Chapter 2.4, Problem 4P

Existence and uniqueness of Solutions. In each of Problem $1$ through $6$, use Theorem $\mathrm{2.4.1}$ to determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist.

$\left(4-t^2\right)y\text{'}+2ty=3t^2,y\left(-3\right)=1$.

Theorem $\mathrm{2.4.1}$:

If the functions $p$ and $g$ are continuous on an open interval $I=\left(\alpha ,\beta \right)$ containing the point $t=t_0$, then there exists a unique function $y=\varphi \left(t\right)$ that satisfies the differential equation

$y\text{'}+p\left(t\right)y=g\left(t\right)$ for each $t$ in $I$ and satisfies the initial condition

$y\left(t_0\right)=y_0$,

where $y_0$ is an arbitrary prescribed initial value.