The solution of cos y + (x sin y - 1) frac{dy | vedclass

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Question

(A) The given differential equation is $\cos y + (x \sin y - 1) \frac{dy}{dx} = 0$. Rearranging the terms,we get $(x \sin y - 1) \frac{dy}{dx} = -\cos

Vedclass · Accepted Answer

$x \sec y = 	an y + C$

Vedclass · Answer

(A) The given differential equation is $\cos y + (x \sin y - 1) \frac{dy}{dx} = 0$. Rearranging the terms,we get $(x \sin y - 1) \frac{dy}{dx} = -\cos y$. Taking the reciprocal,$\frac{dx}{dy} = -\frac{x \sin y - 1}{\cos y} = -x 	an y + \sec y$. This can be written as $\frac{dx}{dy} + x 	an y = \sec y$. This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = 	an y$ and $Q = \sec y$. The integrating factor $(IF)$ is $e^{\int P dy} = e^{\int 	an y dy} = e^{\ln |\sec y|} = \sec y$. The general solution is given by $x(IF) = \int Q(IF) dy + C$. Substituting the values,$x \sec y = \int \sec y \cdot \sec y dy + C$. $x \sec y = \int \sec^2 y dy + C$. $x \sec y = 	an y + C$.

The solution of $\cos y + (x \sin y - 1) \frac{dy}{dx} = 0$ is

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