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

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Accepted Answer

# Explanation:
## Step1: Make substitution
Let $t = x - 5$, then $x=t + 5$.

First, find the 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 $x=t + 5$ into the differential equation:

The coefficient $-20 + 10x-x^{2}=-20+10(t + 5)-(t + 5)^{2}$

$$
\begin{align*}
&=-20+10t+50-(t^{2}+10t + 25)\
&=-20+10t+50 - t^{2}-10t - 25\
&=5 - t^{2}
\end{align*}
$$

The term $-6(x - 5)=-6t$

The differential equation becomes $(5 - t^{2})y''-6ty'-4y = 0$

Assume $y=\sum_{n = 0}^{\infty}c_{n}t^{n}$, then $y'=\sum_{n = 1}^{\infty}nc_{n}t^{n - 1}$, $y''=\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}$

Substitute $y,y',y''$ into $(5 - t^{2})y''-6ty'-4y = 0$

$$
\begin{align*}
(5 - t^{2})\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}-6t\sum_{n = 1}^{\infty}nc_{n}t^{n - 1}-4\sum_{n = 0}^{\infty}c_{n}t^{n}&=0\
5\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n - 2}-\sum_{n = 2}^{\infty}n(n - 1)c_{n}t^{n}-6\sum_{n = 1}^{\infty}nc_{n}t^{n}-4\sum_{n = 0}^{\infty}c_{n}t^{n}&=0
\end{align*}
$$

Re - index the first sum: Let $m=n - 2$, then $n=m + 2$

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

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

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

For $t^{1}$: $30c_{3}-6c_{1}-4c_{1}=0\Rightarrow30c_{3}-10c_{1}=0\Rightarrow c_{3}=\frac{1}{3}c_{1}$

For $t^{n}(n\geq2)$:

$$
\begin{align*}
5(n + 2)(n + 1)c_{n+2}-n(n - 1)c_{n}-6nc_{n}-4c_{n}&=0\
5(n + 2)(n + 1)c_{n+2}-(n^{2}-n + 6n+4)c_{n}&=0\
5(n + 2)(n + 1)c_{n+2}-(n^{2}+5n + 4)c_{n}&=0\
c_{n+2}=\frac{(n + 1)(n + 4)}{5(n + 2)(n + 1)}c_{n}=\frac{n + 4}{5(n + 2)}c_{n}
\end{align*}
$$

Using the initial conditions: When $x = 5$, $t = 0$

$y(5)=0\Rightarrow c_{0}=0$

$y'(5)=4\Rightarrow c_{1}=4$

Since $c_{0}=0$, then $c_{2}=0$, $c_{3}=\frac{4}{3}$, $c_{4}=\frac{4 + 4}{5(4+2)}c_{2}=0$, $c_{5}=\frac{5 + 4}{5(5+2)}c_{3}=\frac{9}{35}\times\frac{4}{3}=\frac{12}{35}$

In general, when $n = 2k$ (even), $c_{2k}=0$

When $n = 2k+1$ (odd), $c_{2k + 1}=\frac{4}{3}\times\frac{5}{5\times4}\times\frac{7}{5\times6}\times\cdots\times\frac{2k + 3}{5\times(2k+2)}$

$$
\begin{align*}
y&=\sum_{n = 0}^{\infty}c_{n}t^{n}\
&=4t+\frac{4}{3}t^{3}+\frac{12}{35}t^{5}+\cdots\
&=4\sum_{k = 0}^{\infty}\frac{\prod_{j = 0}^{k}(2j + 3)}{5^{k}(k + 1)(k+2)\cdots(2k + 2)}t^{2k+1}
\end{align*}
$$

Since $t=x - 5$, $y = 4\sum_{k = 0}^{\infty}\frac{\prod_{j = 0}^{k}(2j + 3)}{5^{k}(k + 1)(k+2)\cdots(2k + 2)}(x - 5)^{2k+1}$

For the interval of convergence, we use the ratio test.

Let $a_{n}$ be the coefficient of $t^{n}$ in the power - series solution.

For the differential equation $(5 - t^{2})y''-6ty'-4y = 0$, the singular points of the original equation (in terms of $x$) are found by setting $-20 + 10x-x^{2}=0$

$x^{2}-10x + 20=0$, $x=\frac{10\pm\sqrt{100 - 80}}{2}=5\pm\sqrt{5}$

The distance from $x = 5$ (where $t = 0$) to the nearest singular point is $\sqrt{5}$. So the interval of convergence is $(5-\sqrt{5},5+\sqrt{5})$

# Answer:
$y(x)=4\sum_{n = 0}^{\infty}\frac{\prod_{j = 0}^{n}(2j + 3)}{5^{n}(n + 1)(n+2)\cdots(2n + 2)}(x - 5)^{2n+1}$, interval of convergence $(5-\sqrt{5},5+\sqrt{5})$

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