For the differential equation in given question find a particular solution satisfying the given condition.

$\frac{d y}{d x} = y t a n x, y = 1 w h e n x = 0$

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Solution

$G i v e n, \frac{d y}{d x} = y t a n x$
On separating the variables, we get $\frac{1}{y} d y = t a n x d x$
On integrating both sides, we get
$\int \frac{1}{d y} = \int t a n x d x \Rightarrow l o g | y | = l o g | s e c x | + l o g | C | \Rightarrow l o g y = l o g | C s e c x | [∵ l o g m + l o g n = l o g m n] \Rightarrow y = C s e c x (∵ l o g m = l o g n \Rightarrow m = n) . . . . (i)$
Now, put y=1 when x=0, we get
$1 = C s e c 0 \Rightarrow 1 = C \times 1 \Rightarrow C = 1$
On substituting C=1 in Eq.(i), we get y=secx
which is the required particular solution.

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