The integrating factor of the differential eq | vedclass

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(A) Given differential equation: $\sin y \frac{d y}{d x} = \cos y (1 - x \cos y)$.Expanding the right side: $\sin y \frac{d y}{d x} = \cos y - x \cos^

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$e^{-x}$

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(A) Given differential equation: $\sin y \frac{d y}{d x} = \cos y (1 - x \cos y)$.Expanding the right side: $\sin y \frac{d y}{d x} = \cos y - x \cos^2 y$.Dividing both sides by $\cos^2 y$: $\frac{\sin y}{\cos^2 y} \frac{d y}{d x} = \frac{1}{\cos y} - x$.This simplifies to: $\sec y 	an y \frac{d y}{d x} = \sec y - x$.Rearranging the terms: $\sec y 	an y \frac{d y}{d x} - \sec y = -x$.Let $v = \sec y$. Then $\frac{d v}{d x} = \sec y 	an y \frac{d y}{d x}$.Substituting into the equation: $\frac{d v}{d x} - v = -x$.This is a linear differential equation of the form $\frac{d v}{d x} + P(x)v = Q(x)$,where $P(x) = -1$.The integrating factor ($I$.$F$.) is given by $e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}$.

The integrating factor of the differential equation $\sin y \left(\frac{d y}{d x}\right) = \cos y (1 - x \cos y)$ is

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