#### Answer

$y_1=0.5, y_2=0.5,y_3=0.625$, $y=\dfrac{1}{-x^2-x+1}$, and $y(0.5)=\dfrac{1}{-(0.5)^2-(0.5)+1} \approx 4$

#### 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)
We are given $y(-1)=1$
This implies $x_0=-1,x_1=-1+0.5=-0.5, x_2=-0.5+0.5=0$
Likewise, by using equation (1), we have
$y_1=0.5, y_2=0.5,y_3=0.625$
In order to determine the exact solution, isolate the x and y terms on one side and integrate both sides.
$\int \dfrac{dy}{y^2}=\int (1+2x) dx$ ...(2)
In order to determine the differential equation, we will have to take the derivative of the differential equation.
Equation (2) gives: $y=\dfrac{1}{-x^2-x+c}$ ...(3)
Apply the initial conditions, to calculate the value of $c$
$1=\dfrac{1}{-(-1)^2-(-1)+c} \implies c=1$
Now, equation (3) becomes:
$y=\dfrac{1}{-x^2-x+1}$
Now, $y(0.5)=\dfrac{1}{-(0.5)^2-(0.5)+1} \approx 4$
The accuracy of the approximation can be found as: $4-0.625=3.375$
Hence, our answers are: $y_1=0.5, y_2=0.5,y_3=0.625$ and $y=\dfrac{1}{-x^2-x+1}$ and the accuracy is: $3.375$