differential equation/Lipschitz I want to show that the $ (t_- , t_+) $ of $$ y'=\cos y \sqrt{t^2+y^2},  y(0)=\pi$$ 
is $ (t_- , t_+) =\mathbb R$
Therefore I want to show, that f is Lipschitz.
$$ \bigg|\cos y_1 \sqrt{t^2+y_1^2} - \cos y_2 \sqrt{t^2+y_2^2}\bigg|\leq \bigg|\sqrt{t^2+y_1^2}+\sqrt{t^2+y_1^2} \leq\bigg| \frac{t^2+y_1^2-t^2-y_2^2}{\sqrt{t^2+y_1^2}-\sqrt{t^2+y_1^2}}\bigg| $$
This leads to nothing, because I have "-" at the denominator. 
Is there an easier way?
 A: The function
$$
 f: \Bbb R \times (0, \infty) \to \Bbb R \, \\
 f(t, y) = \cos y \sqrt{t^2+y^2}
$$
is locally Lipschitz continuous with respect to $y$ because the partial derivative
$$
   \frac{\partial}{\partial y} f(t, y) = -\sin y \sqrt{t^2+y^2} + \cos y \frac{y}{\sqrt{t^2+y^2}}
$$
exists and is continuous in $ \Bbb R \times (0, \infty)$. Therefore the initial value problem
$$
 y' = f(t, y) \, , \quad y(0) = \pi
$$
has a unique solution in a neighbourhood of $t=0$, and that solution has a “maximal interval of existence” $(t_- , t_+)$. If $t_+ < \infty$ then
$$ \tag{*}
 \lim_{t \nearrow t_+} |y(t)| = \infty
$$
must hold, and similarly for the left boundary. (The solution “blows up” at the boundaries of the existence interval. See for example “1.5 Intervals of Existence” in  Christopher P. Grant: Theory of Ordinary Differential Equations.)
Now the initial value problems
$$
 y' = f(t, y) \, , \quad y(0) = \pi/2 \\
 y' = f(t, y) \, , \quad y(0) = 3 \pi/2
$$
have the constant solutions $y_1(t) = \pi/2$ and $y_2(t) = 3\pi/2$, respectively. Because of the local uniqueness of solutions, $y, y_1, y_2$ cannot intersect, so that
$$
   \pi/2 < y(t) < 3 \pi/2
$$
for $t \in (t_- , t_+)$. In other words, $y$ is bounded and $(*)$ cannot hold.
Consequently, the maximal existence interval of $y$ is $(t_- , t_+) = (-\infty, \infty) = \Bbb R$.
Remark: $f(y, t)$ is not globally Lipschitz continuous with respect to $y$ since the partial derivative $\frac{\partial}{\partial y} f(t, y)$ is not globally bounded. Therefore  Lipschitz continuity alone is not sufficient to show that the solution exists on $\Bbb R$.
A: I would write $$\cos(y_1)\sqrt{t^2+y_1^2}-\cos(y_2)\sqrt{t^2+y_1^2}+\cos(y_2)\sqrt{t^2+y_2^2}-\cos(y_2)\sqrt{t^2+y_1^2}$$
The first summand is $$\sqrt{t^2+y_1^2}(\cos(y_1)-\cos(y_2))=\sqrt{t^2+y_1^2}(y_1-y_2)(-\sin(\xi))$$ by the mean value theorem.
And $$\sqrt{t^2+y_1^2}-\sqrt{t^2+y_1^2}=(t^2+y_2^2-t^2-y_1^2)\cdot \frac {1}{2}(t^2+\eta^2)^{-1/2}2\eta$$
