Question

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

Answer 1

\[ \frac{dy}{dx} = 2xy, y\left( 0 \right) = 1\]
\[ \Rightarrow \frac{1}{y}dy = 2x dx\]
Integrating both sides, we get 
\[\int\frac{1}{y}dy = \int2x dx\]
\[\log \left| y \right| = x^2 + C . . . . . (1)\]
\[\text{We know that at }x = 0, y = 1 . \]
Substituting the values of x and y in (1), we get
\[0 = 0 + C\]
\[ \Rightarrow C = 0\]
Substituting the value of C in (1), we get 
\[\log \left| y \right| = x^2 \]
\[ \Rightarrow y = e^{x^2} \]
\[\text{ Hence, }y = e^{x^2}\text{ is the required solution }. \]