Slope Fields and Differential Equations

By: Jonathan Herlong

&

Curt Harper

Slope Fields

• Slope fields are a means to graphically display the solutions of differential equations.

• When solving a differential equation using a slope field it is not necessary for you to solve the equation analytically, you just have to insert the x and y values to find out what the slope of the graph at that point is.

• The resultant lines representing the slopes of a differential equation at certain points on a graph allows the person to visualize the actual graph of the equation.

How to Solve Slope Fields

• The differential equation you are given to solve a slope field will look like:

• The equation you are given is in terms of dy/dx because that represents slope.

• You will then be given a graph with points marked on it that you must insert into the differential equation to find the slope for that particular point.

How to Solve Slope Fields Continued

• The points we are given to insert into the differential equation are:

– (0,0) – (1,0) – (-1,0) – (0,1)– (1,1)– (-1,1)

• After Plugging these points into the equation: you get the values:

– (0,0): 2– (1,0): 3– (-1,0): 1– (0,1): 1– (1,1): (3/2)– (-1,1): (1/2)

How to Solve Slope Fields Continued

• You then must graph the following slopes you just calculated and draw them on the graph that is provided.

Slope Field Practice

• Solve: • =(y^2-2)/(x+2)• Given the points marked in red dots:

Slope Field Practice Continued

• Plug in the points you are given into the differential equation: =(y^2-2)/(x+2)

• (-1,0): -2• (0,0): -1• (1,0): -(2/3)• (2,0): -(1/2)• (-1,1): -1• (0,1): -(1/2)• (1,1): -(1/3)• (2,1): -(1/4)• (-1,2): 2• (0,2): 1• (1,2): (2/3)• (2,2): (1/2)

Try Me!

• Solve the following slope field for the points given on the graph:

• =cos(x)

Solution To Try Me

• =cos(x)

• [-(π/2),0]=0• [-(π/3),0]=(1/2)• [-(π/4),0]=[√(2)/2]• [-(π/6),0]=[√(3)/2]• (0,0)=1• [(π/6),0]=[√(3)/2]• [(π/4),0]=[√(2)/2]• [(π/3,0]=(1/2)• [(π/2,0]=0

• [-(π/2),1]=0• [-(π/3),1]=(1/2)• [-(π/4),1]=[√(2)/2]• [-(π/6),1]=[√(3)/2]• (0,1)=1• [(π/6),1]=[√(3)/2]• [(π/4),1]=[√(2)/2]• [(π/3,1]=(1/2)• [(π/2,1]=0

Solutions To Try Me Slope Field

Try Me!

• Solve the following slope field for the points given on the graph:

• X^2dx=y^3dy

Solution To Try Me

• =(y^3/x^2)

• (-1,0)=0• (1,0)=0• (2,0)=0• (-1,1)=1• (1,1)=1• (2,1)=(1/4)

• (-1,2)=8• (1,2)=8• (2,2)=2• (-1,-1)=-1• (1,-1)=-1• (2,-1)=-(1/4)

Solutions To Try Me Slope Field

Integration Review

• The most commonly used type of integration when solving differential equation is the power rule.

• To integrate using the power rule you must add one to your power and then divide by that number. Next, you add “C” to the end of the solution.

• Equation:

• If the power of the equation you are integrating is -1 you must write “ln” of the equation and balance by taking the derivative of the equation.

• Example: 1/(x)dx=lnIxI+C

Practice Integration

• Problems: – ∫[x^4+4]dx

– Implement the power rule to the “x^4” and the “4.”

(x^5/5)+4x+C– The Resultant values of the power rule are shown and don’t

forget to add “C” to the end of your answer.

– ∫[1/(2x+1)]dx– Use natural log since the power of the variable is -1. Then,

take the derivative of the denominator and balance with a (1/2) on the outside of the integral. Finally, add C to the end.

(1/2)lnI2x+1I+C

How to Apply The Exponential Function to an Equation

• The exponential function must be applied when taking the integral of both sides of a differential equation results with one side being natural log of some value.

• It must be applied to cancel out the natural log because “e” raised to the “ln(x)” results in “x.”

