Show this initial value problem has a unique solution for initial value forall t

We shall use the standard existence and uniqueness results for ODEs. For instance see Theorems 1 and 2 in:
 
https://faculty.math.illinois.edu/~tyson/existence.pdf

Since the $F(u)=\cos(u) \sqrt{1+u^2} ​ +e^{ −u ^2}$ and its derivative are continuous functions on $\mathbb{R}$ (this could be easily verified), by Theorem 2 there exists a unique solution on $(-\delta_1, \delta_1)$ for some $\delta_1 >0$.

Suppose $\alpha\geq \delta_1$ is the largest value so that the equation has a unique solution on $(0, \alpha)$. We claim that $\alpha=\infty$. Suppose $\alpha <\infty$. Note that the solution $u$ is a continuous function on $(0, \alpha)$. In addition $u$ is also monotone on $(\alpha -\epsilon, \alpha)$ for some $\epsilon>0$ , thus $u^*:=\lim _{t \rightarrow \alpha ^-}u(t)$ exists. By Theorem 2, there exists a $\delta_1>0$ so that the ODE with initial condition $u(\alpha)=u^*$ has a unique solution on $(\alpha -\delta_2, \alpha+\delta_2)$. Therefore the ODE has a unique solution on $(0, \alpha+\delta_2)$ which is a contradiction as we had assumed $\alpha$ is the largest number with this property. Therefore $\alpha=\infty$.

Now assume $\beta<0$ is the smallest number so that the ODE has a solution on $(\beta, 0)$. With a similar argument we can show that $\beta=-\infty$. 

So the ODE has a unique solution on $(-\infty,\infty).$