{y}=d\left(xy\right) \\ d\left(\mathcal{l}n\left(x\right)\right)-d\left(\mathcal{l}n\left(y\right)\right)=d\left(xy\right) \\ \mathcal{l}n\left|\frac{x}{y}\right|=xy+c \end{gathered}$$