• Example: =y/x^2 ∫ 1/ydy= ∫ 1/x^2dx lnIyI=-1/x+C e^(lnIyI)=e^(-1/x+C) y=e^(-1/x)e^(C) y=Ce^(-1/x)

Basic Exponential Rules

• Some situations you may encounter:• e^(C)=C

– You are just raising “e” to a number which will result in another number allowing you to write “C” as the answer

• e^(x+C)=e^(x)e^(C)=Ce^(x)– When “e” is raised to some variable plus another number

or variable, you can write it as “e” to the variable times “e” to the other variable or number. In this situation “e^C” is a number so it can be written as “C.”

• e^(0)=1

Differential Equations

• A differential equation is any equation that contains derivatives.

• The two types of differential equations are general differential equations and particular differential equations.– General: contain a “C.” – Particular: the “C” has been solved for by the

given point.

General Differential Equation Explanation

• In general differential equations, the goal is to find the function with a number, C, by integrating both sides of the equation.

• Integrating differential equations can look as if it’s hard, but it is just integration, usually starting with a =f(x)

• The desired outcome of a differential equation is a Y= equation.

dy

dx

General Differential Equation Solved

/dy

x ydx

1. First, cross multiply to put the X function with the dx, and the Y function with the dy.

2. Step two you integrate both sides of the equation.

3. After integration combine the C terms on to the X side of the equation, since C is a number transferring the C from the Y side will not create a wrong answer.

4. Now Solve for Y to get your General Differential Equation.

ydy xdx

ydy xdx 1

2y2 1

2x2

C

2y x C

Try Me!

Consider the equation

and find the general differential equation.

2 / 7dy

x ydx

Solution to Try Me!

2

2

2

2 3

2 3

3

/ 7

7

7

7 / 2 ( / 3)

(2 / 21)

(2 / 21)

dyx y

dx

ydy x dx

ydy x dx

y x C

y x C

y x C

Reminders

• When you integrate, dy or dx can never be in the denominator of an equation.

• C is a number in every equation, so you do not need to say 2C if an equation is multiplied by 2. C is the appropriate value of that number.

• Derivatives and Integration are opposites so don’t get the two confused.

Particular Differential Equation Explanation

• Particular differential equations are the same as general differential equations, except you find C in the equation.

• To do this solve the differential equation just like a general one, and then plug in the X and Y values to find C.

• Once C is found, plug C back into the equation and solve for y.

Particular Differential Equation Solved

For this particular equation the point given is (e,1)

1/5 1/5

1/5 2/5

/ 5

/ / 5

(1/ ) (1/ 5) (1/ )

ln( ) (ln( ) / 5)

ln(1) (ln( ) / 5)

0 (1/ 5)

( 1/ 5)

( 1)

dyy x

dxdy y dx x

y x

y x C

e C

C

C

y xe e

y e xe

1. First cross multiply to put the dy over y and the dx over 5x

2. Next, integrate both sides.

3. Plug the given point into both sides.

4. Solve for C.

5. Apply the exponential function to both sides of the equation, and pull out an 1/5e

Try Me!

Consider the equation

and find the particular differential equation

for the point (1,2)

2 5/dy

x ydx

Solution to Try Me2 5

5 2

5 2

6 3

6 3

6 3

/

/ 6 ( / 3)

(2 / 2) (1 / 3)

32 (1/ 3)

95 / 3

2 190

dyx y

dx

y dy x dx

y dy x dx

y x C

C

C

C

y x

Try Me With Differentials and Slope fields

Solve the slope field for the following

Equation and find the particular differential equation for the point (2,1)

2( ) / 2dy

x ydx

Solution to Try Me (Slope field)

Solution to Try Me (Particular)

3

2

2

3

3

/6

( ) / 2

1/ 2 1/

/ 6 ln

2 / 6 ln1

4 / 3 0

4 / 3

( 4 ) / 3x

dyx y

dx

x dx y

x C y

C

C

C

y e

Sources

• http://calculuslab.deltacollege.edu/ODE/7-1/7-1-1-c-2-mma.html

• http://apcentral.collegeboard.com/apc/public/courses/teachers_corner/11871.html

• http://incompetech.com/graphpaper/leftborder/

• http://tutorial.math.lamar.edu/Classes/DE/Definitions.aspx

• ©Jonathan Herlong and Curt Harper 2011