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 = -2x/y
dy/dx = -2(-1)/(2)
dy/dx = 1

3. Part (a)
dy/dx = -2x/y
dy/dx = -2(0)/(-1)
dy/dx = 0

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

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

6. Part (b)
When we see “tangent to the graph”, we should immediately think “first derivative”.
We know that the derivative represents SLOPE, so the slope of the tangent line at (1,-1) is… dy dx =
-2x y
-2(1)
(-1) = = 2
Tangent line: (y+1) = 2 (x-1) or y = 2x-3

7. Part (b)
f(1.1) is about -0.8
Tangent line: (y+1) = 2 (x-1) or y = 2x-3

8. (Separate the variables) (Integrate both sides)
Part (c)
½y2 = -x2 + C
½(-1)2 = -(1)2 + C
y = x2
dy/dx = -2x/y
½ = -1 + C
(Separate the variables)
This is the only solution because the positive square root will not pass through (1,-1) as it should.
3/2 = C
½y2 = -x2 + 3/2
(Double both sides)
y dy = -2x dx
(Integrate both sides)
y2 = -2x2 + 3
y = x2 + -
½y2 = -x2 + C