I'm stuck in solving the following ptoblem:
Show that for each $\alpha \in \mathbb{R}$, $\alpha \geq 1$ exists $c \in \mathbb{R},$ $ c \geq 0$ so that the solution of the following non linear ODE is equal to $\alpha$ in $1$:
\begin{equation} \begin{cases} y_c'=\frac{\sqrt c}{ \sqrt{1+y_c^2}} \\y_c(0)=1 \end{cases} \end{equation}
I would like to solve it without finding esplicitly the solution. Here I write some thoughts I had about this problem:
For each $c > 0$ the solution is strictly increasing while constant and equal to $1$ if $c=0$
By observing that
$\frac{\sqrt c}{ \sqrt{2}y}=\frac{\sqrt c}{ \sqrt{2y^2}} \leq \frac{\sqrt c}{ \sqrt{1+y^2}}\leq \frac{\sqrt c}{ \sqrt{y^2}}=\frac{\sqrt c}{ y}$
if $y \geq 1 $ and by considering the relative Cauchy problems we deduce that:
$2(\frac{\sqrt c}{ \sqrt{2}}x)^{\frac{1}{2}}\leq y_c(x) \leq 2 (\sqrt{c} x)^{\frac{1}{2}} $
for every $x \in [0,1]$. In particular we deduce that $y_c(1)$ tends to $+ \infty$ as $c$ tends to $+ \infty$.
Can someone help? :)