Question
Find the particular solution of the differential equation $\text{e}^\text{x}\tan\text{y dx}+(2-\text{e}^\text{x})\text{sec}^2\text{y dy}=0,$ given that $\text{y}=\frac{\pi}{4}\ \text{x} = 0.$
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Answer
$\text{e}^\text{x}\tan\text{y dx}+(2-\text{e}^\text{x})\text{sec}^2\text{y dy}=0$
$\text{e}^\text{x}\tan\text{y dx}+(\text{e}^\text{x}-2)\text{sec}^2\text{y dy}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\tan\text{y}}{\text{e}^\text{x}\text{sec}^2\text{y}-2\text{sec}^2\text{y}}$
$\frac{\text{dx}}{\text{dy}}=\frac{\text{e}^\text{x}\text{sec}^2\text{y}-2\text{sec}^2\text{y}}{\text{e}^\text{x}\tan\text{y}}$
$\frac{\text{dx}}{\text{dy}}=\frac{\text{sec}^2\text{y}}{\tan\text{y}}-\frac{2\text{sec}^2\text{y}}{\tan\text{y}}\text{e}^{\text{-x}}$
$\frac{\text{dx}}{\text{dy}}=\frac{\text{sec}^2\text{y}}{\tan\text{y}}\big[1-2\text{e}^{-\text{x}}\big]$
$\int\frac{\text{sec}^2\text{y}}{\tan\text{y}}\text{dy}=\int\frac{1}{1-2\text{e}^{-\text{x}}}\text{dx}$
$\tan\text{y}=\text{t}$
$\text{sec}^2\text{y dy}=\text{dt}$
$\int\frac{\text{dt}}{\text{t}}=\int\frac{\text{e}^\text{x}}{\text{e}^\text{x}-2}\text{dx}$
$\text{e}^\text{x}-2=\text{u}$
$\text{e}^\text{x}\text{dx}=\text{du}$
$\log\text{t}=\log\text{u}+\log\text{C}$
$\log(\tan\text{y})=\log(\text{e}^\text{x}-2)\text{C}$
$\tan\text{y}=\text{C}(\text{e}^\text{x}-2)$
$\text{Put y }=\frac{\pi}{4},\ \text{x}=0\ \ \ \tan\frac{\pi}{4}=\text{C}(1-2)$
$\text{C}=-1$
$\tan\text{y}=-(\text{e}^\text{x}-2)$
$\text{e}^\text{x}\tan\text{y dx}+(\text{e}^\text{x}-2)\text{sec}^2\text{y dy}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\tan\text{y}}{\text{e}^\text{x}\text{sec}^2\text{y}-2\text{sec}^2\text{y}}$
$\frac{\text{dx}}{\text{dy}}=\frac{\text{e}^\text{x}\text{sec}^2\text{y}-2\text{sec}^2\text{y}}{\text{e}^\text{x}\tan\text{y}}$
$\frac{\text{dx}}{\text{dy}}=\frac{\text{sec}^2\text{y}}{\tan\text{y}}-\frac{2\text{sec}^2\text{y}}{\tan\text{y}}\text{e}^{\text{-x}}$
$\frac{\text{dx}}{\text{dy}}=\frac{\text{sec}^2\text{y}}{\tan\text{y}}\big[1-2\text{e}^{-\text{x}}\big]$
$\int\frac{\text{sec}^2\text{y}}{\tan\text{y}}\text{dy}=\int\frac{1}{1-2\text{e}^{-\text{x}}}\text{dx}$
$\tan\text{y}=\text{t}$
$\text{sec}^2\text{y dy}=\text{dt}$
$\int\frac{\text{dt}}{\text{t}}=\int\frac{\text{e}^\text{x}}{\text{e}^\text{x}-2}\text{dx}$
$\text{e}^\text{x}-2=\text{u}$
$\text{e}^\text{x}\text{dx}=\text{du}$
$\log\text{t}=\log\text{u}+\log\text{C}$
$\log(\tan\text{y})=\log(\text{e}^\text{x}-2)\text{C}$
$\tan\text{y}=\text{C}(\text{e}^\text{x}-2)$
$\text{Put y }=\frac{\pi}{4},\ \text{x}=0\ \ \ \tan\frac{\pi}{4}=\text{C}(1-2)$
$\text{C}=-1$
$\tan\text{y}=-(\text{e}^\text{x}-2)$
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