The general solution of the differential equa | vedclass

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Question

(C) Given differential equation: $	an x 	an y \, dx + \cos^2 x \operatorname{cosec}^2 y \, dy = 0$ Divide the entire equation by $\cos^2 x 	an y$:

Vedclass · Accepted Answer

$	an^2 x - \cot^2 y = C$

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(C) Given differential equation: $	an x 	an y \, dx + \cos^2 x \operatorname{cosec}^2 y \, dy = 0$ Divide the entire equation by $\cos^2 x 	an y$: $\frac{	an x}{\cos^2 x} \, dx + \frac{\operatorname{cosec}^2 y}{	an y} \, dy = 0$ $	an x \sec^2 x \, dx + \cot y \operatorname{cosec}^2 y \, dy = 0$ Integrating both sides: $\int 	an x \sec^2 x \, dx + \int \cot y \operatorname{cosec}^2 y \, dy = C_1$ Let $u = 	an x$,then $du = \sec^2 x \, dx$. Let $v = \cot y$,then $dv = -\operatorname{cosec}^2 y \, dy$. $\int u \, du - \int v \, dv = C_1$ $\frac{u^2}{2} - \frac{v^2}{2} = C_1$ $	an^2 x - \cot^2 y = 2C_1 = C$ Thus,the general solution is $	an^2 x - \cot^2 y = C$.

The general solution of the differential equation $\tan x \tan y \, dx + \cos^2 x \operatorname{cosec}^2 y \, dy = 0$ is

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