1. Slope Fields and Euler’s Method
Chapter 7.1

2. Definition of Differential Equations
An equation involving a derivative is called a differential equation. The order of a differential equation is the order of the highest derivative involved in the equation.
Examples: 𝑑𝑦 𝑑𝑥 =2𝑥−1, 𝑥 𝑦 ′ +𝑦=0, 𝑦 ′′ + 𝑦 ′ =1

3. Example 1: Solving a Differential Equation
Find all functions 𝑦 that satisfy 𝑑𝑦 𝑑𝑥 = sec 2 𝑥 +2𝑥+5.

4. Example 1: Solving a Differential Equation
Although Finney delays introducing this, it is helpful to remember that we can write this in differential form: 𝑑𝑦= sec 2 𝑥 +2𝑥+5 𝑑𝑥 Now it is easy to visualize that the solution is 𝑦 and to obtain this solution we must integrate with respect to 𝑥 ∫𝑑𝑦=∫ sec 2 𝑥 +2𝑥+5 𝑑𝑥 The family of functions that represent the solution are 𝑦= tan 𝑥 + 𝑥 2 +5𝑥+𝐶, 𝐶 constant

5. Solving a Differential Equation
The solution in the previous example is a general solution to the differential equation because it represents all possible solutions that differ only by the constant 𝐶
If the general solution to a first-order differential equation is continuous, then knowing the value of the function (that is, the solution) at a single point will give us the particular solution to the differential equation
The known value is called an initial condition and such a differential equation problem is called an initial value problem

6. Example 2: Solving an Initial Value Problem
Find the particular solutions to the equation 𝑑𝑦 𝑑𝑥 = 𝑒 𝑥 −6 𝑥 2 whose graph passes through the point (1,0).

7. Example 2: Solving an Initial Value Problem
Writing this in differential form gives 𝑑𝑦= 𝑒 𝑥 −6 𝑥 2 𝑑𝑥 Integrating both sides ∫𝑑𝑦=∫ 𝑒 𝑥 −6 𝑥 2 𝑑𝑥 Solving (finding an antiderivative on the right) 𝑦= 𝑒 𝑥 −2 𝑥 3 +𝐶 Now, since the point (1,0) is known to be on the graph, we find 𝐶: 0= 𝑒 1 − 𝐶 𝐶=2−𝑒 The particular solution is 𝑦= 𝑒 𝑥 −2 𝑥 3 +2−𝑒

8. Discontinuity in an Initial Value Problem
If the general solution to an initial value problem is a continuous curve, then the initial condition “captures” the curve for the entire domain
However, if the curve is discontinuous, the initial condition “captures” the curve for the domain over which it is continuous
So for these kinds of initial value problems, include the domain over which the function is continuous (as shown in the next example)

9. Example 3: Handling Discontinuity in an Initial Value Problem
Find the particular solution to the equation 𝑑𝑦 𝑑𝑥 =2𝑥− sec 2 𝑥 whose graph passes through the point (0,3).

10. Example 3: Handling Discontinuity in an Initial Value Problem
In differential form: 𝑑𝑦= 2𝑥− sec 2 𝑥 𝑑𝑥 The general solution is : ∫𝑑𝑦=∫ 2𝑥− sec 2 𝑥 𝑑𝑥 𝑦= 𝑥 2 − tan 𝑥 +𝐶 Note that this function is continuous over − 𝜋 2 , 𝜋 2 and 𝑥=3 is contained in this interval. So the particular solution is 3= 0 2 − tan 0 +𝐶⟹𝐶=3 and 𝑦= 𝑥 2 − tan 𝑥 +3, − 𝜋 2 ≤𝑥≤ 𝜋 2 Note that you must stipulate that this is true only for the interval.

11. Example 3: Handling Discontinuity in an Initial Value Problem

12. Example 4: Using the FTC to Solve an Initial Value Problem
Find the solution to the differential equation 𝑓 ′ 𝑥 = 𝑒 − 𝑥 2 for which 𝑓 7 =3.

