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
( (-12 + 8x - x^2)y - 6(x - 4)y - 4y = 0, y(4) = 0, y(4) = 2 )
( y(x)=square )
(type any series in summation notation using ( n ) as the index variable and 0 as the starting index )

Explanation:

Step1: Make substitution

Let \(t=x - 4\), then \(x=t + 4\).
\(y(x)=y(t + 4)\), \(y^{\prime}(x)=\frac{dy}{dt}\), \(y^{\prime\prime}(x)=\frac{d^{2}y}{dt^{2}}\)
\(-12+8x - x^{2}=-12+8(t + 4)-(t + 4)^{2}=-12+8t+32-(t^{2}+8t + 16)=4 - t^{2}\)
\(6(x - 4)=6t\)
The differential equation becomes \((4 - t^{2})y^{\prime\prime}-6ty^{\prime}-4y = 0\)

Step2: Assume power - series solution

Assume \(y=\sum_{n = 0}^{\infty}c_{n}t^{n}\), then \(y^{\prime}=\sum_{n = 1}^{\infty}nc_{n}t^{n - 1}\), \(y^{\prime\prime}=\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}\)
\((4 - t^{2})\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}-6t\sum_{n = 1}^{\infty}nc_{n}t^{n - 1}-4\sum_{n = 0}^{\infty}c_{n}t^{n}=0\)
\(4\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}-\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n}-6\sum_{n = 1}^{\infty}nc_{n}t^{n}-4\sum_{n = 0}^{\infty}c_{n}t^{n}=0\)
Change the index in the first sum: let \(m=n - 2\), then \(n=m + 2\)
\(4\sum_{m = 0}^{\infty}(m + 2)(m + 1)c_{m+2}t^{m}-\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n}-6\sum_{n = 1}^{\infty}nc_{n}t^{n}-4\sum_{n = 0}^{\infty}c_{n}t^{n}=0\)
\(4\times2\times1c_{2}+4\times3\times2c_{3}t+4\sum_{m = 2}^{\infty}(m + 2)(m + 1)c_{m+2}t^{m}-\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n}-6\times1\times c_{1}t-6\sum_{n = 2}^{\infty}nc_{n}t^{n}-4c_{0}-4c_{1}t-4\sum_{n = 2}^{\infty}c_{n}t^{n}=0\)

Step3: Find recurrence relation

For \(t^{0}\): \(8c_{2}-4c_{0}=0\Rightarrow c_{2}=\frac{1}{2}c_{0}\)
For \(t^{1}\): \(24c_{3}-6c_{1}-4c_{1}=0\Rightarrow24c_{3}-10c_{1}=0\Rightarrow c_{3}=\frac{5}{12}c_{1}\)
For \(n\geq2\):
\(4(n + 2)(n + 1)c_{n+2}-n(n - 1)c_{n}-6nc_{n}-4c_{n}=0\)
\(4(n + 2)(n + 1)c_{n+2}-(n^{2}-n + 6n+4)c_{n}=0\)
\(4(n + 2)(n + 1)c_{n+2}-(n^{2}+5n + 4)c_{n}=0\)
\(c_{n+2}=\frac{(n + 1)(n + 4)}{4(n + 2)(n + 1)}c_{n}=\frac{n + 4}{4(n + 2)}c_{n}\)

Step4: Use initial conditions

Since \(y(4)=0\), when \(t = 0\) (i.e., \(x = 4\)), \(y(0)=c_{0}=0\)
Since \(y^{\prime}(4)=2\), when \(t = 0\), \(y^{\prime}(0)=c_{1}=2\)
\(c_{2}=0\), \(c_{3}=\frac{5}{6}\), \(c_{4}=0\), \(c_{5}=\frac{5}{6}\times\frac{5 + 4}{4(5+2)}=\frac{15}{28}\), \(c_{6}=0\), \(c_{7}=\frac{15}{28}\times\frac{7 + 4}{4(7+2)}=\frac{55}{336}\)
\(y=\sum_{n = 0}^{\infty}c_{n}t^{n}=2t+\frac{5}{6}t^{3}+\frac{15}{28}t^{5}+\frac{55}{336}t^{7}+\cdots\)
Substituting back \(t=x - 4\)
\(y(x)=2(x - 4)+\frac{5}{6}(x - 4)^{3}+\frac{15}{28}(x - 4)^{5}+\frac{55}{336}(x - 4)^{7}+\cdots=\sum_{n = 0}^{\infty}\frac{(2n+1)!!}{(2n + 2)!!}(x - 4)^{2n+1}\)

Step5: Find interval of convergence

The singular points of \((4 - t^{2})y^{\prime\prime}-6ty^{\prime}-4y = 0\) are \(t=\pm2\)
The radius of convergence \(R = 2\), and the interval of convergence for the power - series solution about \(t = 0\) (i.e., \(x = 4\)) is \(|x - 4|\lt2\), or \(x\in(2,6)\)

Answer:

\(y(x)=\sum_{n = 0}^{\infty}\frac{(2n+1)!!}{(2n + 2)!!}(x - 4)^{2n+1}\), interval of convergence \((2,6)\)