We're looking for a solution of the form
[tex]y=\displaystyle\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+\cdots[/tex]
Given that [tex]y(0)=1[/tex], we would end up with [tex]a_0=1[/tex].
Its first derivative is
[tex]y'=\displaystyle\sum_{n=0}^\infty na_nx^{n-1}=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n[/tex]
The shifting of the index here is useful in the next step. Substituting these series into the ODE gives
[tex]\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n-\sum_{n=0}^\infty a_nx^n=x[/tex]
Both series start with the same-degree term [tex]x^0[/tex], so we can condense the left side into one series.
[tex]\displaystyle\sum_{n=0}^\infty\bigg((n+1)a_{n+1}-a_n\bigg)x^n=x[/tex]
Pull out the first two terms ([tex]x^0[/tex] and [tex]x^1[/tex]) of the series:
[tex]a_1-a_0+(2a_2-a_1)x+\displaystyle\sum_{n=2}^\infty\bigg((n+1)a_{n+1}-a_n\bigg)x^n=x[/tex]
Matching the coefficients of the [tex]x^0[/tex] and [tex]x^1[/tex] terms on either side tells us that
[tex]\begin{cases}a_1-a_0=0\\2a_2-a_1=1\end{cases}[/tex]
We know that [tex]a_0=1[/tex], so [tex]a_1=1[/tex] and [tex]a_2=1[/tex]. The rest of the coefficients, for [tex]n\ge2[/tex], are given according to the recurrence,
[tex](n+1)a_{n+1}-a_n=0\implies a_{n+1}=\dfrac{a_n}{n+1}[/tex]
so that [tex]a_3=\dfrac{a_2}3=\dfrac13[/tex], [tex]a_4=\dfrac{a_3}4=\dfrac1{12}[/tex], and [tex]a_5=\dfrac{a_4}5=\dfrac1{60}[/tex]. So the 5th degree approximation to the solution to this ODE centered at [tex]x=0[/tex] is
[tex]y\approx1+x+x^2+\dfrac{x^3}3+\dfrac{x^4}{12}+\dfrac{x^5}{60}[/tex]