13. Example 4: Using the FTC to Solve an Initial Value Problem
If we think of this as 𝑑𝑦 𝑑𝑥 = 𝑒 − 𝑥 2 , then 𝑑𝑦= 𝑒 − 𝑥 2 𝑑𝑥 and ∫𝑑𝑦=∫ 𝑒 −𝑥 2 𝑑𝑥 But we want the particular solution and this is set up as 𝑦=𝑓(𝑥)=3+ 7 𝑥 𝑒 − 𝑡 2 𝑑𝑡

14. Example 4: Using the FTC to Solve an Initial Value Problem
𝑓(𝑥)=3+ 7 𝑥 𝑒 − 𝑡 2 𝑑𝑡 We are done. That is, the particular solution is 𝑦=3+ 7 𝑥 𝑒 − 𝑡 2 𝑑𝑡. Does something seem to be missing? If so, what? Can we find 𝑓(7)? How? Can we find 𝑓(−2)? How?

15. Example 4: Using the FTC to Solve an Initial Value Problem
𝑓(𝑥)=3+ 7 𝑥 𝑒 − 𝑡 2 𝑑𝑡 We are done. That is, the particular solution is 𝑦=3+ 7 𝑥 𝑒 − 𝑡 2 𝑑𝑡. Does something seem to be missing? If so, what? You might be thinking that we haven’t found the explicit form of 𝑓. But strictly speaking, we have found the function we are after. Can we find 𝑓(7)? How? 𝑓 7 = 𝑒 − 𝑡 2 𝑑𝑡=3+0=3 Can we find 𝑓(−2)? How? Without an explicit function, we must rely on approximate values. This is most easily carried out using a calculator: 𝑓 −2 ≈1.232

16. Example 5: Graphing a General Solution
Graph the family of functions that solve the differential equations 𝑑𝑦 𝑑𝑥 = cos 𝑥 .

17. Example 5: Graphing a General Solution
Graph the family of functions that solve the differential equations 𝑑𝑦 𝑑𝑥 = cos 𝑥 .
In differential form: 𝑑𝑦= cos 𝑥 𝑑𝑥
Integrate: ∫𝑑𝑦= cos 𝑥 𝑑𝑥⟹𝑦= sin 𝑥 +𝐶
Now, if we change the values of 𝐶, the graphs of the general solution “shift” vertically.

18. Example 5: Graphing a General Solution

19. Slope Fields Let 𝑑𝑦 𝑑𝑥 = cos 𝑥
Since this is a basic derivative, it is easy to find the antiderivative:
𝑦= ∫ cos 𝑥 = sin 𝑥 +𝐶

20. Slope Fields What do the following values mean for the graph of y
𝑑𝑦 𝑑𝑥 = cos 0 =1
𝑑𝑦 𝑑𝑥 = cos 𝜋 2 =0
𝑑𝑦 𝑑𝑥 = cos 𝜋 =−1
𝑑𝑦 𝑑𝑥 = cos 3𝜋 2 =0
These are the slopes of the tangent lines at each of the input values

21. Example 6: Constructing a Slope Field
Construct a slope field for the differential equation 𝑑𝑦 𝑑𝑥 = cos 𝑥 .

22. Example 6: Constructing a Slope Field

23. Example 7: Constructing a Slope Field for a Nonexact Differential Equation
Use a calculator to construct a slope field for the differential equation 𝑑𝑦 𝑑𝑥 =𝑥+𝑦 and sketch a graph of the particular solution that passes through the point (2,0).

