Question

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

Answer 1

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