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

Vedclass · Accepted Answer

$y = \cos x - 2 \cos^2 x$

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(A) The given differential equation is $\frac{dy}{dx} + 2y 	an x = \sin x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2 	an x$ and $Q = \sin x$.First,we find the Integrating Factor $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int 2 	an x dx} = e^{2 \ln |\sec x|} = e^{\ln |\sec^2 x|} = \sec^2 x$.The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + C$.$y \cdot \sec^2 x = \int (\sin x \cdot \sec^2 x) dx + C$.$y \cdot \sec^2 x = \int (	an x \cdot \sec x) dx + C$.$y \cdot \sec^2 x = \sec x + C$  --- $(1)$.Given $y = 0$ when $x = \frac{\pi}{3}$:$0 \cdot \sec^2(\frac{\pi}{3}) = \sec(\frac{\pi}{3}) + C$.$0 = 2 + C \Rightarrow C = -2$.Substituting $C = -2$ into $(1)$:$y \cdot \sec^2 x = \sec x - 2$.Dividing by $\sec^2 x$ (or multiplying by $\cos^2 x$):$y = \cos x - 2 \cos^2 x$.

Find a particular solution satisfying the given condition: $\frac{dy}{dx} + 2y \tan x = \sin x$; $y = 0$ when $x = \frac{\pi}{3}$.

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