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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