91Ó°ÊÓ

Therefore,

$$\begin{gathered} y_0\text{'}\left(x\right)=\frac{d}{dx}\left[1+\underset{k=1}{\overset{∞}{∑}}\frac{x^{3k}}{(2·3)(5·6)···\left(\right(3k-1)·3k)}\right] \\ {y}_0\text{'}\left(x\right)=\frac{d}{dx}\left[1\right]+\underset{k=1}{\overset{∞}{∑}}\frac{1}{(2·3)(5·6)···\left(\right(3k-1)·3k)}\frac{d}{dx}\left[x^{3k}\right] \\ {y}_0\text{'}\left(x\right)=0+\underset{k=1}{\overset{∞}{∑}}\frac{1}{(2·3)(5·6)···\left(\right(3k-1)·3k)}3k{x}^{3k-1} \end{gathered}$$

Hence,

$y_0\text{'}\left(x\right)=\underset{k=1}{\overset{∞}{∑}}\frac{3k}{(2·3)(5·6)···\left(\right(3k-1)·3k)}{x}^{3k-1}$

So,

$$\begin{gathered} y_0\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left[\underset{k=1}{\overset{∞}{∑}}\frac{3k}{(2·3)(5·6)···\left(\right(3k-1)·3k)}{x}^{3k-1}\right] \\ {y}_0\text{'}\text{'}\left(x\right)=\underset{k=1}{\overset{∞}{∑}}\frac{3k(3k-1)}{(2·3)(5·6)···\left(\right(3k-1)·3k)}{x}^{3k-2} \end{gathered}$$

Or,

$y_0\text{'}\text{'}\left(x\right)=x+\underset{k=2}{\overset{∞}{∑}}\frac{x^{3k-2}}{(2·3)(5·6)···\left(\right(3k-4)·(3k-3\left)\right)}$

Finally, calculate $x{y}_0\left(x\right)$

So,

role="math" localid="1668618118815" $$\begin{gathered} x{y}_0\left(x\right)=x\left[1+\underset{k=1}{\overset{∞}{∑}}\frac{x^{3k}}{(2·3)(5·6)···\left(\right(3k-1)·3k)}\right] \\ x{y}_0\left(x\right)=x+\underset{k=1}{\overset{∞}{∑}}\frac{x^{3k+1}}{(2·3)(5·6)···\left(\right(3k-1)·3k)} \end{gathered}$$

Therefore,

$$\begin{gathered} x{y}_0\left(x\right)=x+\underset{k=2}{\overset{∞}{∑}}\frac{x^{3(k-1)+1}}{(2·3)(5·6)···\left(3\right(k-1)-1)·3(k-1))} \\ x{y}_0\left(x\right)=x+\underset{k=2}{\overset{∞}{∑}}\frac{x^{3(k-1)+1}}{(2·3)(5·6)···(3k-4)·(3k-3))} \end{gathered}$$