The general solution of the differential equa | vedclass

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(A) Given the differential equation: $\sec^{2} x 	an y \, dx + \sec^{2} y 	an x \, dy = 0$ Rearranging the terms,we get: $\sec^{2} x 	an y \, dx =

Vedclass · Accepted Answer

$	an x 	an y = c$

Vedclass · Answer

(A) Given the differential equation: $\sec^{2} x 	an y \, dx + \sec^{2} y 	an x \, dy = 0$ Rearranging the terms,we get: $\sec^{2} x 	an y \, dx = -\sec^{2} y 	an x \, dy$ Separating the variables: $\frac{\sec^{2} x}{	an x} \, dx = -\frac{\sec^{2} y}{	an y} \, dy$ Integrating both sides: $\int \frac{\sec^{2} x}{	an x} \, dx = -\int \frac{\sec^{2} y}{	an y} \, dy$ Let $u = 	an x$,then $du = \sec^{2} x \, dx$. Similarly,let $v = 	an y$,then $dv = \sec^{2} y \, dy$. Thus,$\int \frac{1}{u} \, du = -\int \frac{1}{v} \, dv$ $\log |u| = -\log |v| + \log |c|$ $\log |	an x| + \log |	an y| = \log |c|$ $\log |	an x 	an y| = \log |c|$ Therefore,the general solution is $	an x 	an y = c$.

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

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