Answer
$y_1=2, y_2=2.0202,y_3=2.0618$, $y=e^{x^2}+1$, and $y(0.3)=e^{(0.3)^2}+1 \approx 2.0942$
Work Step by Step
Apply the formula to find the approximations as follows:
$y_{n+1}=y_n+f(x_n,y_n) dx$ ....(1)
As we are given $y(0)=2$ This implies $x_0=0,x_1=0+0.1=0.1, x_2=0.1+0.1=0.2$
Likewise by using equation (1), we have
$y_1=2, y_2=2.0202,y_3=2.0618$
In order to determine the exact solution, isolate the x and y terms on one side and integrate both sides.
$\int dy=\int 2xe^{x^2} dx$ ...(2)
In order to determine the differential equation, we will have to take the derivative of the differential equation.
Equation (2) gives: $y=e^{x^2}+c$ ...(3)
Apply the initial conditions, to calculate the value of $c$
$2=e^{0^2}+c \implies c=1$
Now, equation (3) becomes:
$y=e^{x^2}+1$
Now, $y(0.3)=e^{(0.3)^2}+1 \approx 2.0942$
The accuracy of the approximation can be found as: $2.0942-2.0618=0.0324$
Hence, our answers are: $y_1=2, y_2=2.0202,y_3=2.0618$ and $y=e^{x^2}+1$ and the accuracy is: $0.0324$
