We know that the coefficients of the Taylor series are given by $a_{n}={\frac {y^{(n)}(0)}{n!}}$ and that the initial condition is

- $y(0)={\color {MediumVioletRed}4}.$

Since $y$ solves the differential equation $y'+2y=x^{2}$, or equivalently, $y'=x^{2}-2y$, we must have

- $y'(0)=0^{2}-2y(0)=0-2\cdot {\color {MediumVioletRed}4}={\color {Chocolate}-8}.$

By differentiating the equation for $y'$ (with respect to $x$), we obtain $y''=2x-2y'$, which implies that

- $y''(0)=2\cdot 0-2y'(0)=0-2\cdot {\color {Chocolate}-8}={\color {DarkGreen}16}.$

Differentiating the equation once more, we obtain $y'''=2-2y''$, so

- $y'''(0)=2-2y''(0)=2-2\cdot {\color {DarkGreen}16}=-30.$

We conclude that $a_{0}={\frac {y(0)}{0!}}=\color {blue}4$, $a_{1}={\frac {y'(0)}{1!}}=\color {blue}-8$, $a_{2}={\frac {y''(0)}{2!}}=\color {blue}8$, and $a_{3}={\frac {y'''(0)}{3!}}=\color {blue}-5$.