Differential Equations Question 17
Question 17 - 2024 (31 Jan Shift 2)
Let $y=y(x)$ be the solution of the differential equation
$\sec ^{2} x d x+\left(e^{2 y} \tan ^{2} x+\tan x\right) d y=0$
$0<x<\frac{\pi}{2}, y\left(\frac{\pi}{4}\right)=0$. If $y\left(\frac{\pi}{6}\right)=\alpha$,
Then $\mathrm{e}^{8 \alpha}$ is equal to
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Answer (9)
Solution
$\sec ^{2} x \frac{d x}{d y}+e^{2 y} \tan ^{2} x+\tan x=0$
(Put $\tan x=t \Rightarrow \sec ^{2} x \frac{d x}{d y}=\frac{d t}{d y}$ )
$\frac{d t}{d y}+e^{2 y} \times t^{2}+t=0$
$\frac{\mathrm{dt}}{\mathrm{dy}}+\mathrm{t}=-\mathrm{t}^{2} \cdot \mathrm{e}^{2 \mathrm{y}}$
$\frac{1}{t^{2}} \frac{d t}{d y}+\frac{1}{t}=-e^{2 y}$
$\left(\right.$ Put $\left.\frac{1}{\mathrm{t}}=\mathrm{u} \frac{-1}{\mathrm{t}^{2}} \frac{\mathrm{dt}}{\mathrm{dy}}=\frac{\mathrm{du}}{\mathrm{dy}}\right)$
$\frac{-\mathrm{du}}{\mathrm{dy}}+\mathrm{u}=-\mathrm{e}^{2 \mathrm{y}}$
$\frac{d u}{d y}-u=e^{2 y}$
I.F. $=\mathrm{e}^{-\int d y}=\mathrm{e}^{-y}$
$u e^{-y}=\int e^{-y} \times e^{2 y} d y$
$\frac{1}{\tan x} \times e^{-y}=e^{y}+c$
$\mathrm{x}=\frac{\pi}{4}, \mathrm{y}=0, \mathrm{c}=0$
$\mathrm{x}=\frac{\pi}{6}, \mathrm{y}=\alpha$
$\sqrt{3} \mathrm{e}^{-\alpha}=\mathrm{e}^{\alpha}+0$
$\mathrm{e}^{2 \alpha}=\sqrt{3}$
$e^{8 a}=9$
