Question

\[\frac{dy}{dx} = y \sin 2x, y\left( 0 \right) = 1\]

Answer 1

We have, 
\[\frac{dy}{dx} = y \sin2x, y\left( 0 \right) = 1\]
\[ \Rightarrow \frac{1}{y}dy = \sin 2x dx\]
Integrating both sides, we get
\[\int\frac{1}{y}dy = \int\sin 2x dx\]
\[ \Rightarrow \log \left| y \right| = - \frac{\cos 2x}{2} + C . . . . . (1)\]
\[\text{ Given:} x = 0, y = 1 . \]
Substituting the values of x and y in (1), we get
\[\log \left| 1 \right| = - \frac{1}{2} + C\]
\[ \Rightarrow C = \frac{1}{2}\]
Substituting the value of C in (1), we get
\[\log \left| y \right| = - \frac{\cos 2x}{2} + \frac{1}{2}\]
\[ \Rightarrow \log \left| y \right| = \frac{1 - \cos 2x}{2}\]
\[ \Rightarrow \log \left| y \right| = \sin {}^2 x\]
\[ \Rightarrow y = e^{sin^2} x \]
\[\text{ Hence, }y = e^{sin^2} x\text{ is the required solution }.\]