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

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Solution

$\frac{d y}{d x} = y tan x$
$\frac{d y}{y} = d x (tan x)$
On integrating, we get
$log y = - log cos x + log c$
So, $y = \frac{x}{cos x}$
Now, $y (0) = 1$
So, $c = 1$
So, $y = \frac{1}{cos x} = sec x$

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dydx=ytanx; y=1 when x=0.

dydx=ytanxdyy=dx(tanx)On integrating, we getlogy=−logcosx+logcSo, y=xcosxNow, y(0)=1So, c=1So, y=1cosx=secx

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

$\frac{d y}{d x} = y tan x$
$\frac{d y}{y} = d x (tan x)$
On integrating, we get
$log y = - log cos x + log c$
So, $y = \frac{x}{cos x}$
Now, $y (0) = 1$
So, $c = 1$
So, $y = \frac{1}{cos x} = sec x$