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Question

(A) The given differential equation is: $\sec^{2} x 	an y \, dx + \sec^{2} y 	an x \, dy = 0$. Dividing both sides by $	an x 	an y$,we get: $\frac

Vedclass · Accepted Answer

$	an x 	an y = C$

Vedclass · Answer

(A) The given differential equation is: $\sec^{2} x 	an y \, dx + \sec^{2} y 	an x \, dy = 0$. Dividing both sides by $	an x 	an y$,we get: $\frac{\sec^{2} x}{	an x} \, dx + \frac{\sec^{2} y}{	an y} \, dy = 0$. Integrating both sides: $\int \frac{\sec^{2} x}{	an x} \, dx + \int \frac{\sec^{2} y}{	an y} \, dy = C_1$. Let $u = 	an x$,then $du = \sec^{2} x \, dx$. Let $v = 	an y$,then $dv = \sec^{2} y \, dy$. Substituting these into the integral: $\int \frac{1}{u} \, du + \int \frac{1}{v} \, dv = C_1$. $\ln|u| + \ln|v| = C_1$. $\ln|	an x| + \ln|	an y| = C_1$. $\ln|	an x 	an y| = C_1$. Taking the exponential of both sides: $	an x 	an y = e^{C_1} = C$.

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

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