# Solution of a Cauchy ODE

Let $$y(t)$$ be a solution of a Cauchy Problem: $$\dot{y}=\ln\left(\sqrt{1+y^2}\right)$$ with the initial condition $$y(0)=y_0$$.

1. Prove that if $$y_0>0$$, then $$y$$ is a strictly increasing convex ; if $$y_0<0$$, then $$y$$ is strictly increasing and concave
2. Prove that $$y$$ is globally defined

Could someone please help me in understanding how to solve this kind of problems? I would like to be able to solve the whole class of such problems if possible.

To show $$y$$ is convex it suffices to prove $$y'' > 0$$. Note that since $$y_0 > 0$$, you have $$y'(0) = \ln \left(\sqrt{1+y_0^2}\right) = \ln(1+\epsilon) > 0$$ and $$y''(t) = \frac{dy'(t)}{dt} = ...$$ can you argue the result is always positive?
We have that $$\sqrt{1+y_t^2} \geq 1 > 0$$ and as the function $$x \mapsto \ln(x)$$ is (strictly) increasing in $$\mathbb{R}_+$$. Hence, we have : $$y_t' \geq 0$$. In fact, this inequality is a strict inequality if $$y_0 >0$$. Finally, we know that a strictly increasing function is a strictly convex.