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Question

(C) Given the differential equation: $\sec^2 x 	an y \, dx + \sec^2 y 	an x \, dy = 0$ Divide both sides by $	an x 	an y$: $\frac{\sec^2 x}{	an x

Vedclass · Accepted Answer

$	an x = c \cot y$

Vedclass · Answer

(C) Given the differential equation: $\sec^2 x 	an y \, dx + \sec^2 y 	an x \, dy = 0$ Divide both sides by $	an x 	an y$: $\frac{\sec^2 x}{	an x} \, dx + \frac{\sec^2 y}{	an y} \, dy = 0$ Integrating both sides: $\int \frac{\sec^2 x}{	an x} \, dx + \int \frac{\sec^2 y}{	an y} \, dy = \int 0 \, dx$ Using the substitution $u = 	an x$ $(du = \sec^2 x \, dx)$ and $v = 	an y$ $(dv = \sec^2 y \, dy)$,we get: $\ln|	an x| + \ln|	an y| = \ln|c|$ Using the property $\ln A + \ln B = \ln(AB)$: $\ln|	an x 	an y| = \ln|c|$ $	an x 	an y = c$ Since $	an y = \frac{1}{\cot y}$,we can rewrite this as: $	an x = c \cot y$.

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

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