Answer
The approximate $ y$-values for the solution of the initial-value problem
$$y^{\prime}=y-2 x, y(1)=0$$
are
$$
\begin{aligned}
y_{1} &=y_{0}+h F\left(x_{0}, y_{0}\right) \\
&=-1 , \\
y_{2}&=y_{1}+h F\left(x_{1}, y_{1}\right) \\
&=-3 ,\\
y_{3} &=y_{2}+h F\left(x_{2}, y_{2}\right) \\
&=-6.5 ,\\
y_{4} &=y_{3}+h F\left(x_{3}, y_{3}\right) \\
&=-12.25
\end{aligned}$$
Work Step by Step
The approximate $ y$-values for the solution of the initial-value problem
$$y^{\prime}=y-2 x, y(1)=0$$
can be find as follows:
First, We are given that
$$ h=0.5, x_{0}=1, y_{0}=0, \text { and } F(x, y)=y-2 x
$$
Note that
$$
x_{1}=x_{0}+h=1+0.5=1.5, x_{2}=2, \text { and } x_{3}=2.5
$$
So we have
$$
\begin{aligned}
y_{1} &=y_{0}+h F\left(x_{0}, y_{0}\right) \\
&=0+0.5 F(1,0) \\
&=0.5[0-2(1)] \\
&=-1 , \\
y_{2}&=y_{1}+h F\left(x_{1}, y_{1}\right) \\
&=-1+0.5 F(1.5,-1) \\
&=-1+0.5[-1-2(1.5)] \\
&=-3 ,\\
y_{3} &=y_{2}+h F\left(x_{2}, y_{2}\right) \\
&=-3+0.5 F(2,-3) \\
&=-3+0.5[-3-2(2)] \\
&=-6.5 ,\\
y_{4} &=y_{3}+h F\left(x_{3}, y_{3}\right) \\
&=-6.5+0.5 F(2.5,-6.5) \\
&=-6.5+0.5[-6.5-2(2.5)] \\
&=-12.25
\end{aligned}$$
