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(A) Given the differential equation: $\sec^2 x dx + (e^{2y} 	an^2 x + 	an x) dy = 0$.Dividing by $dy$,we get: $\sec^2 x \frac{dx}{dy} + e^{2y} 	an^

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) Given the differential equation: $\sec^2 x dx + (e^{2y} 	an^2 x + 	an x) dy = 0$.Dividing by $dy$,we get: $\sec^2 x \frac{dx}{dy} + e^{2y} 	an^2 x + 	an x = 0$.Let $t = 	an x$,then $\frac{dt}{dy} = \sec^2 x \frac{dx}{dy}$.The equation becomes: $\frac{dt}{dy} + t = -e^{2y} t^2$.Dividing by $t^2$: $t^{-2} \frac{dt}{dy} + t^{-1} = -e^{2y}$.Let $u = t^{-1} = \frac{1}{	an x}$,then $\frac{du}{dy} = -t^{-2} \frac{dt}{dy}$.Substituting this into the equation: $-\frac{du}{dy} + u = -e^{2y}$,or $\frac{du}{dy} - u = e^{2y}$.This is a linear differential equation with Integrating Factor $I.F. = e^{\int -1 dy} = e^{-y}$.The solution is $u e^{-y} = \int e^{2y} e^{-y} dy = \int e^y dy = e^y + C$.So,$\frac{1}{	an x} e^{-y} = e^y + C$.Given $y(\frac{\pi}{4}) = 0$,we have $\frac{1}{	an(\pi/4)} e^0 = e^0 + C \Rightarrow 1 = 1 + C \Rightarrow C = 0$.Thus,$\frac{1}{	an x} e^{-y} = e^y \Rightarrow e^{2y} = \frac{1}{	an x} = \cot x$.For $x = \frac{\pi}{6}$,$e^{2\alpha} = \cot(\frac{\pi}{6}) = \sqrt{3}$.Therefore,$e^{8\alpha} = (e^{2\alpha})^4 = (\sqrt{3})^4 = 9$.

Let $y=y(x)$ be the solution of the differential equation $\sec^2 x dx + (e^{2y} \tan^2 x + \tan x) dy = 0$,where $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{4}) = 0$. If $y(\frac{\pi}{6}) = \alpha$,then $e^{8\alpha}$ is equal to:

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