The general solution of sin y cdot frac{dy}{d | vedclass

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(B) Given differential equation is $\sin y \frac{dy}{dx} = \cos y(1 - x \cos y)$.Dividing both sides by $\cos^2 y$,we get:$\frac{\sin y}{\cos^2 y} \fr

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$\sec y = x + 1 + c e^x$

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(B) Given differential equation is $\sin y \frac{dy}{dx} = \cos y(1 - x \cos y)$.Dividing both sides by $\cos^2 y$,we get:$\frac{\sin y}{\cos^2 y} \frac{dy}{dx} = \frac{1}{\cos y} - x$$\sec y 	an y \frac{dy}{dx} = \sec y - x$Let $\sec y = t$. Then $\sec y 	an y \frac{dy}{dx} = \frac{dt}{dx}$.Substituting this into the equation,we get the linear differential equation:$\frac{dt}{dx} = t - x \implies \frac{dt}{dx} - t = -x$.The integrating factor $(IF)$ is $e^{\int -1 dx} = e^{-x}$.The solution is $t(IF) = \int (-x)(IF) dx + c$.$t e^{-x} = \int -x e^{-x} dx + c$.Using integration by parts for $\int -x e^{-x} dx$:$\int -x e^{-x} dx = x e^{-x} - \int e^{-x} dx = x e^{-x} + e^{-x} + c$.Thus,$t e^{-x} = x e^{-x} + e^{-x} + c$.Multiplying by $e^x$,we get $t = x + 1 + c e^x$.Substituting back $t = \sec y$,we get $\sec y = x + 1 + c e^x$.

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

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