Solving Homogenous First Order ODE Using Substitution Problem
Find the general solution of the following homogeneous equation.
$$ ty' = y + \sqrt { t^2 - y^2} $$
Attempt
I'm following the algorithm provided in section 1.5 of the William Adkins & Mark G. Davidson, Ordinary Differential Equations textbook on page 63, which suggests rearranging the equation into the form:
$$ y' =  ... $$
where $ y = tv $ and $ y' = v + tv' $ (v is the substitution variable).
I haven't been able to get too far in the procedure to make the substitution:
$$ y' = \frac{y + \sqrt { t^2 - y^2}}{t} $$
$$ y' = \frac{\frac{1}{t}}{\frac{1}{t}} \frac{y + \sqrt { t^2 - y^2}}{t} $$
$$ y' = \frac{y}{t} + \frac{\sqrt { t^2 - y^2}}{t} $$
Notes
Can someone please help me reduce the RHS of the above equation to the form $ y' = f(\frac{y}{t}) $ (an equation in terms of $ (\frac{y}{t}) $), using the substitution $ y = tv $? For bonus points, please provide a solution to the ODE so myself and other viewers can check their answers!
Thanks in advance!
Solution
With the help of Isham, I was able to get the solution.
$$ ty' = y + \sqrt { t^2 - y^2 } $$
It is implied that $ t^2 - y^2 \geq 0 $, and so $ t^2 \geq y^2 \gt 0 $ or $ \lt 0 $.
$$ ty' - y = \sqrt { t^2 - y^2} $$
$$ \frac {1}{\sqrt {t^2} } (ty' - y) = \frac {1}{\sqrt {t^2} } \sqrt { t^2 - y^2} $$
$$ \frac {1}{t} (ty' - y) = \sqrt { \frac{t^2}{t^2} - \frac{y^2}{t^2}} $$
$$ y' - \frac{y}{t} = \sqrt { 1 - \frac{y^2}{t^2}} $$
$$ y' - \frac{y}{t} = \sqrt { 1 - (\frac{y}{t})^2} $$
$$ y' = \sqrt { 1 - (\frac{y}{t})^2} + \frac{y}{t} $$
Let $ v = \frac{y}{t} $. Then, $ y = tv $ and so $ y' = \frac{dy}{dt} = v + tv' $. Substituting for $ v $ in the above equation we get:
$$ v + tv' = \sqrt {1 - v^2} + v $$
$$ tv' = \sqrt {1 - v^2} $$
Note, since $ t^2 \geq y^2 $, $ \frac{y^2}{t^2} = v^2 \leq 1 $, so there are no equilibrium cases to consider.
$$ t\frac{dv}{dt} = \sqrt {1 - v^2} $$
$$ \frac{1}{\sqrt { 1 - v^2}}dv = \frac{1}{t}dt $$
$$ \int\frac{1}{\sqrt { 1 - v^2}}dv = \int\frac{1}{t}dt $$
$$ arcsin(v) = ln|t| + C_1, C_1 \in \mathbb R $$
$$ \frac {y}{t} = \sin{(ln|t| + C_1)} $$
$$ y = t \sin{(ln|t| + C_1)} $$

According with the suggested 
$$
y'=v+\sqrt{1-v^2}
$$
with
$$
v = \frac yt
$$
so
$$
y' = v + t v'
$$
then
$$
v + t v' = v +\sqrt{1-v^2}
$$
or
$$
\frac{dv}{\sqrt{1-v^2}} = \frac{dt}{t}
$$
