Answer
The actual result: $4.016$
The approximation: $4.023$
Difference: $4.023-4.016=0.007$
Work Step by Step
We are given $\frac{dy}{dx}=x^{3}$
Using Euler's method $g(x,y)=x^{3}$
Since $x=0, y=4$
$g(x_{0},y_{0})=0$ and $y_{1}=y_{0}+g(x_{0},y_{0})h=4+0\times0.1=4$
Now $x_{1}=0.1, y_{1}=4$ and
$g(x_{1},y_{1})=0.001$
Then $y_{2}=4+(0.001)\times0.1=4.0001$
$y_{3}=4.0001+(0.008)\times0.1=4.0009$
$y_{4}=4.0009+(0.027)\times0.1=4.0036$
$y_{5}=4.0036+(0.064)\times0.1=4.01$
$y_{6}=4.01+(0.125)\times0.1=4.023$
$y(0.5)=4.023$
$\frac{dy}{dx}=x^{3}$
$\int dy=\int x^{3}dx$
$y=\frac{x^{4}}{4}+C$
With $x=0, y=4$
$C=4- \frac{0^{4}}{4}=4$
$y=\frac{x^{4}}{4}+4$
$y(0.5)=\frac{(0.5)^{4}}{4}+4=4.016$
Difference between the result and approximation: $4.023-4.016=0.007$