What is the solution of the differential equation \[\left( {{x^2} - y{x^2}} \right)\dfrac{{dy}}{{dx}} + {y^2} + x{y^2} = 0\]?
A. \[\log \left( {\dfrac{x}{y}} \right) = \dfrac{1}{x} + \dfrac{1}{y} + c\]
B. \[\log \left( {\dfrac{y}{x}} \right) = \dfrac{1}{x} + \dfrac{1}{y} + c\]
C. \[\log \left( {xy} \right) = \dfrac{1}{x} + \dfrac{1}{y} + c\]
D. \[\log \left( {xy} \right) + \dfrac{1}{x} + \dfrac{1}{y} = c\]

Hint: First we separate all variables of the differential equation. Then take integration operations on both sides of the equation and by using integration formulas solve it.

Formula used:
Integration formula:
\[\int {\dfrac{1}{x}dx} = \log x + c\]
\[\int {\dfrac{1}{{{x^2}}}dx} = - \dfrac{1}{x} + c\]

Complete step by step solution:
Given differential equation is \[\left( {{x^2} - y{x^2}} \right)\dfrac{{dy}}{{dx}} + {y^2} + x{y^2} = 0\].
Now we will separate the variable of the differential equation:
\[\left( {{x^2} - y{x^2}} \right)\dfrac{{dy}}{{dx}} = - {y^2}\left( {1 + x} \right)\]
\[ \Rightarrow {x^2}\left( {1 - y} \right)\dfrac{{dy}}{{dx}} = - {y^2}\left( {1 + x} \right)\]
Divide both sides by \[{x^2}{y^2}\]
\[ \Rightarrow \dfrac{{{x^2}\left( {1 - y} \right)}}{{{x^2}{y^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{{y^2}\left( {1 + x} \right)}}{{{x^2}{y^2}}}\]
Cancel out same terms from the denominator and numerator:
\[ \Rightarrow \dfrac{{{x^2}\left( {1 - y} \right)}}{{{x^2}{y^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{{y^2}\left( {1 + x} \right)}}{{{x^2}{y^2}}}\]
\[ \Rightarrow \dfrac{{\left( {1 - y} \right)}}{{{y^2}}}\dfrac{{dy}}{{dx}} = - \dfrac{{\left( {1 + x} \right)}}{{{x^2}}}\]
\[ \Rightarrow \dfrac{{\left( {1 - y} \right)}}{{{y^2}}}dy = - \dfrac{{\left( {1 + x} \right)}}{{{x^2}}}dx\]
Break it into terms:
\[ \Rightarrow \left( {\dfrac{1}{{{y^2}}} - \dfrac{1}{y}} \right)dy = - \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{x}} \right)dx\]
\[ \Rightarrow \left( {\dfrac{1}{{{y^2}}} - \dfrac{1}{y}} \right)dy + \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{x}} \right)dx = 0\]
Taking integration on both sides:
\[ \Rightarrow \int {\dfrac{1}{{{y^2}}}dy} - \int {\dfrac{1}{y}dy} + \int {\dfrac{1}{{{x^2}}}dx} + \int {\dfrac{1}{x}dx} = 0\]
Applying integration formula:
\[ \Rightarrow - \dfrac{1}{y} - \log y - \dfrac{1}{x} + \log x = c\]
Rewrite the above equation:
\[ \Rightarrow \log x - \log y = c + \dfrac{1}{x} + \dfrac{1}{y}\]
Applying quotient formula of logarithm:
\[ \Rightarrow \log \left( {\dfrac{x}{y}} \right) = c + \dfrac{1}{x} + \dfrac{1}{y}\]
Hence option A is the correct option.

Note: Students often do mistake to integrate \[\int {\dfrac{1}{{{x^2}}}dx} \].They apply the formula \[\int {\dfrac{1}{{{x^2}}}dx} = \dfrac{1}{x} + c\] which is incorrect formula. The correct formula is \[\int {\dfrac{1}{{{x^2}}}dx} = - \dfrac{1}{x} + c\].