Slope Fields and Solving Differential Equations

Slope Fields - A graphical representation of a differential equation. They are used to visualize the parallel curves that constitute a family of general solutions to a differential equation without actually solving it.

Examples:

dy/dx = x-y/2



Differential Equations
A differential equation contains one or more terms involving derivatives of one variable with respect to another variable.


Example:



This is a general solution to a differential equation.
Another thing you can do is find the particular solution if you are given an initial condition.

Example:

y(2)=-1



-1=1/3(2)^3 + C
C = -11/3
y = 1/3x^2 -11/3

Integration by Substitution

Integration by Substitution is "sort of the reverse Chain Rule".

To integrate using substitution, step one is to choose "u." "u" is a piece of the equation that you want to substitute. Most often, if there is a "ln(x)," choose that to substitute. For example in the equation below, "u" is the part in the parenthesis.

Step 2: Now that we have our "u," we want to take the derivative of it.

u= x²-1
du = 2x*dx

Now that we have du, we want to isolate dx so we can plug it back into the original.

du = 2x*dx
2x = 2x
dx= du
         2x

Step 3: Plug dx into the original function and replace the part of the equation that you used for "u" with "u." 

If you look closely, it appears that the 2x's cancel out and you are left with:

                                                              ∫u4du

Step 4: Now you know how to take the integral of this kind of function and you get:

Step 5: This may look complete, but the "u" is just something we substituted for, so we 

want to replace "u" with what we previously defined it as. 

Completed, the integral would be:

Properties of Integrals

Properties of Integrals:

Different Properties of Integrals can be used to evaluate certain integral equations.

To evaluate an integral:

1. f(x) must be defined

2. The function must be continuous on the interval

Adding/Subtracting Integrals:

 Integrals with an interval of 0:

 Multiplying Integrals by a constant:

 Adding Integrals of adjacent intervals:

 Negative Integrals:

Example from textbook (pg. 290 #2):

a. (-2) ∫f(x)dx = (-2)(-1) = 2

b. ∫f(x)dx + ∫h(x)dx = 5+4 = 9

c. ∫[(2)∫f(x) - (-3)∫h(x)]dx

    (2)∫f(x)dx - (-3)∫h(x)dx = (2)(5)-(-3)(4) = 10+12 = 22

d. ∫f(x)dx = 1

e. (from 1 to 9)∫f(x)dx + (from 7 to 1)∫f(x)dx =  (from 7 to 9)∫f(x)dx =  

    -1 + -∫f(x)dx = 5

    5 +1 = 6

    -6 = (from 1 to 7)∫f(x)dx

f.  ∫h(x)dx - ∫f(x)dx = 4-5 = -1

*keep in mind that the intervals must be the same as what you are evaluating for as I could not type and show that here

Derivatives of Inverse Functions

Inverse Functions:

Inverse functions can be written as f^(-1)(x) and they occur when the y and x values of a function are switched so that the graph flips over dotted line in the example below.



Finding the Derivative of an Inverse Function:

The derivative of an inverse function is the same as the reciprocal of the derivative of the original function when the x value of the inverse function is equal to the y value of the function. In other words, to find the derivative of the function you must:

1. Find the x value of the original function so that the y value original function is equal to the x value of the inverse function.
2. Find the derivative of the original function when x is equal to the x value you found in step 1.
3. Find the derivative of the inverse function by finding the reciprocal of the answer from step 2. (In other words flip the fraction, but remember that the sign does not change)

Example:
Find (f^-1)'(3) when f(x) = -x^2+5

3 = x+5
5-3 = x
2 = x
f'(x) = -2x^1+0 = -2x
f'(2) = -2(2) = -4
(f^-1)'(3) = -1/4

Sometime you are given a chart instead of an equation.

Example:

Find (g^-1)'(3)

3 = g(x)
x = 4
g'(4) = 1/2
(g^-1)'(3) = 2