Question

Solution of $\left( {1 + xy} \right)ydx + \left( {1 - xy} \right)xdy = 0$ is
A. $\log \dfrac{x}{y} + \dfrac{1}{{xy}} = c$
B. $\log \dfrac{x}{y} = c$
C. $\log \dfrac{x}{y} - \dfrac{1}{{xy}} = c$
D. $\log \dfrac{y}{x} - \dfrac{1}{{xy}} = c$

Answer 1

Hint : First expand the given equation by multiplying the terms inside the bracket with the term outside the bracket. After that put all the like terms together. Then find their integration using the below mentioned formulas.
Formulas used:
1. $\int \dfrac{1}{x}dx = \log x + c$
2. $d\left( {xy} \right) = xdy + ydx$
3. $\int {x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$
4. $\log a - \log b = \log \dfrac{a}{b}$

Complete step-by-step answer :
We are given to find the solution of the equation $\left( {1 + xy} \right)ydx + \left( {1 - xy} \right)xdy = 0$ .
First we are multiplying the terms inside the bracket with the term outside the bracket in the left hand side of the equation.
$\Rightarrow ydx + xy.ydx + xdy - xy.xdy = 0$
$\Rightarrow ydx + x{y^2}dx + xdy - {x^2}ydy = 0$
On putting the linear terms one side and the square terms another side, we get
$\Rightarrow ydx + xdy + x{y^2}dx - {x^2}ydy = 0$
As we can see in the 3rd and 4th terms of the above equation, we can take out $xy$ common
$\Rightarrow ydx + xdy + xy\left( {ydx - xdy} \right) = 0$
On dividing the whole equation (both LHS and RHS) by ${x^2}{y^2}$ , we get
$\Rightarrow \dfrac{{ydx + xdy + xy\left( {ydx - xdy} \right)}}{{{x^2}{y^2}}} = \dfrac{0}{{{x^2}{y^2}}}$
$\Rightarrow \dfrac{{ydx + xdy + xy\left( {ydx - xdy} \right)}}{{{x^2}{y^2}}} = 0$
$\Rightarrow \dfrac{{ydx + xdy}}{{{x^2}{y^2}}} + \dfrac{{xy\left( {ydx - xdy} \right)}}{{{x^2}{y^2}}} = 0$
${x^2}{y^2}$ can also be written as ${\left( {xy} \right)^2}$
$\Rightarrow \dfrac{{ydx + xdy}}{{{x^2}{y^2}}} + \dfrac{{xy\left( {ydx - xdy} \right)}}{{{{\left( {xy} \right)}^2}}} = 0$
$\Rightarrow \dfrac{{ydx + xdy}}{{{x^2}{y^2}}} + \dfrac{{xy}}{{{{\left( {xy} \right)}^2}}}\left( {ydx - xdy} \right) = 0$
$\Rightarrow \dfrac{{ydx + xdy}}{{{x^2}{y^2}}} + \dfrac{1}{{xy}}\left( {ydx - xdy} \right) = 0$
$\Rightarrow \dfrac{{ydx + xdy}}{{{x^2}{y^2}}} + \left( {\dfrac{{ydx}}{{xy}} - \dfrac{{xdy}}{{xy}}} \right) = 0$
$\Rightarrow \dfrac{{ydx + xdy}}{{{{\left( {xy} \right)}^2}}} + \left( {\dfrac{1}{x}dx - \dfrac{1}{y}dy} \right) = 0$ …… equation (1)
Let us consider $xy$ as t, $xy = t$
On differentiating both sides with respect to t, we get
$\Rightarrow \dfrac{d}{{dt}}\left( {xy} \right) = \dfrac{{dt}}{{dt}}$
$\Rightarrow x\dfrac{{dy}}{{dt}} + y\dfrac{{dx}}{{dt}} = 1$
$\Rightarrow xdy + ydx = dt$
On substituting t in the place of $xy$ and $dt$ in the place of $xdy + ydx$ in equation 1, we get
$\Rightarrow \dfrac{{dt}}{{{{\left( t \right)}^2}}} + \left( {\dfrac{1}{x}dx - \dfrac{1}{y}dy} \right) = 0$
Now we are integrating the above equation
$\Rightarrow \int \left[ {\dfrac{{dt}}{{{{\left( t \right)}^2}}} + \left( {\dfrac{1}{x}dx - \dfrac{1}{y}dy} \right)} \right] = 0$
$\Rightarrow \int \dfrac{{dt}}{{{{\left( t \right)}^2}}} + \int \dfrac{1}{x}dx - \int \dfrac{1}{y}dy = 0$
$\Rightarrow \int {\left( t \right)^{ - 2}}dt + \int \dfrac{1}{x}dx - \int \dfrac{1}{y}dy = 0$
We already know that $\int {x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$ , here x is t and n is -2
This gives, $\Rightarrow \dfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + \int \dfrac{1}{x}dx - \int \dfrac{1}{y}dy = 0$
And the value of $\int \dfrac{1}{x}dx = \log x + c$
$\Rightarrow \dfrac{{{t^{ - 1}}}}{{ - 1}} + c + \log x + c - \left( {\log y + c} \right) = 0$
$\Rightarrow - \dfrac{1}{t} + c + \log x + c - \log y - c = 0$
$\Rightarrow - \dfrac{1}{t} + \log x - \log y = c$
Substituting t as $xy$ and the value of $\log a - \log b = \log \dfrac{a}{b}$
This gives
$\Rightarrow - \dfrac{1}{{xy}} + \log \dfrac{x}{y} = c$
$\Rightarrow \log \dfrac{x}{y} - \dfrac{1}{{xy}} = c$
Therefore, the solution of $\left( {1 + xy} \right)ydx + \left( {1 - xy} \right)xdy = 0$ is $\log \dfrac{x}{y} - \dfrac{1}{{xy}} = c$
Hence, the correct option is Option C.
So, the correct answer is “Option C”.

Note : The variable ‘c’ we get after integrating an expression is a constant, so if we add another constant to it then the result will also be a constant. So do not worry about its sign and coefficient. Be careful with the results of differentiation and integration as they both are inverse to each other.