Hello folks, how should I proceed to solve this differential equation through power series? The problem is:
$$
y'-y = x \\   
y(0) = 0
$$
I know I have use the general form and its derivatives
$$
\sum a_n(x-x_0)^n
$$
My problem is with the alone $x$ variable on the right side. Could someone give me any tips?
Thanks in advance!

Because of the initial condition you can use a power series about $\;x=0\;$ , so:
$$y=\sum_{n=0}^\infty a_nx^n\implies y'=\sum_{n=0}^\infty na_nx^{n-1}\implies$$
$$x=y'-y=\sum_{n=0}^\infty \left(a_nnx^{n-1}-a_nx^n\right)=(a_1-a_0)+(2a_2-a_1)x+(3a_3-a_2)x^2+\ldots$$
We can form a recursive sequence, or else observe that
$$a_1-a_0=0\implies a_1=a_0\;;\;\;2a_2-a_1=1\implies a_2=\frac{a_1=a_0}2+\frac12=\frac{a_0+1}2\;;$$
$$\;\;3a_3-a_2=0\implies a_3=\frac13a_2=\frac{a_0+1}6,\;etc. \text{(other relations are }\;a_n-(n+1)a_{n+1}=0)$$
So
$$a_4=\frac{a_0+1}{24}\,,\ldots,a_n=\frac{a_0+1}{n!}\;,\;\;n\ge2$$
Thus, the general solution is
$$y(x)=a_0+a_0x+\frac{a_0+1}2x^2+\frac{a_0+1}{3!}x^3+\ldots+\frac{a_0+1}{n!}x^n+\ldots=$$
$$=a_0\left(\sum_{n=0}^\infty \frac{x^n}{n!}\right)+\sum_{n=2}^\infty\frac{x^n}{n!}=a_0e^x+(e^x-x-1)=Ke^x-x-1,\,K=\text{constant} $$
With the inital condition $\;y(0)=0\;$ we get
$$0=y(0)=K-1\implies K=1\implies y(x)=e^x-x-1$$
If you need the above particular solution as power series just take the corresponding power series for the last function above...
