Solving a differential equation by the method of direction field (continued)

 Problem. Sketch the direction field for the following differential equation: y' = (y² - y - 2)(1 - y)²

As we saw in the previous example the direction field is nothing more than a set of tangents drawn at different points of the coordinate plane. Let's study the derivative and see how the direction field looks like in different intervals. Then we can sketch the direction field for the differential equation knowing the direction field in these different intervals.

Let's study the sign of the derivative: 

y' = 0  (y² - y - 2)(1 - y)² = 0

Solving the equation we find y = -1, y = 2, y = 1. Let's study its sign:

Y               -∞                   -1           1           2            +∞

y² - y - 2                +          0    -              -    0    +

(1 - y)²                   +                 +     0      +          +

Y'                           +          0      -      0      -   0      +

Let's sketch the direction field for the values that annul the derivative. For these values we will only have the tangents parallel. The direction field is sketched below.

When y<-1 the derivatives are positive. The slopes are then positive. We can try to see how the tangent look like in this interval. Let's choose y = -2 and see  the value of the derivative. Let's plug -2 in the derivative. We have: y' = 36. Since the value is high we can say that the slope of the tangents in this region will be very steep.

Let's see what happens when y approaches -1 which is where the tangents are parallel. Let's choose y =-1.25 then y' = 4.11. This slope is positive and has lesser value than when we choose y = -2.  The slopes are still positive but they tend to be less steep. The tangents become less and less flat as we approach -1 and the arrows are oblique. At y = -1 they become parallel. The direction field in this region looks like this:

Let's study the interval -1<y<1:

We are going to use the same strategy as above. In this region the derivative is negative. Therefore the slope is negative. Let's choose y = 0 to see how steep is the slope. For y = 0, we have y' = -2. The slope is not that steep as in the previous region.

Let's see what happens when y approaches 1. Let's choose y = 0.75, y' = -0.136. The slopes approach zero while staying negative. The tangents become flat. The arrows of the tangents are pointing down.

Let's see what happens when y moves away from 0 towards -1. Let's have y = -0.5. The derivative is y' = -2.8 . The slope becomes steeper than it was at y = 0.  

Let's see what happens when y approaches y approaches -1. Let's choose y = -0.75. The derivative is y' = -1.75. The slope approaches zero and flattens. The arrows are still pointing down,

Let's study the direction field in the interval 1<y<2.  Let's choose y = 1.5 y' = -0.3125. The slopes in this region are negative and not steep. 

Let's see what happens near y = 1. Let's choose y = 1.25 y' = -0.10. At this point the slope flattens since the value of y approaches 1.

When y moves away from 1 towards 2 like when y = 1.5 in the first consideration the slope gets steeper.

Let's see what happens near y = 2. Let's choose y = 1.75 y' = -0.38. The slope approaches zero and flattens. The direction field looks like this:

Let's study the direction field when y>2. Let's do the test to see how the slopes of the tangents look like in this region. Let's choose y = 3. Then y' = 16. The slope at this point is steep. 

Let's look at the slope near y = 2. Let's choose y = 2.25 y' = 1.2. The slope starts flat near y = 2 but as we move away from 2 like when y = 2 the slope gets steep.

Here is the  complete direction field for the differential equation:

Following the arrows of the tangents we can sketch the solution curves for the differential equation. We obtain the set of integral curves.