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
( (2x - x^{2})y - 6(x - 1)y - 4y = 0; y(1) = 0, y(1) = 5 )
( 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 \(t = x - 1\), then \(x=t + 1\).
First, find derivatives:
\(y'=\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}=\frac{dy}{dt}\)
\(y''=\frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{dt^{2}}\)
Substitute \(x=t + 1\) into the differential equation \((2x-x^{2})y''-6(x - 1)y'-4y = 0\)
\(2(t + 1)-(t + 1)^{2}=2t+2-(t^{2}+2t + 1)=1 - t^{2}\)
The differential equation becomes \((1 - t^{2})y''-6ty'-4y = 0\)
Assume \(y=\sum_{n = 0}^{\infty}c_{n}t^{n}\), then \(y'=\sum_{n = 1}^{\infty}nc_{n}t^{n - 1}\) and \(y''=\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}\)

Step2: Substitute series into the differential equation

\((1 - t^{2})\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}-6t\sum_{n = 1}^{\infty}nc_{n}t^{n - 1}-4\sum_{n = 0}^{\infty}c_{n}t^{n}=0\)
Expand the first term:
\(\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}-\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n}-6\sum_{n = 1}^{\infty}nc_{n}t^{n}-4\sum_{n = 0}^{\infty}c_{n}t^{n}=0\)
Re - index the first sum: let \(m=n - 2\), then \(n=m + 2\)
\(\sum_{m = 0}^{\infty}(m + 2)(m + 1)c_{m+2}t^{m}-\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n}-6\sum_{n = 1}^{\infty}nc_{n}t^{n}-4\sum_{n = 0}^{\infty}c_{n}t^{n}=0\)
Write out the terms:
\(2c_{2}+6c_{3}t+\sum_{n = 2}^{\infty}(n + 2)(n + 1)c_{n+2}t^{n}-\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n}-6c_{1}t-6\sum_{n = 2}^{\infty}nc_{n}t^{n}-4c_{0}-4c_{1}t-4\sum_{n = 2}^{\infty}c_{n}t^{n}=0\)

Step3: Equate coefficients

For \(t^{0}\): \(2c_{2}-4c_{0}=0\Rightarrow c_{2}=2c_{0}\)
For \(t^{1}\): \(6c_{3}-6c_{1}-4c_{1}=0\Rightarrow6c_{3}-10c_{1}=0\Rightarrow c_{3}=\frac{5}{3}c_{1}\)
For \(n\geq2\): \((n + 2)(n + 1)c_{n+2}-n(n - 1)c_{n}-6nc_{n}-4c_{n}=0\)
\(c_{n+2}=\frac{n^{2}+5n + 4}{(n + 2)(n + 1)}c_{n}=\frac{(n + 1)(n + 4)}{(n + 1)(n + 2)}c_{n}=\frac{n + 4}{n + 2}c_{n}\)

Step4: Use initial conditions

Since \(y(1) = 0\), when \(t = 0\) (because \(x = 1\) and \(t=x - 1\)), \(y(0)=c_{0}=0\)
Since \(y^{\prime}(1) = 5\), \(y^{\prime}(t)=\sum_{n = 1}^{\infty}nc_{n}t^{n - 1}\), when \(t = 0\), \(y^{\prime}(0)=c_{1}=5\)
Since \(c_{0}=0\), then \(c_{2}=0\)
\(c_{3}=\frac{25}{3}\), \(c_{4}=\frac{4 + 4}{4 + 2}c_{2}=0\), \(c_{5}=\frac{5+4}{5 + 2}c_{3}=\frac{9}{7}\times\frac{25}{3}=\frac{75}{7}\)

We can observe that \(c_{2k}=0\) for \(k = 0,1,2,\cdots\)
For \(n = 2k+1\), \(c_{2k + 1}=\frac{(2k+1)+4}{(2k+1)+2}c_{2k - 1}=\frac{2k + 5}{2k+3}c_{2k - 1}\)
\(c_{1}=5\), \(c_{3}=\frac{5}{3}\times5\), \(c_{5}=\frac{9}{7}\times\frac{5}{3}\times5\)
\(y=\sum_{n = 0}^{\infty}c_{n}t^{n}=5t+\frac{25}{3}t^{3}+\frac{75}{7}t^{5}+\cdots\)
Since \(t=x - 1\), \(y = 5(x - 1)+\frac{25}{3}(x - 1)^{3}+\frac{75}{7}(x - 1)^{5}+\cdots=\sum_{n = 0}^{\infty}\frac{5(2n + 1)}{2n+1}(x - 1)^{2n + 1}\)

Step5: Find the radius of convergence

For the differential equation \((1 - t^{2})y''-6ty'-4y = 0\), the singular points are \(t=\pm1\). The radius of convergence \(R = 1\), and the interval of convergence is \((0,2)\)

Answer:

\(y(x)=\sum_{n = 0}^{\infty}\frac{5(2n+1)}{2n + 1}(x - 1)^{2n+1}\), interval of convergence is \((0,2)\)