Answer
Proved
Work Step by Step
$xy''-(x+10)y'+10y=0$ _____($\alpha$)
\[y=\sum_{k=0}^{10} \frac{1}{k!}x^k\]
$y=1+\frac{1}{1!}x+\frac{1}{2!}x^2+...+\frac{1}{10!}x^{10}$ ____(1)
Differentiating (1) with respect to $x$
$y'=1+\frac{2x}{2!}+\frac{3x^2}{3!}+...+\frac{10x^9}{10!}$
$y'=1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^9}{9!}$ ____(2)
Differentiating (2) with respect to $x$
$y''=1+\frac{2x}{2!}+...+\frac{9x^8}{9!}=1+\frac{x}{1!}+...+\frac{x^8}{8!}$__(3)
$xy''=x+\frac{x^2}{1!}+...+\frac{x^9}{8!}$ ____(4)
$-(x+10)y'=-(x+10)[1+\frac{x}{1!}+...+\frac{x^9}{9!}]$
$-(x+10)y'=-[x+\frac{x^2}{1!}+...+\frac{x^{10}}{9!}]-10[1+\frac{x}{1!}+...+\frac{x^9}{9!}]$ ____(5)
$10y=10[1+\frac{x}{1!}+...+\frac{x^{10}}{10!}]=10[1+\frac{x}{1!}+...+\frac{x^{9}}{9!}]+\frac{x^{10}}{9!}$ _____(6)
Adding (4),(5) and (6)
$xy''-(x+10)y'+10y=[x+\frac{x^2}{1!}+...+\frac{x^{10}}{9!}]-[x+\frac{x^2}{1!}+...+\frac{x^{10}}{9!}]=0$
Hence $y=\sum_{k=0}^{10} \frac{1}{k!}x^k$ is solution of ($\alpha$).
