उत्तर (9)

समाधान

$\sec ^{2} x \dfrac{d x}{d y}+e^{2 y} \tan ^{2} x+\tan x=0$

(मान लीजिए $\tan x=t \Rightarrow \sec ^{2} x \dfrac{d x}{d y}=\dfrac{d t}{d y}$ )

$\dfrac{d t}{d y}+e^{2 y} \times t^{2}+t=0$

$\dfrac{dt}{dy}+t=-t^{2} \cdot e^{2 y}$

$\dfrac{1}{t^{2}} \dfrac{d t}{d y}+\dfrac{1}{t}=-e^{2 y}$

(मान लीजिए $\dfrac{1}{t}=u \Rightarrow \dfrac{-1}{t^{2}} \dfrac{dt}{dy}=\dfrac{du}{dy}$)

$\dfrac{-du}{dy}+u=-e^{2 y}$

$\dfrac{d u}{d y}-u=e^{2 y}$

I.F. $=e^{-\int d y}=e^{-y}$

$u e^{-y}=\int e^{-y} \times e^{2 y} d y$

$\dfrac{1}{\tan x} \times e^{-y}=e^{y}+c$

जब $ x=\dfrac{\pi}{4}, y=0\Rightarrow c=0$

$\dfrac{1}{\tan x} \times e^{-y}=e^{y}$

$x=\dfrac{\pi}{6}, y=\alpha$

$\sqrt{3} e^{-\alpha}=e^{\alpha}$

$e^{2 \alpha}=\sqrt{3}$

$e^{8 a}=9$