A function y=f(x) satisfies f(x) sin 2x + sin | vedclass

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(A) The given differential equation is $(1 + \cos^2 x) \frac{dy}{dx} - (\sin 2x) y = \sin x$.Dividing by $(1 + \cos^2 x)$,we get $\frac{dy}{dx} - \lef

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(A) The given differential equation is $(1 + \cos^2 x) \frac{dy}{dx} - (\sin 2x) y = \sin x$.Dividing by $(1 + \cos^2 x)$,we get $\frac{dy}{dx} - \left( \frac{\sin 2x}{1 + \cos^2 x} \right) y = \frac{\sin x}{1 + \cos^2 x}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{\sin 2x}{1 + \cos^2 x}$ and $Q(x) = \frac{\sin x}{1 + \cos^2 x}$.The integrating factor $I.F. = e^{\int P(x) dx} = e^{-\int \frac{\sin 2x}{1 + \cos^2 x} dx}$.Let $u = 1 + \cos^2 x$,then $du = -2 \cos x \sin x dx = -\sin 2x dx$.So,$I.F. = e^{\int \frac{du}{u}} = e^{\ln(u)} = 1 + \cos^2 x$.The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y(1 + \cos^2 x) = \int \left( \frac{\sin x}{1 + \cos^2 x} \right) (1 + \cos^2 x) dx = \int \sin x dx = -\cos x + C$.Given $f(0) = 0$,at $x = 0$,$y(1 + \cos^2 0) = -\cos 0 + C \implies 0(2) = -1 + C \implies C = 1$.Thus,$y(1 + \cos^2 x) = 1 - \cos x$.At $x = \frac{\pi}{2}$,$y(1 + \cos^2 \frac{\pi}{2}) = 1 - \cos \frac{\pi}{2} \implies y(1 + 0) = 1 - 0 \implies y = 1$.

$A$ function $y=f(x)$ satisfies $f(x) \sin 2x + \sin x - (1 + \cos^2 x) f'(x) = 0$ with the condition $f(0) = 0$. Then $f(\frac{\pi}{2})$ is equal to

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