Question

Solution of $\left( {1 + xy} \right)ydx + \left( {1 - xy} \right)xdy = 0$ is
1. $\log \left( {\dfrac{x}{y}} \right) + \left( {\dfrac{1}{{xy}}} \right) = c$
2. $\log \left( {\dfrac{x}{y}} \right) = c$
3. $\log \left( {\dfrac{x}{y}} \right) - \left( {\dfrac{1}{{xy}}} \right) = c$
4. None of these

Answer 1

Hint: In this question, we have to find the solution of the given differential equation i.e., $\left( {1 + xy} \right)ydx + \left( {1 - xy} \right)xdy = 0$. The first step is to multiply the terms and then take the terms common outside where the differential terms will be inside the bracket. Solve further using product rule and then integrate the required function.

Formula Used:
Product rule of differentiation –
$\dfrac{d}{{dx}}\left( {xy} \right) = x\dfrac{d}{{dx}}\left( y \right) + y\dfrac{d}{{dx}}\left( x \right)$
Integration formula –
$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$, $\int {\dfrac{1}{x}dx = \log x} $
Logarithm formula –
$\log x - \log y = \log \dfrac{x}{y}$

Complete step by step Solution:
Given that,
$\left( {1 + xy} \right)ydx + \left( {1 - xy} \right)xdy = 0$
\[ydx + x{y^2}dx + xdy - {x^2}ydy = 0\]
It can be written as, \[\left( {ydx + xdy} \right) + xy\left( {ydx - xdy} \right) = 0\]
Using product rule of differentiation where $d = \dfrac{d}{{dx}}$,
\[d\left( {xy} \right) + xy \times \dfrac{{xy}}{{xy}}\left( {ydx - xdy} \right) = 0\]
\[d\left( {xy} \right) + {\left( {xy} \right)^2}\left( {\dfrac{{ydx - xdy}}{{xy}}} \right) = 0\]
\[{\left( {xy} \right)^{ - 2}}d\left( {xy} \right) + \dfrac{{dx}}{x} - \dfrac{{dy}}{y} = 0\]
Integrating both the sides,
\[\dfrac{{{{\left( {xy} \right)}^{ - 1}}}}{{ - 1}} + \log x - \log y = c\]
It implies that, \[\log \dfrac{x}{y} - \dfrac{1}{{xy}} = c\]

Hence, the correct option is 3.

Note: To solve such problems, one should have a good knowledge of the basics and formulas of differentiation and integration. Also, the differential equation is a mathematical equation that contains one or more terms and the derivatives of one variable (the dependent variable) with respect to another variable (i.e., the independent variable). This question can also be solved by converting this to the general differential equation $\dfrac{{dy}}{{dx}} + Px = Q$ and then finding the integral factor, then putting it into the formula of the differential equation.