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प्रश्न
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उत्तर
We have,
\[\frac{dy}{dx} = y \tan 2x, y\left( 0 \right) = 2\]
\[ \Rightarrow \frac{1}{y}dy = \tan 2x dx\]
Integrating both sides, we get
\[\int\frac{1}{y}dy = \int\tan 2x dx\]
\[ \Rightarrow \log \left| y \right| = \frac{1}{2}\log \left| \sec 2x \right| + \frac{1}{2}\log C\]
\[ \Rightarrow y^2 = C \sec 2x . . . . . \left( 1 \right)\]
It is given that at x = 0, y = 2 .
\[ \therefore C = 4\]
Substituting the value of C in (1), we get
\[ \therefore y^2 = \frac{4}{\cos 2x}\]
\[ \Rightarrow y = \frac{2}{\sqrt{\cos 2x}} \]
\[\text{ Hence, }y = \frac{2}{\sqrt{\cos 2x}} \text{ is the required solution }.\]
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