The solution of (1 + xy)y,dx + (1 - xy)x,dy = | vedclass

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(B) Given the differential equation: $(1 + xy)y\,dx + (1 - xy)x\,dy = 0$. Expanding the terms: $ydx + xy^2\,dx + x\,dy - x^2y\,dy = 0$. Rearranging th

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$\log \left( \frac{x}{y} \right) = \frac{1}{xy} + k$

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(B) Given the differential equation: $(1 + xy)y\,dx + (1 - xy)x\,dy = 0$. Expanding the terms: $ydx + xy^2\,dx + x\,dy - x^2y\,dy = 0$. Rearranging the terms: $(ydx + x\,dy) + (xy^2\,dx - x^2y\,dy) = 0$. Dividing the entire equation by $x^2y^2$: $\frac{ydx + x\,dy}{x^2y^2} + \frac{xy^2\,dx}{x^2y^2} - \frac{x^2y\,dy}{x^2y^2} = 0$. This simplifies to: $\frac{d(xy)}{(xy)^2} + \frac{dx}{x} - \frac{dy}{y} = 0$. Integrating both sides: $\int \frac{d(xy)}{(xy)^2} + \int \frac{dx}{x} - \int \frac{dy}{y} = \int 0$. $-\frac{1}{xy} + \log|x| - \log|y| = C$. Using the property of logarithms: $\log\left(\frac{x}{y}\right) = \frac{1}{xy} + C$.

The solution of $(1 + xy)y\,dx + (1 - xy)x\,dy = 0$ is

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