The general solution of the differential equa | vedclass

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Question

(B) Given differential equation is $	an(y) dx + \sec^2(y) 	an(x) dy = 0$. Rearranging the terms to separate variables,we get: $\sec^2(y) 	an(x) dy

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$\sin(x) 	an(y) = c$

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(B) Given differential equation is $	an(y) dx + \sec^2(y) 	an(x) dy = 0$. Rearranging the terms to separate variables,we get: $\sec^2(y) 	an(x) dy = -	an(y) dx$ $\frac{\sec^2(y)}{	an(y)} dy = -\frac{1}{	an(x)} dx$ $\frac{\sec^2(y)}{	an(y)} dy = -\cot(x) dx$ Integrating both sides: $\int \frac{\sec^2(y)}{	an(y)} dy = -\int \cot(x) dx$ Let $u = 	an(y)$,then $du = \sec^2(y) dy$. $\int \frac{1}{u} du = -\ln|\sin(x)| + C$ $\ln|	an(y)| = -\ln|\sin(x)| + C$ $\ln|	an(y)| + \ln|\sin(x)| = C$ $\ln|\sin(x) 	an(y)| = C$ Taking exponential on both sides,we get $\sin(x) 	an(y) = e^C = c$.

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

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