Question

Question :

$\begin{gathered}\sf{y(1 + xy + x^{2}y^{2})dx + x(1 - xy + x^{2}y^{2})dy = 0} \\ \\ \end{gathered}$
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Answer 1

Answer:

$\orange{xy - \frac{1}{xy} + ln( \frac{x}{y} ) = c}$

Step-by-step explanation:

Arrange given equation as following

$(ydx + xdy) + ( {x}^{2} {y}^{3}dx \\ + {x}^{3} {y}^{2} dy) + (x {y}^{2} dx - {x}^{2} ydy) = 0$

Now we will try to make this exact differential . Notice first terms (ydx + xdy ) looks like d(xy) so it can be d(xy) or some power multiply of xy with it .

Noticing all three terms we can make some guesses .

Here is what will work

divide whole equation by x^2.y^2

$\frac{(ydx + xdy) }{ {x}^{2} {y}^{2} } + \\ \frac{ ( {x}^{2} {y}^{3}dx + {x}^{3} {y}^{2} dy) } { {x}^{2} {y}^{2} }+ \\ \frac{ (x {y}^{2} dx - {x}^{2} ydy) }{ {x}^{2} {y}^{2}} = 0 \\ \\ \\ \\ d( \frac{ - 1}{ xy} ) + (ydx + xdy) + \frac{ydx - xdy}{xy} = 0 \\ \\ \\ d( - \frac{1}{xy} ) + d(xy) + \frac{dx}{x} - \frac{dy}{y} = 0 \\ \\ \int \: d( - \frac{1}{xy} ) + \int d(xy) + \int \frac{dx}{x} - \int\frac{dy}{y} = 0 \\ \\ \frac{ - 1}{xy} + xy + ln(x) - ln(y) = c \\ \\ xy - \frac{1}{xy} + ln( \frac{x}{y} ) = c$