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Qualitative methods for differential equations

12.2 Visualizing solutions with slope fields

6 problems · hints, answers and solutions shown beside each one

Q1

Slope fields. Consider the differential equations given below. In each case, draw a slope field, determine the values of yy for which no change takes place - such values are called steady states - and use your slope field to predict what would happen starting from an initial value y(0)=1y(0)=1.

  1. dydt=−0.5y\displaystyle \frac{dy}{dt}= -0.5 y

  2. dydt=0.5y(2−y)\displaystyle \frac{dy}{dt}= 0.5 y (2-y)

  3. dydt=y(2−y)(3−y)\displaystyle \frac{dy}{dt} = y(2-y)(3-y)

Answer

  1. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

    Steady state: y=0y=0. If y(0)=1y(0)=1, y→0y \to 0.

  2. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

    Steady states: y=0, 2y=0,\,2. If y(0)=1y(0)=1, y→2y \to 2.

  3. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

    Steady states: y=0, 2, 3.y=0,\,2,\,3. If y(0)=1y(0)=1, y→2y \to 2.

Q2

Drawing slope fields. Draw a slope field for each of the given differential equations:

  1. dydt=2+3y\frac{dy}{dt} = 2+3y

  2. dydt=−y(2−y)\frac{dy}{dt} = -y(2-y)

  3. dydt=2−3y+y2\frac{dy}{dt} = 2-3y+y^{2}

  4. dydt=−2(3−y)2\frac{dy}{dt} = -2(3-y)^{2}

  5. dydt=y2−y+1\frac{dy}{dt} = y^{2}-y+1

  6. dydt=y3−y\frac{dy}{dt} = y^{3}-y

  7. dydt=y(y−2)(y−3)2\frac{dy}{dt} = \sqrt{y}(y-2)(y-3)^{2}, y≥0y \ge 0.

Answer

  1. dydt=2+3y\frac{dy}{dt} = 2+3y:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  2. dydt=−y(2−y)\frac{dy}{dt} = -y(2-y):

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  3. dydt=2−3y+y2\frac{dy}{dt} = 2-3y+y^{2}:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  4. dydt=−2(3−y)2\frac{dy}{dt} = -2(3-y)^{2}:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  5. dydt=y2−y+1\frac{dy}{dt} = y^{2}-y+1:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  6. dydt=y3−y\frac{dy}{dt} = y^{3}-y:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  7. dydt=y(y−2)(y−3)2\frac{dy}{dt} = \sqrt{y}(y-2)(y-3)^{2}, y≥0y \ge 0:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

Q3

Using slope fields. For each of the differential equations (a) to (g) in Exercise 2, plot dydt\frac{dy}{dt} as a function of yy, draw the motion along the yy-axis, identify the steady state(s) and indicate if the motions are toward or away from the steady state(s).

Answer

  1. dydt=2+3y\frac{dy}{dt} = 2+3y:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  2. dydt=−y(2−y)\frac{dy}{dt} = -y(2-y):

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  3. dydt=2−3y+y2\frac{dy}{dt} = 2-3y+y^{2}:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  4. dydt=−2(3−y)2\frac{dy}{dt} = -2(3-y)^{2}:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  5. dydt=y2−y+1\frac{dy}{dt} = y^{2}-y+1:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  6. dydt=y3−y\frac{dy}{dt} = y^{3}-y:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

  7. dydt=y(y−2)(y−3)2\frac{dy}{dt} = \sqrt{y}(y-2)(y-3)^{2}, y≥0y \ge 0:

    Figure from prac_s12.2, line 2

    Figure from prac_s12.2, line 2

Q4

Slope field. The slope field shown in the figure below corresponds to which differential equation?

  1. dydt=ry(y+1)\displaystyle \diff{y}{t}=ry(y+1)

  2. dydt=r(y−1)(y+1)\displaystyle \diff{y}{t}=r(y-1)(y+1)

  3. dydt=−r(y−1)(y+1)\displaystyle \diff{y}{t}=-r(y-1)(y+1)

  4. dydt=ry(y−1)\displaystyle \diff{y}{t}=ry(y-1)

  5. dydt=−ry(y+1)\displaystyle \diff{y}{t}=-ry(y+1)

Figure from prac_s12.2, line 347

Figure from prac_s12.2, line 347

Answer

(B)

Q5

Shown below is the slope field for t differential equation dydt=y(1−y2)\diff{y}{t}=y(1-y^2).

Figure from prac_s12.2, line 2

Figure from prac_s12.2, line 2

Use the field to sketch solutions to the differential equation with the passing through the indicated point.

  1. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

  2. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

  3. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

  4. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

  5. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

  6. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

Hint

Extrapolate in both directions to get as much of the domain as will fit on the field.

Answer

  1. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

  2. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

  3. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

  4. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

  5. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

  6. Figure from prac_s12.2, line 1

    Figure from prac_s12.2, line 1

Q6

Below is the slope field for a differential equation.

Figure from prac_s12.2, line 2

Figure from prac_s12.2, line 2

For which initial values do we expect the solution to tend towards a constant real number? For which initial values do we expect the solution to tend towards positive resp. negative infinity?

Answer

For initial values in the range [0,∞)[0,\infty), we expect the solution to have a finite limit. For negative initial values, we expect the limit of the solution as t→∞t \to\infty to be −∞-\infty. There are no initial values that create a solution tending towards positive infinity.

Full solution

For initial values in the range [0,∞)[0,\infty), we expect the solution to have a finite limit. For negative initial values, we expect the limit of the solution as t→∞t \to\infty to be −∞-\infty. There are no initial values that create a solution tending towards positive infinity.

With practice, you can read the above off of the slope field directly. If you're having a hard time seeing it, though, we've drawn in a few explicit solutions, to help.

Figure from prac_s12.2, line 2

Figure from prac_s12.2, line 2

Source

Questions 1-4 in this section are from Keshet, Chapter 13. The others are original content.

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From the UBC Math 100 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.