= Solution
This is a <separable differential equation>. On the solution with $y<1$, integration from the <initial condition> gives
$$
\int_0^y\frac{dv}{\sqrt{1-v^2}}=\int_0^x\frac{u\,du}{\sqrt{1-u^2}},\qquad \arcsin y=1-\sqrt{1-x^2}.
$$
The right side lies in $[0,1)$, so taking its sine stays on the required branch. \b[The solution is]
$$
\boxed{y(x)=\sin\left(1-\sqrt{1-x^2}\right),\quad 0\le x<1.}
$$
Near the origin, $1-\sqrt{1-x^2}=x^2/2+O(x^4)$, so $y=x^2/2+O(x^4)$ and the <tangent line> is horizontal. At the other end, $y\to\sin1$, while
$$
y'=\frac{x\cos(1-\sqrt{1-x^2})}{\sqrt{1-x^2}}\longrightarrow+\infty.
$$
Thus the curve has a vertical limiting <tangent line> as $x\uparrow1$, even though $x=1$ is excluded from the domain.
For the <direction field>, the slope is nonnegative everywhere in the square. The segments are horizontal on $x=0$ and on $y=1$, become steeper as $x$ increases at fixed $y<1$, and become flatter as $y$ increases at fixed $x$. Near $x=1$ they are almost vertical unless $y$ is correspondingly close to one. The excluded corner $(1,1)$ has no single limiting slope: the ratio $(1-y^2)/(1-x^2)$ depends on the path of approach.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-2-direction-field.png]
{title=Positive slope field and the solution from the origin, with its horizontal and vertical limiting tangents}
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