Prove existence and uniqueness for a Cauchy problem I need a help in proving the existence and uniqueness for the following Cauchy problem:
\begin{cases}y''+e^{x}y=0  \\ y(0)=1 \\ y'(0)=0\end{cases}
This can be recasted as a first order system where  $f$ is defined as $$f(x,y)=[-e^x y , y]^T$$
In order to prove (local) existence and uniqueness, I need to show that $f$ is locally Lipschitz w.r.t $y$, (it is the RHS of an ODE)
I compute:
$$\left|| f(x,y_1)-f(x,y_2) \right|| = \left|| [e^x (y_2 - y_1),y_1 - y_2]^T \right|| = (y_2 - y_1)^2 \Bigl( 1 + e^{2x} \Bigr) = \left|| y_1 - y_2\right|| \Bigl( 1 + e^{2x} \Bigr) $$
So, for $|x| < a$ (i.e. in a neigbourhood of $x_0=0$ I have $$\Bigl( 1 + e^{2x} \Bigr)\leq1+e^{2a}$$, so it's locally Lipschitz (but no globally)
Is everything correct?
 A: You got the function $f(x,y)$ wrong. What you need to do is define a third variable to serve as the first derivative of $y$. The function you want is $$f([y,y']^T,x) = [y',-e^xy]^T$$. This is the function you want to show is Lipschitz.
A: $$\frac{d^2y}{dx^2}+e^x y=0$$
Change of variable :$\quad e^x=t\quad\implies\quad \frac{dt}{dx}=t$
$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=t\frac{dy}{dt}$
$\frac{d^2y}{dx^2}=\frac{d\frac{dy}{dx}}{dt}\frac{dt}{dx}=(\frac{dy}{dt}+t\frac{d^2y}{dt^2})t=t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt}$
$$t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt}+ty=0$$
$$\frac{d^2y}{dt^2}+\frac{1}{t}\frac{dy}{dt}+\frac{1}{t}y=0$$
This is a Bessel equation which solution is well known. See Eq.(6) and (7) in : https://mathworld.wolfram.com/BesselDifferentialEquation.html
$$y(t)=c_1J_0\big(2\sqrt{t}\big)+c_2Y_0\big(2\sqrt{t}\big)$$
$J_0$ and $Y_0$ are the Bessel functions of the first and second kind respectively.
The general solution of the ODE is :
$$y(x)=c_1J_0\big(2e^{x/2}\big)+c_2Y_0\big(2e^{x/2}\big)$$
The coefficients $c_1$ and $c_2$ are determined according to the conditions
$y(0)=1$ and $y'(0)=0$ which leads to the unique solution :
$$y(x)=\frac{Y_1(2)J_0\big(2e^{x/2}\big)-J_1(2)Y_0\big(2e^{x/2}\big)}{Y_1(2)J_0(2)-J_1(2)Y_0(2)}$$
