Answer
The solution is $y_{10}=4.89$
Work Step by Step
We have
$$y'=4y-1$$
Setting $h=0.05$ in equation $y'=4y-1$
$$y_{n+1}=y_n+0.05(4y_n-1)$$
Hence,
$y_1=y_0+0.05(4y_0-1)=1+0.05(4\times1-1)=1.15$
$y_2=y_1+0.05(4y_0-1)=1.15+0.05(4\times1.15-1)=1.33$
$y_3=y_2+0.05(4y_0-1)=1.33+0.05(4\times1.33-1)=1.546$
$y_4=y_3+0.05(4y_0-1)=1.546+0.05(4\times1.546-1)=1.8052$
$y_5=y_4+0.05(4y_0-1)=1.8052+0.05(4\times1.8052-1)=2.116$
$y_6=y_5+0.05(4y_0-1)=2.116+0.05(4\times2.116-1)=2.4892$
$y_7=y_6+0.05(4y_0-1)=2.4892+0.05(4\times2.4892-1)=2.937$
$y_8=y_7+0.05(4y_0-1)=2.937+0.05(4\times2.937-1)=3.4744$
$y_9=y_8+0.05(4y_0-1)=3.4744+0.05(4\times3.4744-1)=4.119$
$y_{10}=y_9+0.05(4y_0-1)=4.119+0.05(4\times4.119-1)=4.89$
If we increase the step size to $h = 0.05$, the corresponding approximation to $y(0.5)$ becomes $y_{10}=4.89$
