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(A) The given differential equation is $\frac{dy}{dx} - 3y \cot x = \sin 2x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py =

Vedclass · Accepted Answer

$y = 4 \sin^3 x - 2 \sin^2 x$

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(A) The given differential equation is $\frac{dy}{dx} - 3y \cot x = \sin 2x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -3 \cot x$ and $Q = \sin 2x$. Integrating Factor $(I.F.)$ $= e^{\int P dx} = e^{-3 \int \cot x dx} = e^{-3 \ln |\sin x|} = e^{\ln |\sin x|^{-3}} = \frac{1}{\sin^3 x}$. The general solution is $y(I.F.) = \int (Q 	imes I.F.) dx + C$. $y \cdot \frac{1}{\sin^3 x} = \int \left( \sin 2x \cdot \frac{1}{\sin^3 x} ight) dx + C$. Since $\sin 2x = 2 \sin x \cos x$,we have $y \cdot \frac{1}{\sin^3 x} = \int \frac{2 \sin x \cos x}{\sin^3 x} dx + C = 2 \int \frac{\cos x}{\sin^2 x} dx + C$. Let $u = \sin x$,then $du = \cos x dx$. The integral becomes $2 \int u^{-2} du = -2u^{-1} = -\frac{2}{\sin x}$. So,$\frac{y}{\sin^3 x} = -\frac{2}{\sin x} + C$. Multiplying by $\sin^3 x$,we get $y = -2 \sin^2 x + C \sin^3 x$. Given $y = 2$ at $x = \frac{\pi}{2}$,we have $2 = -2 \sin^2(\frac{\pi}{2}) + C \sin^3(\frac{\pi}{2}) \Rightarrow 2 = -2(1) + C(1) \Rightarrow C = 4$. Thus,the particular solution is $y = 4 \sin^3 x - 2 \sin^2 x$.

Find a particular solution satisfying the given condition: $\frac{dy}{dx} - 3y \cot x = \sin 2x$; $y = 2$ when $x = \frac{\pi}{2}$.

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