Let $y=y(x)$ be the solution curve of the differential equation, $\left(y^{2}-x\right) \frac{d y}{d x}=1$, satisfying $y(0)=1$. This curve intersects the $x$-axis at a point whose abscissa is:
Correct Option: 1
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)
$0=1-2+2+\frac{C}{e}$
$C=-e$
$y=0, x=0-0+2+(-e)\left(e^{-0}\right)$
$x=2-e$