Existence and Uniqueness for a nonlinear ODE Suppose I have the first-order nonlinear ODE: $$(y’)^2+y^2=1$$
Upon inspection, we see that $y(t)=\pm1$ and $y(t)=\pm\sin(t+a)$. ‘Another’ solution one sees upon inspection is $y(t)=\pm\cos(t+a),$ but this is contained in the other solution ($\pm\sin(t+a)$). 
My question is: 
Are these solutions unique? Are there any others and how do we know? Thanks in advance. 
 A: There are also "piecewise" solutions such as 
$$ y(t) = \cases{-1 & if $t \le -\pi/2$ \cr 
                \sin(t) & if $-\pi/2 \le t \le \pi/2$\cr
                 1  & if $t \ge \pi/2$\cr} $$
For each initial condition $y(t_0) = y_0$ with $-1 < y_0 < 1$, there are two solutions, one increasing and one decreasing, as long as $|y|$ stays less than $1$.  But once you hit $y = \pm 1$, there is no more uniqueness.
A: You need to write the ODE in the normal form $y'=f(y)$. Once it is written in this form, check if $f$ is Lipschitz. If $f$ is Lipschitz, the initial value problem (i.e., when you specify the value $y(0)$) has a unique solution.
In your example, we have 2 normal forms, or 2 branches to consider:
$$
f(y)=\pm\sqrt{1-y^2}.
$$
Hence, in the best case scenario, we would get exactly 2 solutions for a given initial data $y(0)$. Sticking to reals for simplicity, we have the following.


*

*Exactly 2 local solutions if $-1<y(0)<1$.

*The branches of $f$ are non-Lipschitz at $y=\pm1$, so more than 2 solutions if $y(0)=\pm1$.

*Different solutions can be glued together at $t$ whenever $y(t)=\pm1$.

