ODE with 2 parameterized families

[MisterX]

Homework Statement

problem:
Find a 1-parameter family of solutions of each of the following equa-
tions. Assume in each case that the coefficient of $dy \neq 0$.

$(x + \sqrt{ y^2 - xy}) \mathop{dy} - y \mathop{dx} = 0$

answer:
$y = ce^{-2\sqrt{1 - x/y}}, \;\;\; y >0, \, x< y; \;\;\; y = ce^{2\sqrt{1 - x/y}}, \;\;\; y<0, \,x >y$

Edit: This is Exercise 7 - 3 from Tennenbaum & Pollard's Ordinary Differential Equations.

Homework Equations

The Attempt at a Solution

my work:
$y\left(\frac{x}{y} + \sqrt{ 1 - \frac{x}{y}}\right) \mathop{dy} - y \mathop{dx} = 0, \;\;\;\; \frac{x}{y} < 1, \, y \neq 0$
$x = udy \;\;\; dx =y\mathop{du}+ u\mathop{dy}$
$y\left(u + \sqrt{ 1 - u}\right)\mathop{dy} - y \left(y\mathop{du}+ u\mathop{dy}\right) = 0$
$\sqrt{ 1 - u}\mathop{dy} - ydu = 0$
$\int\frac{dy}{y} = \int\frac{du}{\sqrt{ 1 - u}} = -\int\frac{dv}{\sqrt{v}}$
$ln(y) = -2\sqrt{v} + c = -2\sqrt{1-u} + c = -2\sqrt{1 - x/y} + c$
$y = c_1e^{-2\sqrt{1 - x/y}}$

I understand how $\frac{x}{y} < 1, \, y \neq 0$ is the same as $y<0,\, x > y$ or $y >0,\, x< y$. However I am not sure how to discover the 2nd 1-parameter solution with the positive exponent. I think I'm failing to consider something crucial. Also I may not feel as confident as I'd like about discovering multiple solutions to separable first order ODEs that are not part of the same family (generated by changing a parameter). So general advice or techniques might be appreciated.

 

[MisterX]

I think the two solutions split off before the first step when I factor out a y from within the square root.

$y\frac{x}{y}\mathop{dy} + \sqrt{ y^2\left(1 - \frac{x}{y}\right)} \mathop{dy} - y \mathop{dx} = 0, \;\;\;\; \frac{x}{y} < 1, \, y \neq 0$

Because square roots of positive numbers are always positive, if we factor out a y which happens to be negative, we'd need to have a minus sign.

$y\left(\frac{x}{y} + \sqrt{ 1 - \frac{x}{y}}\right) \mathop{dy} - y \mathop{dx} = 0, \;\;\;\; y > 0, \, x<y$
OR
$y\left(\frac{x}{y} - \sqrt{ 1 - \frac{x}{y}}\right) \mathop{dy} - y \mathop{dx} = 0, \;\;\;\; y < 0, \, x>y$The lesson I learned is to be mindful that $\sqrt{y^2(\dots)} = \left|y\right|\sqrt{\dots}$