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

Vedclass · Accepted Answer

$Y^{\prime}\left(\frac{\pi}{4}\right)-Y^{\prime}\left(\frac{-\pi}{4}\right)=\pi-\sqrt{2}$

Vedclass · Answer

(D) The given differential equation is $\frac{dy}{dx}+y 	an x=2x+x^2 	an x$.This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=	an x$ and $Q=2x+x^2 	an x$.The integrating factor $IF = e^{\int 	an x dx} = e^{\ln |\sec x|} = \sec x$.The solution is given by $y \cdot IF = \int Q \cdot IF dx + C$.$y \sec x = \int (2x+x^2 	an x) \sec x dx + C$.$y \sec x = \int 2x \sec x dx + \int x^2 \sec x 	an x dx + C$.Using integration by parts on the second integral: $\int x^2 \sec x 	an x dx = x^2 \sec x - \int 2x \sec x dx$.Substituting this back: $y \sec x = \int 2x \sec x dx + x^2 \sec x - \int 2x \sec x dx + C$.$y \sec x = x^2 \sec x + C$,which simplifies to $y = x^2 + C \cos x$.Given $Y(0)=1$,we have $1 = 0^2 + C \cos(0) \implies C=1$.So,$Y(x) = x^2 + \cos x$.Then $Y'(x) = 2x - \sin x$.Calculating $Y'(\frac{\pi}{4}) - Y'(-\frac{\pi}{4}) = (2(\frac{\pi}{4}) - \sin(\frac{\pi}{4})) - (2(-\frac{\pi}{4}) - \sin(-\frac{\pi}{4}))$.$= (\frac{\pi}{2} - \frac{1}{\sqrt{2}}) - (-\frac{\pi}{2} + \frac{1}{\sqrt{2}}) = \pi - \frac{2}{\sqrt{2}} = \pi - \sqrt{2}$.Thus,option $D$ is correct.

Let $y=Y(x)$ be the solution of the differential equation $\frac{dy}{dx}+y \tan x=2x+x^2 \tan x$,$x \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$,such that $Y(0)=1$,then

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