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 $e^{8 \alpha}$ 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 $\tan x=t \Rightarrow \sec ^{2} x \frac{d x}{d y}=\frac{d t}{d y}$ )
$\frac{d t}{d y}+e^{2 y} \times t^{2}+t=0$
$\frac{dt}{dy}+t=-t^{2} \cdot e^{2 y}$
$\frac{1}{t^{2}} \frac{d t}{d y}+\frac{1}{t}=-e^{2 y}$
$\left(\right.$ Put $\left.\frac{1}{t}=u \frac{-1}{t^{2}} \frac{dt}{dy}=\frac{du}{dy}\right)$
$\frac{-du}{dy}+u=-e^{2 y}$
$\frac{d u}{d y}-u=e^{2 y}$
I.F. $=e^{-\int d y}=e^{-y}$
$u e^{-y}=\int e^{-y} \times e^{2 y} d y$
$\frac{1}{\tan x} \times e^{-y}=e^{y}+c$
$x=\frac{\pi}{4}, y=0, c=0$
$x=\frac{\pi}{6}, y=\alpha$
$\sqrt{3} e^{-\alpha}=e^{\alpha}+0$
$e^{2 \alpha}=\sqrt{3}$
$e^{8 a}=9$
