The solution of dy = cos x(2 - y csc x)dx whe | vedclass

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Question

(A) Given the differential equation: $dy = \cos x(2 - y \csc x)dx$Dividing by $dx$,we get: $\frac{dy}{dx} = 2 \cos x - y \cot x$Rearranging the terms:

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$y = \sin x + \csc x$

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(A) Given the differential equation: $dy = \cos x(2 - y \csc x)dx$Dividing by $dx$,we get: $\frac{dy}{dx} = 2 \cos x - y \cot x$Rearranging the terms: $\frac{dy}{dx} + y \cot x = 2 \cos x$This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = 2 \cos x$.The integrating factor $(I.F.)$ is given by $e^{\int P dx} = e^{\int \cot x dx} = e^{\ln(\sin x)} = \sin x$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.Substituting the values: $y \sin x = \int (2 \cos x \cdot \sin x) dx + c = \int \sin(2x) dx + c$.$y \sin x = -\frac{\cos(2x)}{2} + c$.Alternatively,using $2 \cos x \sin x = \sin(2x)$ or integrating $2 \sin x \cos x$ directly gives $y \sin x = \sin^2 x + c$.Given $y = 2$ when $x = \frac{\pi}{2}$:$2 \sin(\frac{\pi}{2}) = \sin^2(\frac{\pi}{2}) + c \implies 2(1) = (1)^2 + c \implies c = 1$.Thus,$y \sin x = \sin^2 x + 1$.Dividing by $\sin x$,we get $y = \sin x + \csc x$.

The solution of $dy = \cos x(2 - y \csc x)dx$ where $y = 2$ when $x = \frac{\pi}{2}$ is

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