The solution of the differential equation cos | vedclass

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(D) Given the differential equation: $\cos y \log(\sec x + 	an x) dx = \cos x \log(\sec y + 	an y) dy$. Separating the variables,we get: $\frac{\log

Vedclass · Accepted Answer

None of these

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(D) Given the differential equation: $\cos y \log(\sec x + 	an x) dx = \cos x \log(\sec y + 	an y) dy$. Separating the variables,we get: $\frac{\log(\sec x + 	an x)}{\cos x} dx = \frac{\log(\sec y + 	an y)}{\cos y} dy$. Since $\frac{1}{\cos x} = \sec x$,the equation becomes: $\sec x \log(\sec x + 	an x) dx = \sec y \log(\sec y + 	an y) dy$. Integrating both sides: $\int \sec x \log(\sec x + 	an x) dx = \int \sec y \log(\sec y + 	an y) dy$. Let $u = \log(\sec x + 	an x)$,then $du = \frac{1}{\sec x + 	an x} (\sec x 	an x + \sec^2 x) dx = \frac{\sec x(	an x + \sec x)}{\sec x + 	an x} dx = \sec x dx$. Similarly,for the right side,let $v = \log(\sec y + 	an y)$,then $dv = \sec y dy$. The integral becomes: $\int u du = \int v dv$. This yields $\frac{u^2}{2} = \frac{v^2}{2} + C_1$,or $u^2 - v^2 = C$. Substituting back,we get: $[\log(\sec x + 	an x)]^2 - [\log(\sec y + 	an y)]^2 = C$. Comparing this with the given options,none of them match. Therefore,the correct option is $(d)$.

The solution of the differential equation $\cos y \log(\sec x + \tan x) dx = \cos x \log(\sec y + \tan y) dy$ is

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