If cos x frac{dy}{dx} - y sin x = 6x for 0

Question

(D) The given differential equation is $\cos x \frac{dy}{dx} - y \sin x = 6x$.Dividing by $\cos x$,we get $\frac{dy}{dx} - y 	an x = 6x \sec x$.This

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The given differential equation is $\cos x \frac{dy}{dx} - y \sin x = 6x$.Dividing by $\cos x$,we get $\frac{dy}{dx} - y 	an x = 6x \sec x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -	an x$ and $Q = 6x \sec x$.The integrating factor is $IF = e^{\int P dx} = e^{-\int 	an x dx} = e^{-\ln(\sec x)} = \frac{1}{\sec x} = \cos x$.The general solution is $y \cdot IF = \int Q \cdot IF dx + C$.$y \cos x = \int (6x \sec x) \cdot \cos x dx = \int 6x dx = 3x^2 + C$.Given $y(\frac{\pi}{3}) = 0$,we have $0 \cdot \cos(\frac{\pi}{3}) = 3(\frac{\pi}{3})^2 + C$,which gives $0 = \frac{\pi^2}{3} + C$,so $C = -\frac{\pi^2}{3}$.Thus,$y \cos x = 3x^2 - \frac{\pi^2}{3}$.At $x = \frac{\pi}{6}$,$\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$.$y \cdot \frac{\sqrt{3}}{2} = 3(\frac{\pi}{6})^2 - \frac{\pi^2}{3} = 3(\frac{\pi^2}{36}) - \frac{\pi^2}{3} = \frac{\pi^2}{12} - \frac{4\pi^2}{12} = -\frac{3\pi^2}{12} = -\frac{\pi^2}{4}$.$y = -\frac{\pi^2}{4} \cdot \frac{2}{\sqrt{3}} = -\frac{\pi^2}{2\sqrt{3}}$.

If $\cos x \frac{dy}{dx} - y \sin x = 6x$ for $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{3}) = 0$,then $y(\frac{\pi}{6})$ is equal to:

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