Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (c)/Solution 1

Let us denote $f(y)=(y-1)(3-y)$. At $t=0$, we have $y_0=y(0)=2$ and $f(y_0)=f(2)=(2-1)(3-2)=1$. Hence, at our next time step, we have $y_1=1+(1/2)=5/2$ and $f(y_1)=f(5/2)=(3/2)(1/2)=3/4$. At our final time step, we have $y_2=5/2+(1/2)(3/4)=23/8$, which is our approximation.