Question

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

Answer 1

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