Tenenbaum & Pollard Pages 53-54: Apparently does not make sense

Hi. I'm reading the bookhttp://amzn.com/0486649407 [Broken],

in self-study mode.
In page 53 and 54, below:

Apparently does not make sense, because, If the differential equation is:
$$ \frac{\mathrm{d}y }{\mathrm{d} x} = x\frac{\sqrt{1-y}}{\sqrt{1-x^{2}}} $$

then dy/dx = ∞ when x = 1, and y < 0
however, in Image:

dy / dx is not ∞ in point marked in green.

Then, in this point, the graph changes radically the slope? Apparent, does not make sense, It seems that the correct curve would go up until tangent to x = 1, while the particular solution says otherwise.

Why the differential equation "fails", and the solution of the differential equation not? I did not catch that.

I am studying alone, thank you for your patience.

 

Yes, if the dy/dx go to ∞ when x go to 1, then i guess in particular solution this should be the trend.
It is as if the particular solution and differential equation were conflicting.
Imagine that someone will draw the solution curves, from the dy / dx, at each point, does not match the presented solution.
However, the method of solution seems correct, that leaves me confused!

YES, when y<1. Sorry.

 

Thank You!!!

The solution:
$$ \sqrt{1-x^{2}} - 2\sqrt{1-y} = C $$
Results:
$$ y = \frac{3}{4} + \frac{x^{2}}{4} + \frac{C\sqrt{1-x^{2}}}{2} - \frac{C^{2}}{4} $$
(in your solution, yourC = C/2)

Matlab Code

PHP:

x = linspace(-1,1,1000);
clf;
grid on;
hold on;
c=0;
y=3/4+(x.^2)/4 +(c/2)*sqrt(1-(x.^2))-(c/2)^2;plot(x, y, 'r');
c=-0.5;
y=3/4+(x.^2)/4 +(c/2)*sqrt(1-(x.^2))-(c/2)^2;plot(x, y, 'b');
c=-1;
y=3/4+(x.^2)/4 +(c/2)*sqrt(1-(x.^2))-(c/2)^2;plot(x, y, 'g');
 

Matlab Result


c = 0
c = -0.5
c = -1


Thank you, Helped me a lot!