I'm not sure I'm understanding the question quite right.
What I did is: \begin{gather} y'=x+y, \quad\quad y'(0) = 0 + y(0) = 2\\ y''=1+y',\quad\quad y''(0) = 1+y'(0) = 3\\ y'''=y'',\quad\quad y'''(0) = y''(0) = 3\\ y''''=y''', \quad\quad y''''(0) = y'''(0) = 3\\ \end{gather}
Is this right or does it satisfy the question?
All help is appreciated.
Suppose that $y$ is a (smooth) solution of $y'=x+y$ with initial condition $y(0)=2$. The Taylor expansion of $y$ at the point $0$ (also called the Maclaurin series of $y$) is given by $$y(x)=\frac{y(0)}{0!}(x-0)^{0}+\frac{y'(0)}{1!}(x-0)^{1}+\frac{y''(0)}{2!}(x-0)^{2}+\frac{y'''(0)}{3!}(x-0)^{3}+\ldots.$$ Hence with what you already computed, you can compute the first four terms of the expansion above. For example, the third term is given by $$\frac{y''(0)}{2!}(x-0)^{2}=\frac{3}{2}x^{2}.$$ Note that you only need $y''''(0)$ for the fifth term.