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

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(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 	an x$ and $Q = x \sec x$. The

Vedclass · Accepted Answer

$\frac{\pi}{12} - \frac{\sqrt{3}}{2} \log_e\left(\frac{2}{e\sqrt{3}}\right)$

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(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 	an x$ and $Q = x \sec x$. The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int 	an x dx} = e^{\ln|\sec x|} = \sec x$. The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$. $y \sec x = \int (x \sec x) \cdot \sec x dx = \int x \sec^2 x dx$. Using integration by parts,$\int x \sec^2 x dx = x 	an x - \int 	an x dx = x 	an x - \ln|\sec x| + C$. So,$y \sec x = x 	an x - \ln|\sec x| + C$. Given $y(0) = 1$,we have $1 \cdot \sec(0) = 0 \cdot 	an(0) - \ln|\sec(0)| + C \Rightarrow 1 = 0 - 0 + C \Rightarrow C = 1$. Thus,$y \sec x = x 	an x - \ln|\sec x| + 1$. At $x = \frac{\pi}{6}$,$\sec\left(\frac{\pi}{6}ight) = \frac{2}{\sqrt{3}}$ and $	an\left(\frac{\pi}{6}ight) = \frac{1}{\sqrt{3}}$. $y \cdot \frac{2}{\sqrt{3}} = \frac{\pi}{6} \cdot \frac{1}{\sqrt{3}} - \ln\left(\frac{2}{\sqrt{3}}ight) + 1$. $y = \frac{\sqrt{3}}{2} \left( \frac{\pi}{6\sqrt{3}} - \ln\left(\frac{2}{\sqrt{3}}ight) + 1 ight) = \frac{\pi}{12} - \frac{\sqrt{3}}{2} \ln\left(\frac{2}{\sqrt{3}}ight) + \frac{\sqrt{3}}{2}$. Since $1 = \ln e$,we have $y = \frac{\pi}{12} + \frac{\sqrt{3}}{2} \left( 1 - \ln\left(\frac{2}{\sqrt{3}}ight) ight) = \frac{\pi}{12} + \frac{\sqrt{3}}{2} \ln\left(\frac{e\sqrt{3}}{2}ight) = \frac{\pi}{12} - \frac{\sqrt{3}}{2} \ln\left(\frac{2}{e\sqrt{3}}ight)$.

If $y=y(x)$ is the solution curve of the differential equation $\frac{dy}{dx} + y \tan x = x \sec x$,$0 \leq x \leq \frac{\pi}{3}$,with $y(0)=1$,then $y\left(\frac{\pi}{6}\right)$ is equal to

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