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प्रश्न
उत्तर
\[\frac{dy}{dx} = y \tan x, y\left( 0 \right) = 1\]
\[ \Rightarrow \frac{1}{y}dy = \tan x dx\]
Integrating both sides, we get
\[\int\frac{1}{y}dy = \int\tan x dx\]
\[ \Rightarrow \log \left| y \right| = \log \left| \sec x \right| + C . . . . . (1)\]
We know that at x = 0 and y = 1 .
Substituting the values of x and y in (1), we get
\[\log \left| 1 \right| = \log \left| 1 \right| + C\]
\[ \Rightarrow C = 0\]
Substituting the value of C in (1), we get
\[\log \left| y \right| = \log \left| \sec x \right| + 0\]
\[ \Rightarrow y = \sec x\]
\[\text{ Hence, }y = \sec x,\text{ where }x \in \left( \frac{- \pi}{2}, \frac{\pi}{2} \right),\text{ is the required solution .}\]
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