Solve the differential equation $\frac{dy}{dx} = 1 + x + y^2 + xy^2$, when $y = 0$ and $x = 0$.

$y = \tan \left( x + \frac{x^2}{2} \right)$

The correct answer is Option (2) → $y = \tan \left( x + \frac{x^2}{2} \right)$ ##

Given that, $\frac{dy}{dx} = 1 + x + y^2 + xy^2$

$\Rightarrow \frac{dy}{dx} = (1 + x) + y^2(1 + x)$

$\Rightarrow \frac{dy}{dx} = (1 + y^2)(1 + x)$

$\Rightarrow \frac{dy}{1 + y^2} = (1 + x) dx$

$\text{[applying variable separable method]}$

On integrating both sides, we get

$\tan^{-1} y = x + \frac{x^2}{2} + C \quad \dots (i)$

When $y = 0$ and $x = 0$, then substituting these values in Eq. (i), we get

$\tan^{-1} (0) = 0 + 0 + C$

$\Rightarrow C = 0$

$\Rightarrow \tan^{-1} y = x + \frac{x^2}{2}$

$\Rightarrow y = \tan \left( x + \frac{x^2}{2} \right)$