The solution of $(1+x y) y \mathrm{~d} x+(1-x y) x \mathrm{~d} y=0$ is.
The solution of $(1+x y) y \mathrm{~d} x+(1-x y) x \mathrm{~d} y=0$ is.
- $\log \left(\frac{x}{y}\right)+\frac{1}{x y}=\mathrm{k}$, where $\mathrm{k}$ is constant of integration.
- $\log \left(\frac{x}{y}\right)=\frac{1}{x y}+\mathrm{k}$, where $\mathrm{k}$ is constant of integration.
- $\log \left(\frac{x}{y}\right)+x y=\mathrm{k}$, where $\mathrm{k}$ is constant of integration.
- $\log \left(\frac{x}{y}\right)=x y+\mathrm{k}$, where $\mathrm{k}$ is constant of integration.
Solution
$(1+x y) y \mathrm{~d} x+(1-x y) x \mathrm{~d} y=0$
$\Rightarrow y \mathrm{~d} x+x \mathrm{~d} y+x y^2 \mathrm{~d} x-x^2 y \mathrm{~d} y=0$
$\Rightarrow \frac{y \mathrm{~d} x+x \mathrm{~d} y}{x^2 y^2}+\frac{\mathrm{d} x}{x}-\frac{\mathrm{d} y}{y}=0$
$\Rightarrow \frac{\mathrm{d}(x y)}{x^2 y^2}+\frac{\mathrm{d} x}{x}-\frac{\mathrm{d} y}{y}=0$
Integrating on both sides, we get
$-\frac{1}{x y}+\log x-\log y=\mathrm{k}$
$\Rightarrow \log \left(\frac{x}{y}\right)=\frac{1}{x y}+\mathrm{k}$
Asked in: MHT CET 2023 (13 May Shift 2)
Practice more Differential Equations questions on Aicharya