Solution of \[\left( {1 + xy} \right)ydx + \left( {1 - xy} \right)xdy = 0\] is
1.\[\log \dfrac{x}{y} + \dfrac{1}{{xy}} = c\]
2.\[\log \dfrac{x}{y} = c\]
3.\[\log \dfrac{x}{y} - \dfrac{1}{{xy}} = c\]
4.\[\log \dfrac{y}{x} - \dfrac{1}{{xy}} = c\]
Answer
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Hint: In the given question, we have been given that there is a differentiation equation in two variables. We have to calculate the value of the differentiation equation. For doing that we are going to need to first of all, rearrange the terms with different differentiating factors. Then, we are going to have to apply the required formulae and solve for the given variables. Here, the terms are in the differentiating form, so we are going to have to apply integration on the two sides of the equality for solving this question.
Complete step-by-step answer:
The given equation is \[\left( {1 + xy} \right)ydx + \left( {1 - xy} \right)xdy = 0\].
First, we are going to rearrange this equation,
\[ydx + xdy + x{y^2}dx - {x^2}ydy = 0\]
Now, dividing both sides by \[{x^2}{y^2}\],
$\Rightarrow$ \[\dfrac{1}{{{x^2}{y^2}}}\left( {ydx + xdy + x{y^2}dx - {x^2}ydy} \right) = \dfrac{0}{{{x^2}{y^2}}}\]
or \[\dfrac{{ydx + xdy}}{{{x^2}{y^2}}} + \dfrac{{dx}}{x} - \dfrac{{dy}}{y} = 0\]
Now, we are going to apply the
\[d\left( { - \dfrac{1}{{xy}}} \right) = \dfrac{{ydx + xdy}}{{{x^2}{y^2}}}\]
\[d\left( {\log x} \right) = \dfrac{{dx}}{x}\]
\[d\left( {\log y} \right) = \dfrac{{dy}}{y}\]
Now, substituting these in \[\dfrac{{ydx + xdy}}{{{x^2}{y^2}}} + \dfrac{{dx}}{x} - \dfrac{{dy}}{y} = 0\], we get:
\[d\left( { - \dfrac{1}{{xy}}} \right) + d\left( {\log x} \right) - d\left( {\log y} \right) = 0\]
Now, integrating both sides, we get:
$\Rightarrow$ \[\int {d\left( { - \dfrac{1}{{xy}}} \right) + d\left( {\log x} \right) - d\left( {\log y} \right)} = \int 0 \]
$\Rightarrow$ \[ - \dfrac{1}{{xy}} + \log x - \log y = c\]
We know, \[\log \dfrac{a}{b} = \log a - \log b\]
Hence, \[ - \dfrac{1}{{xy}} + \log x - \log y = \log \dfrac{x}{y} - \dfrac{1}{{xy}}\]
Hence, \[\log \dfrac{x}{y} - \dfrac{1}{{xy}} = c\]
Thus, the correct option is (3).
Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we think about the formulae which contain the known and the unknown and pick the one which is the most suitable and the most effective for finding the answer of the given question. Then we put in the knowns into the formula, evaluate the answer and find the unknown. It is really important to follow all the steps of the formula to solve the given expression very carefully and in the correct order, because even a slightest error is going to make the whole expression awry and is going to give us an incorrect answer.
Complete step-by-step answer:
The given equation is \[\left( {1 + xy} \right)ydx + \left( {1 - xy} \right)xdy = 0\].
First, we are going to rearrange this equation,
\[ydx + xdy + x{y^2}dx - {x^2}ydy = 0\]
Now, dividing both sides by \[{x^2}{y^2}\],
$\Rightarrow$ \[\dfrac{1}{{{x^2}{y^2}}}\left( {ydx + xdy + x{y^2}dx - {x^2}ydy} \right) = \dfrac{0}{{{x^2}{y^2}}}\]
or \[\dfrac{{ydx + xdy}}{{{x^2}{y^2}}} + \dfrac{{dx}}{x} - \dfrac{{dy}}{y} = 0\]
Now, we are going to apply the
\[d\left( { - \dfrac{1}{{xy}}} \right) = \dfrac{{ydx + xdy}}{{{x^2}{y^2}}}\]
\[d\left( {\log x} \right) = \dfrac{{dx}}{x}\]
\[d\left( {\log y} \right) = \dfrac{{dy}}{y}\]
Now, substituting these in \[\dfrac{{ydx + xdy}}{{{x^2}{y^2}}} + \dfrac{{dx}}{x} - \dfrac{{dy}}{y} = 0\], we get:
\[d\left( { - \dfrac{1}{{xy}}} \right) + d\left( {\log x} \right) - d\left( {\log y} \right) = 0\]
Now, integrating both sides, we get:
$\Rightarrow$ \[\int {d\left( { - \dfrac{1}{{xy}}} \right) + d\left( {\log x} \right) - d\left( {\log y} \right)} = \int 0 \]
$\Rightarrow$ \[ - \dfrac{1}{{xy}} + \log x - \log y = c\]
We know, \[\log \dfrac{a}{b} = \log a - \log b\]
Hence, \[ - \dfrac{1}{{xy}} + \log x - \log y = \log \dfrac{x}{y} - \dfrac{1}{{xy}}\]
Hence, \[\log \dfrac{x}{y} - \dfrac{1}{{xy}} = c\]
Thus, the correct option is (3).
Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we think about the formulae which contain the known and the unknown and pick the one which is the most suitable and the most effective for finding the answer of the given question. Then we put in the knowns into the formula, evaluate the answer and find the unknown. It is really important to follow all the steps of the formula to solve the given expression very carefully and in the correct order, because even a slightest error is going to make the whole expression awry and is going to give us an incorrect answer.
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