Answer
$y_1=4.2, y_2=6.216,y_3=9.6969$, $y=3 e^{x^2+2x}$, and $y(0.6)=3 e^{(0.6)^2+2(0.6)} \approx 14.2765 $
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 $x_0=0,x_1=0+0.2=0.2, x_2=0.2+0.2=0.4$
Likewise, by using equation (1), we have
$y_1=4.2, y_2=6.216,y_3=9.6969$
In order to determine the exact solution, isolate the x and y terms on one side and integrate both sides.
$\int \dfrac{dy}{y}=\int 2(x+1) dx$ ...(2)
In order to determine the differential equation, we will have to take the derivative of the differential equation.
Equation (2) gives: $\ln |y|=x^2+2x+c$ ...(3)
Apply the initial conditions to calculate the value of $c$
$\ln |3|=(0)^2+2(0)+c \implies c=\ln 3$
Now, equation (3) becomes:
$\ln |y|=x^2+2x+\ln 3$
This implies that $y=e^{x^2+2x+\ln 3} \implies y=3 e^{x^2+2x}$
Now, $y(0.6)=3 e^{(0.6)^2+2(0.6)} \approx 14.2765 $
The accuracy of the approximation can be found as: $14.2765-96969=4.5796$
Hence, our answers are: $y_1=4.2, y_2=6.216,y_3=9.6969$ and $y=3 e^{x^2+2x}$ and the accuracy is: $4.5796$