Question

\[\frac{dy}{dx} = 1 + x + y^2 + x y^2\] when y = 0, x = 0

Answer 1

\[\frac{dy}{dx} = 1 + x + y^2 + x y^2\]
\[\Rightarrow \frac{dy}{dx} = 1 + x + y^2 \left( 1 + x \right)\]
\[ \Rightarrow \frac{dy}{dx} = \left( 1 + x \right)\left( 1 + y^2 \right)\]
\[ \Rightarrow \frac{dy}{\left( 1 + y^2 \right)} = \left( 1 + x \right)dx\]
\[ \Rightarrow \int\frac{dy}{\left( 1 + y^2 \right)} = \int\left( 1 + x \right)dx\]
\[ \Rightarrow \tan^{- 1} y = x + \frac{x^2}{2} + C . . . . . \left( 1 \right)\]
\[\text{ Now, } \tan^{- 1} 0 = 0 + 0 + C ..........\left[\because y = 0, x = 0 \right]\]
\[ \Rightarrow C = 0\]
Substituting the value of C in (1), we get
\[ \tan^{- 1} y = x + \frac{x^2}{2}\]
\[ \Rightarrow y = \tan\left( x + \frac{x^2}{2} \right)\]