If frac{dy}{dx} + 2y tan x = sin x,0

Question

(A) The given linear differential equation is $\frac{dy}{dx} + (2 	an x)y = \sin x$. The integrating factor $(I.F.)$ is $e^{\int 2 	an x \, dx} = e^

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(A) The given linear differential equation is $\frac{dy}{dx} + (2 	an x)y = \sin x$. The integrating factor $(I.F.)$ is $e^{\int 2 	an x \, dx} = e^{2 \ln|\sec x|} = \sec^2 x$. Multiplying both sides by $I.F.$,we get $\frac{d}{dx}(y \sec^2 x) = \sin x \sec^2 x = \sec x 	an x$. Integrating both sides,$y \sec^2 x = \int \sec x 	an x \, dx + C = \sec x + C$. Thus,$y = \cos x + C \cos^2 x$. Given $y(\frac{\pi}{3}) = 0$,we have $0 = \cos(\frac{\pi}{3}) + C \cos^2(\frac{\pi}{3}) = \frac{1}{2} + C(\frac{1}{4})$,which gives $C = -2$. So,$y = \cos x - 2 \cos^2 x$. Let $u = \cos x$. Since $0 $y = u - 2u^2 = -2(u^2 - \frac{1}{2}u) = -2(u - \frac{1}{4})^2 + \frac{1}{8}$. The maximum value occurs at $u = \frac{1}{4}$,which is $\frac{1}{8}$.

If $\frac{dy}{dx} + 2y \tan x = \sin x$,$0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{3}) = 0$,then the maximum value of $y(x)$ is.

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