24. Example 7: Constructing a Slope Field for a Nonexact Differential Equation
This procedure can only be performed on the TI89.
From the Home screen, select MODE, then Graph, then 6:DIFF EQUATIONS
Select Y=; you must use 𝑡 in place of 𝑥 and 𝑦1 in place of 𝑦
Enter 𝑡+𝑦1 in 𝑦1′
Set the window by choosing F2 (Zoom) and 5: ZoomSqr

25. Example 7: Constructing a Slope Field for a Nonexact Differential Equation

26. Example 7: Constructing a Slope Field for a Nonexact Differential Equation
To find the graph of the particular solution that passes through (2,0), press 2nd F8
Enter 2 for 𝑡 and 0 for 𝑦1; the calculator graphs first from left to right, then from right to left, beginning each time at the chosen point

27. Example 7: Constructing a Slope Field for a Nonexact Differential Equation

28. Example 7: Constructing a Slope Field for a Nonexact Differential Equation
The following slides highlight the these facts about the slope field.
The slopes are zero along the line 𝑥+𝑦=0, or 𝑦=−𝑥
The slopes are −1 along the line 𝑥+𝑦=−1 or 𝑦=−𝑥−1
The slopes get steeper as 𝑥 increases. We can see that this follows from 𝑑𝑦 𝑑𝑥 =𝑥+𝑦: as 𝑥 increases for particular values of 𝑦, the slope increases to infinity, so the lines tend to become vertical
The slopes get steeper as 𝑦 increases, for the same reason as in 3.

29. Example 7: Constructing a Slope Field for a Nonexact Differential Equation

30. Example 7: Constructing a Slope Field for a Nonexact Differential Equation

31. Example 7: Constructing a Slope Field for a Nonexact Differential Equation

32. Example 7: Constructing a Slope Field for a Nonexact Differential Equation

33. Example 7: Constructing a Slope Field for a Nonexact Differential Equation
A note about the general solution to the unintimidating differential equation 𝑑𝑦 𝑑𝑥 =𝑥+𝑦 Though it is a simple equation, solving it requires advanced methods that you will learn about in a class on differential equations (sometimes called Diff E Q). The solution is 𝑦=𝐶 𝑒 𝑥 −𝑥−1, where 𝐶 is a constant. Although we can’t yet use the methods for solving, we can take the derivative and show that the derivative actually is 𝑥+𝑦.

34. Example 8: Matching Slope Fields with Differential Equations
Use slope analysis to match each of the following differential equations with one of the slope fields (a) through (d). Do not use your graphing calculator.
𝑑𝑦 𝑑𝑥 =𝑥−𝑦
𝑑𝑦 𝑑𝑥 =𝑥𝑦
𝑑𝑦 𝑑𝑥 = 𝑥 𝑦
𝑑𝑦 𝑑𝑥 = 𝑦 𝑥

35. Example 8: Matching Slope Fields with Differential Equations
The key to matching the graphs is to pick a slope value that is easy to identify (for example, 0, 1, −1).
If 𝑑𝑦 𝑑𝑥 =0, the lines are horizontal and this occurs where 𝑥−𝑦=0 or along the line 𝑦=𝑥

36. Example 8: Matching Slope Fields with Differential Equations

37. Example 8: Matching Slope Fields with Differential Equations
The key to matching the graphs is to pick a slope value that is easy to identify (for example, 0, 1, −1).
If 𝑥=0, then the slopes are 𝑑𝑦 𝑑𝑥 =0 along the y-axis (where 𝑥=0); also, if 𝑦=0, then the slopes are also zero along the x-axis (where 𝑦=0)

38. Example 8: Matching Slope Fields with Differential Equations

39. Example 8: Matching Slope Fields with Differential Equations
The key to matching the graphs is to pick a slope value that is easy to identify (for example, 0, 1, −1).
If 𝑥=0, then 𝑑𝑦 𝑑𝑥 =0. The lines along the line 𝑥=0 (the y-axis) are horizontal. If 𝑦=0, then the field lines are vertical (undefined slope) along the x-axis.

40. Example 8: Matching Slope Fields with Differential Equations

41. Example 8: Matching Slope Fields with Differential Equations
The key to matching the graphs is to pick a slope value that is easy to identify (for example, 0, 1, −1).
The slopes are zero when 𝑦=0 (along the x-axis). The slopes are vertical when 𝑥=0 (along the y-axis).

