If y=y(x) is the solution of the differential | vedclass

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(D) 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 =

Vedclass · Accepted Answer

$\frac{1}{8}$

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(D) 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 $Q = \sin x$.The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P dx} = e^{\int 2 	an x dx} = e^{2 \ln |\sec x|} = \sec^2 x$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.$y \sec^2 x = \int \sin x \cdot \sec^2 x dx = \int \sec x 	an x dx = \sec x + C$.Given $y(\frac{\pi}{3}) = 0$,we substitute $x = \frac{\pi}{3}$ and $y = 0$:$0 \cdot \sec^2(\frac{\pi}{3}) = \sec(\frac{\pi}{3}) + C \implies 0 = 2 + C \implies C = -2$.Thus,$y \sec^2 x = \sec x - 2$,which simplifies to $y = \frac{\sec x - 2}{\sec^2 x} = \cos x - 2 \cos^2 x$.Let $t = \cos x$. Since $-1 \le \cos x \le 1$,we have $y = t - 2t^2$. To find the maximum,we differentiate with respect to $t$:$\frac{dy}{dt} = 1 - 4t = 0 \implies t = \frac{1}{4}$.Since $t = \frac{1}{4}$ is within the range $[-1, 1]$,the maximum value is $y = \frac{1}{4} - 2(\frac{1}{4})^2 = \frac{1}{4} - \frac{2}{16} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$.

If $y=y(x)$ is the solution of the differential equation $\frac{dy}{dx} + 2y \tan x = \sin x$ with the condition $y(\frac{\pi}{3}) = 0$,then the maximum value of the function $y(x)$ over $\mathbb{R}$ is equal to:

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