Let y=y(x) be the solution curve of the differential equation,

Solution:

The given differential equation is $\frac{d x}{d y}+x=y^{2}$\

Comparing with $\frac{d x}{d y}+P x=Q$, where $P=1, Q=y^{2}$

Now, I.F. $=e^{\int 1 \cdot d y}=e^{y}$

$x \cdot e^{y}=\int\left(y^{2}\right) e^{y} \cdot d y=y^{2} \cdot e^{y}-\int 2 y \cdot e^{y} \cdot d y$

$=y^{2} e^{y}-2\left(y \cdot e^{y}-e^{y}\right)+C$

$\Rightarrow x . e^{y}=y^{2} e^{y}-2 y e^{y}+2 e^{y}+C$

$\Rightarrow x=y^{2}-2 y+2+C . e^{-y}$........(1)

As $y(0)=1$, satisfying the given differential eqn,

$\therefore$ put $x=0, y=1$ in eqn. (1)

$0=1-2+2+\frac{C}{e}$

$C=-e$

$y=0, x=0-0+2+(-e)\left(e^{-0}\right)$

$x=2-e$