I need soem help in the following exercise:
Consider the following Cauchy problem:
\begin{cases}y''+e^{x}y=0 \\ y(0)=1 \\ y'(0)=0\end{cases} Prove that $$|y(x)| \leq 1 \quad \text{for all }x \in [0,+\infty)$$
I started noticing that $y''(x)= -e^x y$ and hence $y''(0) = -1 <0$ Also, since $y'(0)=0$, the solution has a local maximum at $0$ and it starts its trajectory "decreasing".
Now I'd like to compute the stationary solutions,but I never faced such an analysis for a second order ODE, so I'm a bit puzzled. Of course, I could reduce everything to the first order by setting $$z'(x) = y(x)$$
hence I have $$-e^x y =0$$ and $$y=0$$
and the stationary solution is $y=0$, but I don't see how this helps me.
UNIQUENESS
I set $f(x,y)=[-e^x y , y]^T$ Then, I check that $f$ is Lipschitz in $y$, uniformly in $x$:
$\left|| f(x,y_1)-f(x,y_2) \right|| = \left|| [e^x (y_2 - y_1),y_1 - y_2]^T \right|| = (y_2 - y_1)^2 \Bigl( 1 + e^{2x} \Bigr) = \left|| y_1 - y_2\right|| \Bigl( 1 + e^{2x} \Bigr) $
This is strange, because I can't bound $\Bigl( 1 + e^{2x} \Bigr) < 0$ for every $x>0$...