how to prove that a solution exists for a fourth order ode Does there exists a solution of the initial value problem
$$(x^2 − 4)\frac {d^4y}{dx^4} + 2x \frac {d^2y}{dx^2} + (\sin x)y(x) = 0$$
where $y(0) = 0, y'(0) = 1, y''(0) = 1, y'''(0) = −1$
So far I have $y_1=y$, $y_2= \frac {dy}{dx}$ and so on til $y_n = \frac {d^{n-1}y}{dx^{n-1}}$ from the Existence and Uniqueness Theory. i dont know how to proceed further. i tried to solve it through the existence and uniqueness theorem but this is all i have right now

I do not guarantee that no mistakes have been made.
Let
\begin{equation*}
y_{n}=\partial _{x}^{n-1}y
\end{equation*}
Then
\begin{eqnarray*}
\partial _{x}y_{1} &=&y_{2} \\
\partial _{x}y_{2} &=&y_{3} \\
\partial _{x}y_{3} &=&y_{4} \\
\partial _{x}y_{4} &=&y_{5}
\end{eqnarray*}
and
\begin{eqnarray*}
(x^{2}-4)\partial _{x}^{4}y+2x\partial _{x}^{2}y+\sin xy &=&0 \\
(x^{2}-4)y_{5}+2xy_{3}+\sin xy_{1} &=&0 \\
y_{5}+\frac{2x}{x^{2}-4}y_{3}+\frac{\sin x}{x^{2}-4}y_{1} &=&0 \\
\partial _{x}y_{4} &=&-\frac{2x}{x^{2}-4}y_{3}-\frac{\sin x}{x^{2}-4}y_{1}
\end{eqnarray*}
Now let
\begin{equation*}
\mathbf{y}(x)=\left(
\begin{array}{c}
y_{1} \\
y_{2} \\
y_{3} \\
y_{4}%
\end{array}
\right) ,\;\mathbf{y}(0)=\left(
\begin{array}{c}
0 \\
1 \\
1 \\
-1
\end{array}
\right)
\end{equation*}
Then
\begin{equation*}
\partial _{x}\mathbf{y}=\left(
\begin{array}{c}
y_{2} \\
y_{3} \\
y_{4} \\
-\frac{2x}{x^{2}-4}y_{3}-\frac{\sin x}{x^{2}-4}y_{1}%
\end{array}
\right) =\left(
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-\frac{\sin x}{x^{2}-4} & 0 & -\frac{2x}{x^{2}-4} & 0
\end{array}
\right) \mathbf{y}=\mathsf{M}\cdot \mathbf{y}
\end{equation*}
or
\begin{equation*}
\mathbf{y}(x)=\exp [x\mathsf{M}]\cdot \mathbf{y}(0)
\end{equation*}
Suppose that $\mathsf{M}\cdot \mathbf{u}=0$. Then
\begin{equation*}
\left(
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-\frac{\sin x}{x^{2}-4} & 0 & -\frac{2x}{x^{2}-4} & 0
\end{array}
\right) \left(
\begin{array}{c}
u_{1} \\
u_{2} \\
u_{3} \\
u_{4}
\end{array}
\right) =\left(
\begin{array}{c}
u_{2} \\
u_{3} \\
u_{4} \\
-\frac{\sin x}{x^{2}-4}u_{1}-\frac{2x}{x^{2}-4}u_{3}
\end{array}
\right) =0
\end{equation*}
and it follows that $\mathbf{u}$ must vanish. Thus $\mathsf{M}$ has an empty
null-space and is well-behaved.
