I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take [tex]z=y'[/tex], so that [tex]z'=y''[/tex] and we're left with the ODE linear in [tex]z[/tex]:
[tex]y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2[/tex]
Now suppose [tex]y[/tex] has a power series expansion
[tex]y=\displaystyle\sum_{n\ge0}a_nx^n[/tex]
[tex]\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}[/tex]
[tex]\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}[/tex]
Then the ODE can be written as
[tex]\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0[/tex]
[tex]\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0[/tex]
[tex]\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0[/tex]
All the coefficients of the series vanish, and setting [tex]x=0[/tex] in the power series forms for [tex]y[/tex] and [tex]y'[/tex] tell us that [tex]y(0)=a_0[/tex] and [tex]y'(0)=a_1[/tex], so we get the recurrence
[tex]\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}[/tex]
We can solve explicitly for [tex]a_n[/tex] quite easily:
[tex]a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}[/tex]
and so on. Continuing in this way we end up with
[tex]a_n=\dfrac{a_1}{n!}[/tex]
so that the solution to the ODE is
[tex]y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x[/tex]
We also require the solution to satisfy [tex]y(0)=a_0[/tex], which we can do easily by adding and subtracting a constant as needed:
[tex]y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}[/tex]
