[tex]y''-y'=0[/tex]
We can reduce the order of the ODE to make solving a bit less taxing. Let [tex]z=y'[/tex] so that [tex]z'=y''[/tex].
[tex]z'-z=0[/tex]
Then assume a solution of the form
[tex]z=\displaystyle\sum_{n=0}^\infty a_nx^n[/tex]
[tex]\implies z'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n[/tex]
Substituting into the ODE gives
[tex]\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n-\sum_{n=0}^\infty a_nx^n=0[/tex]
[tex]\displaystyle\sum_{n=0}^\infty\bigg((n+1)a_{n+1}-a_n\bigg)x^n=0[/tex]
The coefficients in the series are given according to
[tex]\begin{cases}a_0=z(0)=y'(0)\\(n+1)a_{n+1}=a_n&\text{for }n\ge0\end{cases}[/tex]
We can solve the recurrence exactly:
[tex]a_{n+1}=\dfrac{a_n}{n+1}=\dfrac{a_{n-1}}{(n+1)n}=\dfrac{a_{n-2}}{(n+1)n(n-1)}=\cdots=\dfrac{a_0}{(n+1)!}[/tex]
[tex]\implies a_n=\dfrac{a_0}{n!}[/tex]
Then
[tex]z(x)=\displaystyle y'(0)\sum_{n=0}^\infty\frac{x^n}{n!}[/tex]
and integrating both sides to solve for [tex]y(x)[/tex] gives
[tex]y(x)=\displaystyle y(0)+\int_{y(0)}^x z(t)\,\mathrm dt[/tex]
[tex]y(x)=\displaystyle y(0)+y'(0)\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}[/tex]
[tex]y(x)=\displaystyle y(0)+y'(0)\sum_{n=1}^\infty\frac{x^n}{n!}[/tex]
[tex]\implies y(x)=y(0)+y'(0)(e^x-1)[/tex]
(I don't know what method is apparently being used in Section 4.3...)
