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

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

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$\log \left(\frac{x}{y}\right) = \frac{1}{xy} + k$,where $k$ is the constant of integration.

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(B) Given differential equation: $(1+xy)y \, dx + (1-xy)x \, dy = 0$ Expanding the terms: $y \, dx + xy^2 \, dx + x \, dy - x^2y \, dy = 0$ Rearranging: $(y \, dx + x \, dy) + xy(y \, dx - x \, dy) = 0$ Dividing by $x^2y^2$: $\frac{y \, dx + x \, dy}{x^2y^2} + \frac{y \, dx - x \, dy}{xy} = 0$ This can be written as: $\frac{d(xy)}{(xy)^2} - \left(\frac{x \, dy - y \, dx}{xy}\right) = 0$ Integrating both sides: $\int \frac{d(xy)}{(xy)^2} - \int d\left(\log \frac{x}{y}\right) = \int 0$ $-\frac{1}{xy} - \log \left(\frac{x}{y}\right) = -k$ $\log \left(\frac{x}{y}\right) = \frac{1}{xy} + k$

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

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