We continue to work with y' = y – 2t, with initial condition y(1) = 5 using a step size of h = 1 on the interval from 1 to 5. Our initial condition gives us our starting y value (in this case y(1) = 5), so we can fill this into the first row, second column of our table.
We will also fill in a y' value. We will get this from the right hand side of our differential equation. (We must make sure the equation is in the form y' = f(t, y) before we do this. In our example, our f(t, y) = y – 2t.) Remember that t is whatever t is on the current row, and y is whatever the y value is on the current row.
In our example, this means that since we start at t = 1, we have y'(1) = y – 2t become y'(1) = 5 – 2(1) or y'(1) = 3. We enter this in our table as a formula. In our example, we have the current t value in cell A2, and the current y value in cell B2, so we would enter =B22*A2 in cell C2. So our table now looks like the following:
A 
B 
C 

1 
t values 
y values  y' values 
2 
1 
5 
3 
3 
2 

4 
3 

5 
4 

6 
5 
Remember, the first y value comes from the initial condition and the y' value is a formula based on the differential equation.
We are now ready to enter the second row formula and copy down.