Solve intial value problem using power series $xy''+y'+2y=0$ with y(1) =2, y'(1) =4.
What I tried:
\begin{equation}
 xy'' +y'+2y = 0
 \end{equation}
    Let $y=\sum_{k=0}^{\infty}c_kx^k$, $y^\prime=\sum_{k=1}^{\infty}kc_kx^{k-1}$, $y^{\prime\prime}=\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}$
Then
\begin{equation}
 x\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}+\sum_{k=1}^{\infty}kc_{k+1}x^{k-1}+2\sum_{k=0}^{\infty}c_kx^k=0
 \end{equation}
\begin{equation}
\sum_{k=2}^{\infty}k(k+1)c_kx^{k-1}+\sum_{k=1}^{\infty}kc_kx^{k-1}+\sum_{k=0}^{\infty}2c_kx^k=0
 \end{equation}
\begin{equation}
  \sum_{k=0}^{\infty}k(k+1)c_{k+1}x^{k}+\sum_{k=0}^{\infty}(k+1)c_{k+1}x^{k}+\sum_{k=0}^{\infty}2kc_kx^{k}=0
  \end{equation}
\begin{equation}
\sum_{k=0}^{\infty}[k(k+1)c_{k+1}+(k+1)c_{k+1}+2kc_k]x^k=0
\end{equation}
Therefore
\begin{equation}
k(k+1)c_{k+1}+(k+1)c_{k+1}+2kc_k=0 \text{ for }k\ge0
\end{equation}
So
\begin{equation}
c_{k+1}=\frac{-2k}{(k+1)^2}c_k
\end{equation}

This is how much I have tried. The answer is given to be:
  $y=2+4(x-1)-4(x-1)^2+\frac{4}{5}(x-1)^3-\frac{1}{3}(x-1)^4+\frac{2}{15}(x-1)^5+.... $

How do I obtain this form. Pls help. 
Question link: http://imgur.com/a/gTdCYbk
 A: $$
\begin {cases}
xy''+y'+2y=0, \\
y(1) =2, y'(1) =4
\end{cases}
$$
Hint
There are some mistakes in your calculations. And you should use this serie instead since you are given initial conditions at $x=1$ and not at $x=0$:
$$y=\sum_{k=0}^{\infty}c_k(x-1)^k,$$
$$ y'(x)=\sum_{k=0}^{\infty}(k+1)c_{k+1}(x-1)^{k}$$
$$ y''(x)=\sum_{k=1}^{\infty}k(k+1)c_{k+1}(x-1)^{k-1}$$
And you have to use the initial conditions. $y(1)=2$ means that $c_0=2$ and $y'(1)=4$ means that $c_1=4$.
Be carrefull because the equation becomes
$$xy''+y'+2y=0, $$
$$(x-1)y''+y''+y'+2y=0$$
I got this:
$$(k+1)^2c_{k+1}+2c_k+(k+1)(k+2)c_{k+2}=0$$
And also that
$$2c_2+c_1+2c_0=0 \implies c_2=-4$$
For $c_3$ I got $4/3$ and not $4/5$ as stated in your book.
And my serie looks like this :
$$y=2+4(x-1)-4(x-1)^2+\frac{4}{3}(x-1)^3-\frac{1}{3}(x-1)^4++....$$
A: We propose to solve $$xy''+y'+2y=0 ~~~(1)$$ by taylor series about $x=1$ as
$$y(x)=\sum_{k=0}^{\infty}~ y^{(k)}(1) \frac{(x-1)^k}{k!}, y^{(k)}(1)~ \text{denotes
$k$th derivative of $y(x)$ at $x=1$}~~~~(2)$$
solution where we are
given that $y(1)=2, y'(1)=4$. We differentiate (1) $n$ times as
$$xy^{(n+2)}+n y^{(n+1)}+2y^{(n)}=0 \implies y^{(n+2)}(1)=-(n+1)y^{(n+1)}(1)-2y^{(n)}(1)=0$$
$$\implies y^{(n)}(1)=-(n-1)y^{(n-1)}(1)-2y^{(n-2)}(1)$$
$$\implies y^{(2)}(1)=-8, y^{(3)}(1)=8, y^{(4)}(1)=-8, y^{{5}}(1)=16,...$$
Inserting these values in (2), we get
$$y(x)=2+4(x-1)-8\frac{(x-1)^2}{2!}+8\frac{(x-1)^3}{3!}-8\frac{(x-1)^{4}}{4!}+16 \frac{(x-1)^5}{5!}+.....$$
