Match the differential equations given under List I
\( \mathrm{P} \)
with their respective solutions given under List II.
\begin{tabular}{|l|l|l|l|}
\hline \multicolumn{2}{|c|}{ List-I } & \multicolumn{2}{c|}{ List-II } \\
\hline A. & \begin{tabular}{l}
\( (1+x y) x d y+ \) \\
\( (1-x y) y d x=0 \)
\end{tabular} & i. & \begin{tabular}{l}
\( \frac{x^{3}}{3}+\frac{y^{3}}{3}-2 x^{2} y \) \\
\( -2 x y^{2}=C \)
\end{tabular} \\
\hline B. & \begin{tabular}{l}
\( \left(x^{2}-4 x y-2 y^{2}\right) d x \) \\
\( +\left(y^{2}-4 x y-\right. \) \\
\( \left.2 x^{2}\right) d y=0 \)
\end{tabular} & ii. & \( \frac{-1}{x y}+\log \left(\frac{y}{x}\right)=C \) \\
\hline C. & \begin{tabular}{l}
\( e^{y} d x+\left(x e^{y}-2 y\right) \) \\
\( d y=0 \)
\end{tabular} & iii. & \( \sqrt{x^{2}+y^{2}}-\frac{x}{y}=C \) \\
\hline D. & \( \frac{x d x+y d y}{\sqrt{x^{2}+y^{2}}}=\frac{y d x-x d y}{y^{2}} \) & iv. & \( x e^{y}-y^{2}=C \) \\
\hline
\end{tabular}
A B
(1) iii iv
(2) ii
(3) iii
(4) iv
C
i
iv
iv
ii
D
ii
iii
ii
iii
0