Sufficient condition for existence and uniqueness of solution of $1^{st}$ order linear ODE

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Note: Through out the discussion we assume the following hypothesis:

$H_{IVP}$: Let $f:\mathbb{I}\times (\Omega \subseteq \mathbb{R}^n) \rightarrow (\Omega \subseteq \mathbb{R}^n)$ be a continuous function with $\mathbb{I} \subseteq \mathbb{R}$ containing $x_{0} \in \mathbb{R}$. Here $n \in \mathbb{N}$.

Now let's formalise the condition:

Theorem (Picard-Lindelof):
Consider the following IVP:
$$\begin{equation} \frac{dy}{dx}=f(x,y) \quad y(x_0)=y_0 \end{equation}$$
where $f$ satisfies the $H_{IVP}$, where $(x_0,y_0) \in \mathbb{I} \times \Omega$ being interior point of the set along with $\frac{\partial f}{\partial y}$ also satisfies $H_{IVP}$. Also consider the rectangle:
$$R=\{(x,y)\in \mathbb{I} \times \Omega: |x-x_0|\leq a, ||y-y_0||\leq b, \quad a,b \in \mathbb{R}_+ \} \subseteq \mathbb{I}\times \Omega$$

Let $$M=\operatorname{Sup}_{x\in R}f$$ then (1) has a solution in a interval contained in $(x_0-\delta,x_0+\delta)$, where $\delta =\operatorname{min}\{a,\frac{b}{M}\}$.

Note: A more general condition can be imposed on f as f being Lipsticz continuous on $y$ variable instead of the condition $\frac{\partial f}{\partial y}$ satisfying $H_{IVP}$.

Lipsticz Continous: Let $f:X\subseteq \mathbb{R}^m \rightarrow Y\subseteq \mathbb{R}^n$ is called a Lipsticz fucntion if $$||f(x)-f(y)||_n \leq k ||x-y||_m$$ where $k \in \mathbb{R}_+$.

*Lipsticz Continous (on $y$-variable): Let $f:\mathbb{I}\times (\Omega \subseteq \mathbb{R}^m) \rightarrow (\Omega \subseteq \mathbb{R}^m)$ is called a Lipsticz fucntion if $$||f(x,y_1)-f(x,y_2)||_m \leq k ||y_1-y_2||_m \quad \text{for each fixed } x \in \mathbb{R}$$ where $k \in \mathbb{R}_+$, [where k doesn't depend on x].$k$ could very well depend on $x$ in general but or our purpose we enforce the independece of k on x.

Now, if we assume that $\frac{\partial f}{\partial y}$ is continous then on the closed bounded set $R$ in above theorem $\exists M \geq 0$ such that $||\frac{\partial f}{\partial y}(x,y)||_{op} \leq M \quad (x,y) \in R$. Now, by Lagrange's Mean Value Theorem we get the following inequality:
$$\begin{equation} f(x,y_1)-f(x,y_2)=\frac{\partial f}{\partial y}(x,\zeta)(y_1-y_2) \end{equation}$$
where $\zeta \in (y_1,y_2)$, [where $(y_1,y_2):=\{y_1(1-t)+y_2t: t\in (0,1)\}$]
Now taking norm both side we get,
$||f(x,y_1)-f(x,y_2)||_m=||\frac{\partial f}{\partial y}(x,\zeta)(y_1-y_2)||_m$
$\implies ||f(x,y_1)-f(x,y_2)||_m=||\frac{\partial f}{\partial y}(x,\zeta)||_{op}||(y_1-y_2)||_m \leq M||y_1-y_2||_m$
That is f becomes Lipscitz.

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Handling $n^{th}$ order.

I am here showing a particular $2^{nd}$-order case. General case will be similar.

Consider the following IVP:
$$\begin{equation} a_0(x)y''+a_1(x)y'+a_2(x)y=r(x) \quad y(0)=c_1,y'(0)=c_2 \end{equation}$$
where $c_1,c_2 \in \mathbb{R}$ and $(a,b) \subseteq \mathbb{R}$ such that $0 \in (a,b)$ and $a_i(x),r(x)$ are continous functions on $(a,b)$ with $a_0(x)\neq 0 \quad \forall x \in (a,b)$.

Then (3) has a unique solution in a interval containg 0 and contained in $(a,b)$.

Solution:
The trick is to convert the second order ODE to first order ODE and use our known method.

Let $Y=(y,y')$
we have $y''=\frac{r}{a_0}-\frac{a_1}{a_0}y'-\frac{a_2}{a_0}y$

Then we have,
$$\begin{equation} \frac{dY}{dx}=f(x,Y) \quad Y(0)=(y(0),y'(0))=(c_1,c_2)=Y_0 \end{equation}$$
where $f(x,Y)=(y',\frac{r}{a_0}-\frac{a_1}{a_0}y'-\frac{a_2}{a_0}y)$

Now, $f$ is continous function on $(a,b) \times \mathbb{R}^2$. Here we don't even need to consider a subset $\Omega \subseteq \mathbb{R}^2$ containing $(c_1,c_2)$ as the $f$ is linear in both $y,y'$ and hence continous on whole of $\mathbb{R}^2$ and $a_0 \neq 0$ in $(a,b)$ hence $\frac{r}{a_0},\frac{a_1}{a_0},\frac{a_2}{a_0}$ are continous on $(a,b)$.

Also, $$\frac{\partial f}{\partial Y}=\begin{pmatrix} 0 & 1 \\ -\frac{a_2}{a_0} & -\frac{a_1}{a_0} \end{pmatrix}$$ is also continous on $(a,b) \times \mathbb{R}^2$ for same reason above. Hence by Theorem (Picard-Lindelof) we have a unique solution to the ODE $(4) \implies (3)$.