### Home > APCALC > Chapter 10 > Lesson 10.1.8 > Problem10-95

10-95.

Let $\frac { d y } { d x } = 0.25 y ( 3 - y )$.

1. Let $∆x = 0.5$. Plot four different curves with four different starting points using Euler’s Method: $(0,1),(0,1.5),$$(0,3),(0,5)$. Plot points until you get to $x = 3$.

Since $dx = 0.5$, $dy = 0.25y(3 - y)(0.5)$.
Starting at $(0, 1)$, the points are:
$(0 + 0.5, 1 + 0.25(1)(3 - 1)(0.5)) = (0.5, 1.25)$
$(0.5 + 0.5, 1.25 + 0.25(1.25)(3 - 1.25)(0.5)) ≈ (1, 1.523)$
$(1 + 0.5, 1.523 + 0.25(1.523)(3 - 1.523)(0.5)) ≈ (1.5, 1.805)$
$...$

2. What do you notice about all four curves? What is it about $\frac { d y } { d x }$ which causes this to happen?

What are the zeros of $dy/dx$? How are these zeros shown on a graph of the slope field?