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Question:

$\frac{d y}{d x}=y \tan x ; y=1$ when $x=0$

Solution:

$\frac{d y}{d x}=y \tan x$

$\Rightarrow \frac{d y}{y}=\tan x d x$

Integrating both sides, we get:

$\int \frac{d y}{y}=-\int \tan x d x$

$\Rightarrow \log y=\log (\sec x)+\log \mathrm{C}$

$\Rightarrow \log y=\log (\mathrm{C} \sec x)$

$\Rightarrow y=\mathrm{C} \sec x$          …(1)

Now, $y=1$ when $x=0$.

$\Rightarrow 1=\mathrm{C} \times \mathrm{sec} 0$

$\Rightarrow 1=\mathrm{C} \times 1$

$\Rightarrow \mathrm{C}=1$

Substituting C = 1 in equation (1), we get:

y = sec x

 

 

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