42. Example 8: Matching Slope Fields with Differential Equations

43. Euler’s Method
Euler’s Method provides a means of finding an approximate graph for a particular solution to a differential equation
The idea is to take advantage of local linearity and creating short segments of a defined length, starting at the initial value point
Then use the value of x at the endpoint of the short segment to create another short segment
Repeat this process
In this way, the graph approximation will lie to the right of the actual graph; the smaller the segments, the more accurate the graph

44. Euler’s Method
Euler’s Method for Graphing a Solution to an Initial Value Problem
Begin at the point (𝑥,𝑦) specified by the initial condition. This point will be on the graph as required.
Use the differential equation to find the slope 𝑑𝑦 𝑑𝑥 at the point.
Increase x by a small amount Δ𝑥. Increase y by a small amount Δ𝑦, where Δ𝑦= 𝑑𝑦 𝑑𝑥 ⋅Δ𝑥. This defines a new point (𝑥+Δ𝑥,𝑦+Δ𝑦) that lies along the linearization.
Using this new point, return to step 2. Repeating the process constructs the graph to the right of the initial point.
To construct the graph moving to the left from the initial point, repeat the process using negatives values for Δ𝑥.

45. Example 9: Applying Euler’s Method
Let 𝑓 be the function that satisfies the initial value problem in Example 6, 𝑑𝑦 𝑑𝑥 =𝑥+𝑦 and 𝑓 2 =0. Use Euler’s Method and increments of Δ𝑥=0.2 to approximate 𝑓(3).

46. Example 9: Applying Euler’s Method
Construct a table like the one below.
(𝒙,𝒚)
𝒅𝒚 𝒅𝒙 =𝒙+𝒚 𝚫𝒙
𝚫𝒚= 𝒅𝒚 𝒅𝒙 ⋅𝚫𝒙
(𝒙+𝚫𝒙, 𝒚+𝚫𝒚)
(2,0) 2
0.2
0.4
(2.2,0.4)

47. Example 9: Applying Euler’s Method
Construct a table like the one below.
(𝒙,𝒚)
𝒅𝒚 𝒅𝒙 =𝒙+𝒚 𝚫𝒙
𝚫𝒚= 𝒅𝒚 𝒅𝒙 ⋅𝚫𝒙
(𝒙+𝚫𝒙, 𝒚+𝚫𝒚)
(2,0) 2
0.2
0.4
(2.2,0.4)
(2.2, 0.4)
2.6
0.52
(2.4, 0.92)
3.32
0.664
(2.6, 1.584)
4.184
0.8368
(2.8, )
5.2208
(3, )

48. Example 9: Applying Euler’s Method

49. Example 10: Moving Backward with Euler’s Method
If 𝑑𝑦 𝑑𝑥 =2𝑥−𝑦 and if 𝑦=3 when 𝑥=2, use Euler’s Method with five equal steps to approximate 𝑦 when 𝑥=1.5.

50. Example 10: Moving Backward with Euler’s Method
Create the following table.
(𝒙,𝒚)
𝒅𝒚 𝒅𝒙 =𝟐𝒙−𝒚 𝚫𝒙
𝚫𝒚= 𝒅𝒚 𝒅𝒙 ⋅𝚫𝒙
(𝒙+𝚫𝒙, 𝒚+𝚫𝒚)
(2,3) 1
−0.1
(1.9, 2.9)

51. Example 10: Moving Backward with Euler’s Method
Create the following table.
(𝒙,𝒚)
𝒅𝒚 𝒅𝒙 =𝟐𝒙−𝒚 𝚫𝒙
𝚫𝒚= 𝒅𝒚 𝒅𝒙 ⋅𝚫𝒙
(𝒙+𝚫𝒙, 𝒚+𝚫𝒚)
(2,3) 1
−0.1
(1.9, 2.9)
0.9
−0.09
(1.8, 2.81)
0.79
−0.079
(1.7, 2.731)
0.669
−0.0669
(1.6, )
0.5359
−0.5359
(1.5, )

52. Example 10: Moving Backward with Euler’s Method