Question

solve the initial value problem. first make a substitution of the form ( t = x - a ), then find a solution ( sum c_{n}t^{n} ) of the transformed differential equation. state the guaranteed interval of convergence
( left(-6 + 6x - x^{2}
ight)y - 6(x - 3)y - 4y = 0, y(3) = 0, y(3) = 7 )
( y(x)=7sum_{n = 0}^{infty}\frac{2n + 3}{3left(3^{n}
ight)}(x - 3)^{2n + 1} )
(type any series in summation notation using ( n ) as the index variable and 0 as the starting index)
the series converges if ( <x< )

Explanation:

Step1: Analyze the substitution \(t = x - 3\)

Let \(t=x - 3\), then \(x=t + 3\).
The differential equation \((-6 + 6x-x^{2})y''-6(x - 3)y'-4y = 0\) can be rewritten.
First, \(-6+6x - x^{2}=-6+6(t + 3)-(t + 3)^{2}=-6+6t+18-(t^{2}+6t + 9)=3 - t^{2}\)

Step2: Find the singular points of the original differential equation

The original differential equation \(y''+\frac{-6(x - 3)}{-6 + 6x-x^{2}}y'+\frac{-4}{-6 + 6x-x^{2}}y = 0\)
We solve \(-6 + 6x-x^{2}=0\), \(x^{2}-6x + 6=0\) using the quadratic formula \(x=\frac{6\pm\sqrt{36-24}}{2}=3\pm\sqrt{3}\)

Step3: Determine the radius of convergence

The power - series solution about \(x = 3\) (since \(t=x - 3\)) has a radius of convergence \(R\) equal to the distance from \(x = 3\) to the nearest singular point.
The distance from \(x = 3\) to \(x=3+\sqrt{3}\) and \(x=3-\sqrt{3}\) is \(\sqrt{3}\)

Answer:

The series converges if \(3-\sqrt{3}