Answer
The actual result: $1.075$;
The approximation: $1.094$;
Difference: $1.094-1.075=0.019$
Work Step by Step
We are given $\frac{dy}{dx}=x^{2}y$
Using Euler's method $g(x,y)=x^{2}y$
Since $x=0, y=1$
$g(x_{0},y_{0})=0$ and $y_{1}=y_{0}+g(x_{0},y_{0})h=1+0\times0.1=1$
Now $x_{1}=0.1, y_{1}=1$ and
$g(x_{1},y_{1})=0.01$
Then $y_{2}=1+(0.01)\times0.1=1.001$
$y_{3}=1.001+(0.04)\times0.1=1.005$
$y_{4}=1.005+(0.09)\times0.1=1.014$
$y_{5}=1.014+(0.162)\times0.1=1.0302$
$y_{6}=1.0302+(0.258)\times0.1=1.056$
$y_{7}=1.056+(0.38)\times0.1=1.095$
$y(0.6)=y_{7}=1.094$
$\frac{dy}{dx}=x^{2}y$
$\int \frac{1}{y}dy=\int x^{2} dx$
$\ln |y| =\frac{1}{3}x^{3}+C$
With x=0, y=1
$C=\ln |1| - \frac{1}{3}(0)^{3}=0$
$y(0.6) \rightarrow \ln|y|=\frac{1}{3}(0.6)^{3}=0.072$
$|y|=e^{0.072}=1.075$
Difference between the result and approximation: $1.094-1.075=0.019$
