If cos x frac{dy}{dx} - y sin x = 6x,where 0 | vedclass

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Question

(C) Given the differential equation: $\cos x \frac{dy}{dx} - y \sin x = 6x$. This can be written as the derivative of a product: $\frac{d}{dx}(y \cos

Vedclass · Accepted Answer

$\frac{-\pi^2}{2 \sqrt{3}}$

Vedclass · Answer

(C) Given the differential equation: $\cos x \frac{dy}{dx} - y \sin x = 6x$. This can be written as the derivative of a product: $\frac{d}{dx}(y \cos x) = 6x$. Integrating both sides with respect to $x$: $y \cos x = \int 6x \, dx = 3x^2 + C$. Using the initial condition $y(\frac{\pi}{3}) = 0$: $0 \cdot \cos(\frac{\pi}{3}) = 3(\frac{\pi}{3})^2 + C \Rightarrow 0 = \frac{\pi^2}{3} + C \Rightarrow C = \frac{-\pi^2}{3}$. Thus,the general solution is $y \cos x = 3x^2 - \frac{\pi^2}{3}$. Now,to find $y(\frac{\pi}{6})$,substitute $x = \frac{\pi}{6}$: $y(\frac{\pi}{6}) \cos(\frac{\pi}{6}) = 3(\frac{\pi}{6})^2 - \frac{\pi^2}{3}$. $y(\frac{\pi}{6}) \cdot \frac{\sqrt{3}}{2} = 3(\frac{\pi^2}{36}) - \frac{\pi^2}{3} = \frac{\pi^2}{12} - \frac{4\pi^2}{12} = \frac{-3\pi^2}{12} = \frac{-\pi^2}{4}$. Therefore,$y(\frac{\pi}{6}) = \frac{-\pi^2}{4} \cdot \frac{2}{\sqrt{3}} = \frac{-\pi^2}{2 \sqrt{3}}$.

If $\cos x \frac{dy}{dx} - y \sin x = 6x$,where $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{3}) = 0$,then find $y(\frac{\pi}{6})$.

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