How to fill in the table if $dy/dt=0.5t$? Fill in the missing values in the table below, given that $dy/dt=0.5t$. Assume that  the rate of growth, given by $dy/dt$, is approximately constant over each unit time interval.

What exactly am I supposed to do here? I don't think it's just solving the equation and finding the appropriate $C$, and then evaluating at $t=1,2,3,4$. And I need to use the condition that $dy/dt$ is approx. constant over each unit time interval somehow.
 A: The solution of the differential equation
$$\frac{dy}{dt} = 0.5 t$$
is
$$y(t) = a + 0.25 t^2,$$
for a suitable real parameter $a$. If you assume that $y(0) = 8$, then:
$$y(0) = a + 0.25 \cdot 0^2 = 8 \Rightarrow a = 8 \Rightarrow y(t) = 8 + 0.25 t^2.$$

Anyway, if you want to approximate the solution by assuming that the increase is constant in a given time interval, consider that you can use the Euler method:
$$y(t)  = y(t-\tau) + \tau\frac{dy}{dt},$$
where $\tau = 1$ ($\tau$ is the intersample distance).
Therefore:
$$y(t) = y(t-1) + 0.5 t,$$
and hence the approximate solution is:
$$
\begin{array}{cllll}
y(0) & & & = & 8\\
y(1) & = & y(0) + 0.5 \cdot 1 = 8 + 0.5 & = & 8.5\\
y(2) & = & y(1) + 0.5 \cdot 2 = 8.5 + 1 & = & 9.5\\
y(3) & = & y(2) + 0.5 \cdot 3 = 9.5 + 1.5 & = & 11\\
\ldots
\end{array}
$$
and so on.
Notice that the real solution $y(t) = 8 + 0.25 t^2$ gives instead the following values:
$$
\begin{array}{cllll}
y(0) & = & 8 + 0.25 \cdot 0^2 & = & 8\\
y(1) & = & 8 + 0.25 \cdot 1^2 & = & 8.25\\
y(2) & = & 8 + 0.25 \cdot 2^2 & = &9\\
y(3) & = & 8 + 0.25 \cdot 3^2 &= & 10.25\\
\ldots
\end{array}
$$
and so on.
A: You are given:$$y'(t)=0.5t$$
By integrating, you have:
$$y(t)=\dfrac{0.5}{2}t^2+K$$
where K is a constant that is is defined from:
$$y(0)=K=8$$
Thus,
$$y(t)=0.25t^2+8$$
