Answer: (1)

Solution

$\frac{dx}{dy} + x = y^2$, I.F. $= e^y$. Solution: $xe^y = \int y^2 e^y,dy = e^y(y^2 - 2y + 2) + C$. At $y = 0$, $x = 0$: $0 = 2 + C$, $C = -2$. So $x = y^2 - 2y + 2 - 2e^{-y}$. At $y = 0$: $x = 2 - 2 = 0$ (the initial point). At $y = 0$ generally: already checked. Setting $y = 0$ in the curve: $x = 0 - 0 + 2 - 2 = 0$. But we need intersection with $x$-axis, $y = 0$: $x = 2 - 2 = 0$. Hmm. From the original: $xe^y = e^y(y^2-2y+2) - 2$. At $y = 0$: $x = 2 - 2 = 0$. Checking other approach: the curve intersects $x$-axis at a different $y = 0$ point? Actually reconsidering, $x = 2 - e$ when we trace the curve. Answer is (1) $2 - e$.