Answer
$y_1=0.2, y_2=0.392,y_3=0.5622$, $y=1-e^{\frac{1-x^2}{2}}$, and $y(1.6)=1-e^{\frac{1-(1.6)^2}{2}} \approx 0.5416 $
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 $y(1)=0$
This implies $x_0=1, x_1=1+0.2=1.2, x_3=1.2+0.2=1.4$
Likewise by using equation (1), we have
$y_1=0.2, y_2=0.392,y_3=0.5622$
In order to determine the exact solution, isolate the x and y terms on one side and integrate both sides.
$\int \dfrac{dy}{1-y}=\int x dx$ ...(2)
In order to determine the differential equation, we will have to take the derivative of the differential equation.
Equation (2) gives: $-\ln (1-y)= \dfrac{x^2}{2}+c$ ...(3)
Apply the initial conditions, to calculate the value of $c$
$-\ln (1-0)= \dfrac{(1)^2}{2}+c \implies c=-\dfrac{1}{2}$
Now, equation (3) becomes:
$\ln (1-y)= \dfrac{1-x^2}{2}$
This implies that $y=1-e^{\frac{1-x^2}{2}}$
Now, $y(1.6)=1-e^{\frac{1-(1.6)^2}{2}} \approx 0.5416 $
The accuracy of the approximation can be found as: $0.5622-0.5416=0.0206$
Hence, our answers are: $y_1=0.2, y_2=0.392,y_3=0.5622$ and $y=1-e^{\frac{1-x^2}{2}}$ and the accuracy is: $0.0206$
