Question

\[\frac{dy}{dx} = 2 e^x y^3 , y\left( 0 \right) = \frac{1}{2}\]

Answer 1

We have, 
\[\frac{dy}{dx} = 2 e^x y^3 , y\left( 0 \right) = \frac{1}{2}\]
\[ \Rightarrow \frac{1}{y^3}dy = 2 e^x dx\]
Integrating both sides, we get 
\[\int\frac{1}{y^3}dy = \int2 e^x dx\]
\[ \Rightarrow - \frac{1}{2 y^2} = 2 e^x + C . . . . . (1)\]
\[\text{ Given: at }x = 0, y = \frac{1}{2}\]
Substituting the values of x and y in (1), we get
\[ - \frac{1}{2 \times \frac{1}{4}} = 2 e^0 + C\]
\[ \Rightarrow C = - 2 - 2\]
\[ \Rightarrow C = - 4\]
Substituting the value of C in (1), we get
\[ \Rightarrow - \frac{1}{2 y^2} = 2 e^x - 4\]
\[ \Rightarrow y^2 \left( 8 - 4 e^x \right) = 1\]
\[\text{ Hence, }y^2 \left( 8 - 4 e^x \right) = 1 \text{ is the required solution }.\]