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.
( ( - 20 + 10x - x^{2})y - 6(x - 5)y - 4y = 0;y(5)=0,y(5)=4 )
( y(x)=square )
(type any series in summation notation using ( n ) as the index variable and 0 as the starting index)

Explanation:

Step1: Substitute \(t = x - 5\)

Let \(x=t + 5\). Then \(y(x)=y(t + 5)\), \(y^{\prime}(x)=\frac{dy}{dt}\), \(y^{\prime\prime}(x)=\frac{d^{2}y}{dt^{2}}\).

The equation \((-20 + 10x-x^{2})y^{\prime\prime}-6(x - 5)y^{\prime}-4y = 0\) becomes:

First, \(-20+10x - x^{2}=-20+10(t + 5)-(t + 5)^{2}=-20 + 10t+50-(t^{2}+10t + 25)=5 - t^{2}\)

The equation is \((5 - t^{2})\frac{d^{2}y}{dt^{2}}-6t\frac{dy}{dt}-4y = 0\)

Step2: Assume a power - series solution \(y=\sum_{n = 0}^{\infty}c_{n}t^{n}\)

\(y^{\prime}=\sum_{n = 1}^{\infty}nc_{n}t^{n - 1}\), \(y^{\prime\prime}=\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}\)

Substitute into \((5 - t^{2})y^{\prime\prime}-6ty^{\prime}-4y = 0\)

\((5 - t^{2})\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}-6t\sum_{n = 1}^{\infty}nc_{n}t^{n - 1}-4\sum_{n = 0}^{\infty}c_{n}t^{n}=0\)

\(5\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}-\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n}-6\sum_{n = 1}^{\infty}nc_{n}t^{n}-4\sum_{n = 0}^{\infty}c_{n}t^{n}=0\)

Change the index in the first sum: Let \(m=n - 2\), then \(n=m + 2\)

\(5\sum_{m = 0}^{\infty}(m + 2)(m + 1)c_{m+2}t^{m}-\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n}-6\sum_{n = 1}^{\infty}nc_{n}t^{n}-4\sum_{n = 0}^{\infty}c_{n}t^{n}=0\)

\(5\times2\times1c_{2}+5\times3\times2c_{3}t+5\sum_{m = 2}^{\infty}(m + 2)(m + 1)c_{m+2}t^{m}-\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n}-6c_{1}t-6\sum_{n = 2}^{\infty}nc_{n}t^{n}-4c_{0}-4c_{1}t-4\sum_{n = 2}^{\infty}c_{n}t^{n}=0\)

For \(t^{0}\): \(10c_{2}-4c_{0}=0\Rightarrow c_{2}=\frac{2}{5}c_{0}\)

For \(t^{1}\): \(30c_{3}-6c_{1}-4c_{1}=0\Rightarrow c_{3}=\frac{1}{3}c_{1}\)

For \(t^{n}(n\geq2)\): \(5(n + 2)(n + 1)c_{n+2}-n(n - 1)c_{n}-6nc_{n}-4c_{n}=0\)

\(c_{n+2}=\frac{n^{2}+5n + 4}{5(n + 2)(n + 1)}c_{n}=\frac{(n + 1)(n + 4)}{5(n + 2)(n + 1)}c_{n}=\frac{n + 4}{5(n + 2)}c_{n}\)

Using the initial conditions: \(y(5)=0\) (when \(t = 0\), \(y(0)=c_{0}=0\)) and \(y^{\prime}(5)=4\) (when \(t = 0\), \(y^{\prime}(0)=c_{1}=4\))

Since \(c_{0}=0\), then \(c_{2}=0\), \(c_{3}=\frac{4}{3}\), \(c_{4}=\frac{6}{5}\times\frac{4}{3}=\frac{8}{5}\), \(c_{5}=\frac{7}{5}\times\frac{8}{5}=\frac{56}{25}\), \(c_{6}=\frac{8}{5}\times\frac{56}{25}=\frac{448}{125}\)

\(y=\sum_{n = 1}^{\infty}c_{n}t^{n}\), \(t=x - 5\)

The recurrence relation \(c_{n+2}=\frac{n + 4}{5(n + 2)}c_{n}\)

We can also write the solution as \(y = 4t+\frac{4}{3}t^{3}+\frac{8}{5}t^{4}+\frac{56}{25}t^{5}+\cdots\)

The differential equation \((5 - t^{2})y^{\prime\prime}-6ty^{\prime}-4y = 0\) has a singular point at \(t=\pm\sqrt{5}\). The radius of convergence is \(R = \sqrt{5}\), and the interval of convergence is \((5-\sqrt{5},5+\sqrt{5})\)

Answer:

\(y(x)=4(x - 5)+\frac{4}{3}(x - 5)^{3}+\frac{8}{5}(x - 5)^{4}+\frac{56}{25}(x - 5)^{5}+\cdots=\sum_{n = 1}^{\infty}c_{n}(x - 5)^{n}\) (where \(c_{1} = 4\), \(c_{3}=\frac{4}{3}\), \(c_{4}=\frac{8}{5}\), \(c_{5}=\frac{56}{25}\), \(c_{n+2}=\frac{n + 4}{5(n + 2)}c_{n}\) for \(n\geq1\)) and the interval of convergence is \((5-\sqrt{5},5+\sqrt{5})\)