Answer
$y(0.5)=5.7285$
Work Step by Step
We have
$$y'=4y-1$$
Setting $h=0.05$ in equation $y'=4y-1$
$$y_{n+1}=y_n+0.025(4y_n-1+4y_{n+1}^*-1)$$
Hence,
$y_1=y_0+0.025(4y_0-1+4y_{1}^*-1)=1.165$
$y_2=y_1+0.025(4y_1-1+4y_{2}^*-1)=1.366$
$y_3=y_2+0.025(4y_2-1+4y_{3}^*-1)=1.612$
$y_4=y_3+0.025(4y_3-1+4y_{4}^*-1)=1.9115$
$y_5=y_4+0.025(4y_4-1+4y_{5}^*-1)=2.277$
$y_6=y_5+0.025(4y_5-1+4y_{6}^*-1)=2.723$
$y_7=y_6+0.025(4y_6-1+4y_{7}^*-1)=3.267$
$y_8=y_7+0.025(4y_7-1+4y_{8}^*-1)=3.93$
$y_9=y_8+0.025(4y_8-1+4y_{9}^*-1)=4.74$
$y_{10}=y_9+0.025(4y_9-1+4y_{10}^*-1)=5.7285$
If we increase the step size to $h = 0.05$, the corresponding approximation to $y(0.5)$ becomes $y_{10}=5.7285$
