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

Vedclass · Accepted Answer

$\left(\frac{1}{2\sqrt{2}}\right) \log_{e} 2$

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(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 	an x$ and $Q(x) = \sin x$.First,we find the integrating factor ($I$.$F$.):$I.F. = e^{\int 	an x dx} = e^{\ln|\sec x|} = \sec x$.The general solution is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y \sec x = \int \sin x \cdot \sec x dx + C$$y \sec x = \int 	an x dx + C$$y \sec x = \ln|\sec x| + C$.Given $y(0) = 0$,we substitute $x = 0$ and $y = 0$:$0 \cdot \sec(0) = \ln|\sec(0)| + C$$0 = \ln(1) + C \Rightarrow C = 0$.Thus,the particular solution is $y \sec x = \ln|\sec x|$,which simplifies to $y = \cos x \ln|\sec x|$.To find $y\left(\frac{\pi}{4}ight)$:$y\left(\frac{\pi}{4}ight) = \cos\left(\frac{\pi}{4}ight) \ln\left|\sec\left(\frac{\pi}{4}ight)ight|$$y\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}} \ln(\sqrt{2})$$y\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}} \ln(2^{1/2}) = \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \ln(2) = \frac{1}{2\sqrt{2}} \log_{e} 2$.

If $y = y(x)$ is the solution of the differential equation $\frac{dy}{dx} + (\tan x)y = \sin x, 0 \leq x \leq \frac{\pi}{3},$ with $y(0) = 0,$ then $y\left(\frac{\pi}{4}\right)$ is equal to:

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