The general solution of the differential equa | vedclass

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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

Vedclass · Accepted Answer

$	an x 	an y = c$

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: $\int \frac{\sec^2 x}{	an x} \, dx + \int \frac{\sec^2 y}{	an y} \, dy = C_1$. 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: $\int \frac{1}{u} \, du + \int \frac{1}{v} \, dv = C_1$. $\ln|u| + \ln|v| = C_1$. $\ln|	an x| + \ln|	an y| = C_1$. Using the property $\ln a + \ln b = \ln(ab)$: $\ln|	an x 	an y| = C_1$. Taking the exponential of both sides: $|	an x 	an y| = e^{C_1}$. Let $e^{C_1} = c$,so $	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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