Introduction

Proof. Let $$V=\{y:I\to\mathbb{R}\mid y \text{ satisfies the differential equation}\}.$$ Since the equation is linear and homogeneous, $V$ is a vector space.

Fix a point $x_0\in I$. Define the map $$T:V\longrightarrow\mathbb{R}^n$$ by $$T(y)=\bigl(y(x_0),y'(x_0),\ldots,y^{(n-1)}(x_0)\bigr).$$

Clearly, $T$ is linear.

We first show that $T$ is injective. Suppose $$T(y)=0.$$ Then $$y(x_0)=y'(x_0)=\cdots=y^{(n-1)}(x_0)=0.$$ By the uniqueness theorem for linear differential equations, the initial value problem $$L(y)=0,\qquad y(x_0)=y'(x_0)=\cdots=y^{(n-1)}(x_0)=0$$ has a unique solution. Since the zero function satisfies these conditions, it follows that $$y\equiv0.$$ Hence, $$\ker(T)=\{0\},$$ so $T$ is injective.

Next, we show that $T$ is surjective. Let $$(c_0,c_1,\ldots,c_{n-1})\in\mathbb{R}^n$$ be arbitrary. By the existence theorem for linear differential equations, there exists a solution $y\in V$ satisfying $$y(x_0)=c_0,\qquad y'(x_0)=c_1,\qquad \ldots,\qquad y^{(n-1)}(x_0)=c_{n-1}.$$ Therefore, $$T(y)=\bigl(c_0,c_1,\ldots,c_{n-1}\bigr),$$ so every element of $\mathbb{R}^n$ lies in the image of $T$. Thus $T$ is surjective.

Since $T$ is both injective and surjective, it is a vector space isomorphism between $V$ and $\mathbb{R}^n$. Hence, $$V\cong\mathbb{R}^n,$$ and therefore $$\dim(V)=\dim(\mathbb{R}^n)=n.$$ ◻

Proof. Suppose, for the sake of contradiction, that $$\dim(V)=k<n.$$

Let $\{y_1,y_2,\ldots,y_k\}$ be a basis for $V$. Then every solution $y\in V$ can be written uniquely as $$y=c_1y_1+c_2y_2+\cdots+c_ky_k,$$ where $c_1,\ldots,c_k\in\mathbb{R}$.

Consequently, the initial value vector of any solution at a fixed point $x_0\in I$ is given by $$\bigl(y(x_0),y'(x_0),\ldots,y^{(n-1)}(x_0)\bigr) = \sum_{i=1}^{k} c_i \bigl(y_i(x_0),y_i'(x_0),\ldots,y_i^{(n-1)}(x_0)\bigr), i \in \{1,...k\}$$

Hence every initial value vector belongs to the span of the $k$ vectors $$\bigl(y_i(x_0),y_i'(x_0),\ldots,y_i^{(n-1)}(x_0)\bigr), \qquad i=1,\ldots,k.$$

Therefore, the set of all possible initial value vectors has dimension at most $k<n$, but this is a contradiction since we are essentially saying that $R^n$ can be spanned using $k<n$ vectors which means that the dimension of $R^n$ is k. $$\therefore \quad dim(V) = k \ge n$$ But the dimension of $R^n$ is $n$ ,i.e, dimension of V cannot exceed $n$ either since even if it did, we can reduce it to $n$. $$\therefore \quad dim(V) = n$$ ◻