1. Part (a)
Keep in mind that dy/dx is the SLOPE! We simply need to substitute x and y into the differential equation and represent each answer as a slope on the graph.

2. Part (a)
dy/dx = x2 (y-1)
dy/dx = -12 (3-1)
dy/dx = 2

3. Part (a)
dy/dx = x2 (y-1)
dy/dx = 02 (1-1)
dy/dx = 0

4. Part (a)
dy/dx = x2 (y-1)
dy/dx = 12 (2-1)
dy/dx = 1

5. Part (a)
Repeating this process on the other nine points gives us the following slope field…

6. “Where is the slope positive?”
Part (b)
dy/dx = x2 (y-1)
“Where is the slope positive?”

7. Part (b)
dy/dx = x2 (y-1)
x2 can’t be negative, but it can be zero. This occurs when x=0 (or on the y-axis). For our slopes to be positive, we must stay off of the y-axis.

8. (Above this dotted line)
Part (b)
dy/dx = x2 (y-1)
For the entire derivative to be positive, (y-1) will need to be positive. This occurs when y>1.
(Above this dotted line)

9. Part (b)
dy/dx = x2 (y-1)
Therefore, the derivative will be positive at any point (x,y) where x = 0 and y > 1.

10. (Separate the variables) (Integrate both sides)
Part (c)
ln y-1 = 1/3 x3 + C
dy/dx = x2 (y-1)
(Separate the variables)
(y-1)-1 dy = x2 dx
(Integrate both sides)
ln y-1 = 1/3 x3 + C

11. (Exponentiate both sides) (Using the laws of exponents from Algebra)
Part (c)
ln y-1 = 1/3 x3 + C
(Exponentiate both sides)
y-1 = e1/3 x3 + C
(Using the laws of exponents from Algebra)
y = 1 + (e1/3 x3) (eC)
y-1 = +(e1/3 x3) (eC)

12. Part (c)
eC = +2
2 = + (e0) (eC)
3 = 1 + (e1/3 03) (eC)
y = 1 + (e1/3 x3) (eC)

13. Part (c)
eC must be +2, because -2 doesn’t work for the given initial condition.
eC = 2
y = 1 + (e1/3 x3) (2)
y = 1 + (e1/3 x3) (eC)
y = 1 + 2e1/3 x3