Let $$y = y(x)$$ be the solution curve of the differential equation, $$(y^2 - x)\frac{dy}{dx} = 1$$, satisfying $$y(0) = 1$$. This curve intersects the X-axis at a point whose abscissa is

We are given the differential equation

$$ (y^2 - x)\,\frac{dy}{dx}=1 $$

together with the initial condition

$$ y(0)=1. $$

Our aim is to find the point where the solution curve meets the X-axis, i.e. the point where $$y=0$$, and hence obtain the required abscissa (the $$x$$-coordinate).

First, we rewrite the differential equation so that $$x$$ becomes the dependent variable and $$y$$ the independent variable. We have

$$ (y^2 - x)\,\frac{dy}{dx}=1. $$

Dividing both sides by $$\dfrac{dy}{dx}$$ (which is non-zero along the smooth solution curve) gives

$$ y^2 - x = \frac{dx}{dy}. $$

So we can write

$$ \frac{dx}{dy} = y^2 - x. $$

This is a first-order linear differential equation in the variable $$x(y)$$. The standard linear form is

$$ \frac{dx}{dy} + x = y^2. $$

For a linear equation of the form $$\dfrac{dx}{dy}+P(y)\,x=Q(y)$$ the integrating factor is $$\mu(y)=e^{\int P(y)\,dy}$$. In our case, $$P(y)=1$$, so

$$ \mu(y)=e^{\int 1\,dy}=e^{y}. $$

Multiplying the entire equation by this integrating factor, we obtain

$$ e^{y}\,\frac{dx}{dy} + e^{y}\,x = y^2\,e^{y}. $$

The left-hand side is the derivative of the product $$e^{y}x$$, because

$$ \frac{d}{dy}\bigl(e^{y}x\bigr)=e^{y}\frac{dx}{dy}+e^{y}x. $$

Therefore, our equation becomes

$$ \frac{d}{dy}\bigl(e^{y}x\bigr)=y^2\,e^{y}. $$

We now integrate both sides with respect to $$y$$:

$$ e^{y}x = \int y^2\,e^{y}\,dy + C, $$

where $$C$$ is the constant of integration.

To evaluate the integral $$\displaystyle\int y^2 e^{y} \, dy$$, we use repeated integration by parts. The result, which can be verified step by step, is

$$ \int y^2 e^{y}\,dy = e^{y}\,(y^2 - 2y + 2) + C_1, $$

where $$C_1$$ is another constant. We may absorb $$C_1$$ into the general constant $$C$$ already present, so substituting the evaluated integral gives

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

Dividing by $$e^{y}$$ (which is never zero) yields the explicit relation between $$x$$ and $$y$$:

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

We now determine the constant $$C$$ from the initial condition $$y(0)=1$$. At $$x=0$$ and $$y=1$$ we substitute into the above expression:

$$ 0 = (1)^2 - 2(1) + 2 + C\,e^{-1}. $$

Simplifying inside the parentheses, we get

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

$$ 0 = 1 + \frac{C}{e}. $$

Therefore,

$$ \frac{C}{e} = -1 \quad\Longrightarrow\quad C = -e. $$

Substituting this value of $$C$$ back into the relation for $$x$$ gives the particular solution:

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

Since $$e\,e^{-y}=e^{1-y}$$, we can write more compactly

$$ x = y^2 - 2y + 2 - e^{1-y}. $$

To find where the curve crosses the X-axis, we set $$y=0$$ (because any point on the X-axis has ordinate zero):

$$ x = (0)^2 - 2(0) + 2 - e^{1-0}. $$

$$ x = 2 - e^{1}. $$

$$ x = 2 - e. $$

Thus, the required abscissa is $$2 - e$$.

Hence, the correct answer is Option A.