The largest open interval $I$ containing $x = 0$ on which $\exists !$solution $y(x)$ to this problem? (Understanding the solution)

Question

Consider the initial value problem:

$(\sin(x)-1)y''' + (x^2-x)y'' + 1\frac{1}{(x-1)}y' + x^5y = e^{x^3}$

$y(0) = 1, y'(0) = 5, y''(0) = 2$

What is the largest open interval $I$ containing $x = 0$ on which $\exists!$ solution $y(x)$ to this problem?

Solution:

After division by $a(x) = \sin(x) − 1$, the coeffs of the linear DE are continuous everywhere except at $x = 1$ and at roots of $a(x)$, namely $x=\frac{\pi}{2}\, k\in\Bbb Z$ Then
$I = \left(−\frac{3\pi}{2}, 1\right) $
is the largest open interval containing $0$ that avoids these. Existence
& uniqueness of solutions is guaranteed on this interval.

I dont understand:

1. Why do you have to get $y'''$ alone to find the interval, (why y''')?
2. When do i use: $y(0) = 1, y'(0), y''(0)=2$ in this this question?
3. Why is existence and uniqueness of solutions guaranteed on this interval?

Answer 1

Your equation is linear. A standard result on existence says that if you write the equation in normal form $$ Y^{(n)}=a_{n-1}(x)Y^{(n-1)}+\dots+b(x) $$ Then solutions starting at some point, $x=0$ in your case, can be extended as long as the coefficients are continuous. The reason for singling out the highest order derivatives is related to the proof which relies on reduction to a first order system. The smallest interval containing zero where your coefficients remain continuous is the one indicated in the answer.

You do not need the initial values here, just the starting point $x=0$.

Uniqueness is not an issue for linear equations.