math

Linear 1st Order Differential Equations:

Lets say you come across an equation where you can't seperate the variables then what do you do? Steps:

1) Put equation in the following form:

2) Identify a(x)

A) Integrate a(x)

B) Then exponentiate the intergal:

*These two steps give you what is called the multiplicative factor

3) The original Equation becomes:

4) Integrate both sides

4a) if IVP plug in variables & solve for unknowns

5) Solve for y

Existence Uniqueness Theorum:

Consider the IVP y' = f (x , y) ; where y(x _not)=y_not.

If there exists a rectangle R in the plane that contains the initial

point (x_not, y_not), as an interior point such that both functions

f(x , y) & df/dy are continous on R then the IVP has only 1 solution

Steps:

1) Put into the following form and identify f(x , y)

2) Take the derivative of f with respect to y [ ie: df/dy ].

3) Find where the function is discontinuous & graph it.

4) Graph the initial values (x_not , y_not)

5) If you can draw a box arround (x_not , y_not) without touching

discontinuities, then there is 1 unique solution.

In the above example the discontinuity is at x = -1 (the red line). The initial point was at ( 0,1) and because we can draw a bow arround it without touching the discontinuity we know that there exists 1 unique solution to this IVP.

If our IVP was dy/dx = y^(1/3) and our initial point was (0,0). differentiating in this case gives us 1/(3y^(2/3)). The discontinuity in the graph is where y = 0 {because the denominator goes to 0 which is undefined}. This is the x-axis. Since our initial point is on the x-axis we can not draw a box arround it without crossing the discontinuity. This tells us that this particular IVP has no solution.

This theorum works as well if our discontinuity were any other line, such as a parabola, if we can box in the initial values without crossing the discontinuity then there will be 1 unique solution.

Differential Equations for various situations:

The Runge-Kutta method:

The RK methold is used to give you an approximation of a solution at a particular point to a differential equation. My teacher was generous and after doing it a few times told us to get a program on our calculators that could do it for us. Below is the text of the program written out. I’ll eventually put a link to directly download it to a TI calculator. Some of the HPs out there have it allready built in, there are also better versions of this I’ve seen, but this one is functional in case you need it. To use this program go INTO the program and change RIGHT SIDE of the equation after the Define to match your DE. Then go and run the program by putting the variables in in the following order:

rk(initial y value, initial x value, final x value)

If using a TI-89 make sure you hit the little diamond button before hitting enter if you don’t it will give you the answer in fraction format and will take about 5 minutes to do so. Just hit enter after selecting step size or num steps to get each coorisponding step value.

Program:

rk(y,x,x1)

Prgm

ClrIO

Define f(x,y)= 2*x/(1+2*y)

PopUp {“NUM STEPS”,”STEP SIZE”},g

If g=1 Then

Input “NUM STEPS:”,n

(x1-x)/n↦h

ELSE

Input “STEP SIZE:”,h

(x1-x)/h↦n

EndIf

For i, 1, n, 1

x↦x0

y↦y0

f(x,y) ↦k1

x0+h/2↦x

y0+h*k1/2↦y

f(x,y) ↦k2

y0+h8k2/2↦y

f(x,y) ↦k3

x0+h↦x

y0+h*k3↦y

f(x,y) ↦k4

(k1+2*k2+2*k3+k4)/6 ↦k

y0+h*k ↦y

Disp x,y

EndFor

EndPrgm

Linear Algebra

Matrices:

Matrices are useful for solving several equations that have several variables in them such as:

2a + 3b - c = 0

a - 5b - 18b = 6

8a + c = 12

This type of grouping of equations is called a Linear System of Equations. Putting this linear system into a matrix makes it much easier to solve then other methods of getting the same answer. There is a lot you can do with matrices to extract information from a particular system. Unlike most math, matrices I used right away in my Circuit Analysis class.

To put the above given linear system into a Matrix we look at the number of variables in the equations. Each equation has 3 unknown variables “a”,”b”, and “c”. Even though there is no “b” in the last equation we can take to to actually be 0*b. Since there are 3 equations and 3 unknowns we will have a 3x3 Matrix. Likewise, if we had 2 equations and 4 unknowns we would have a 2x4 matrix. For our example the variables would be entered into matrix as follows:

a b c

Notice that the variable “a” is always put into the first column, the “b” variable is always in the second column, and the “c” variable is always in the third. Once you’ve assigned a variable to a column that column is to hold ONLY that variable, so a “b” variable and a “c” variable cannot occupy the same column.

Row Operations & RREF Form:

Once we have the equation entered into the matrix we use row operations to reduce the matrix down to the simplest form we can. This will tell us what the answers to “a”, “b”, and “c” will be for this system. Row Operations are simple mathematical computations we use to “change” a matrix’s rows. (These changes will give us a different matrix but that’s okay if you can do row operations to obtain another matrix the two matricies are called “Row Equivalent” and they will yeild the same answers for a system. Given a matrix A and another row equivalent matrix B we can be write A~B) There are 3 types of row opterations we can use on a matrix:

1. Switch any two rows of the same matrix

2. Multiply any row by a non-zero scalar

3. Add a multiple of one row to another

Using our example matrix a series of row operations would be:

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Teachers will sometimes ask for Echelon Form (EF) or Row Reduced Echelon Form (RREF). These are the forms that give you your systems answers. Using Echelon Form you will have to use back substituation to obtain a final answer so it is good practice to always put the matrix into RREF so you don’t have to do additional algebra to figure out your problems final answers. RREFing our example matrix can be done by following these steps:

******************************

This matrix has now been RREFed.

A matrix in RREF has all zero rows on the bottom, the leading entry (first # on each non-zero row) is a 1 and every other number above and below a 1 is a zero. Also the farthest left 1 has to be on the top, the next farthest left 1 below the first, etc. The following are all examples of matricies in RREF.

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Once in RREF we just remember what variable goes with what column and read off our answer. For our RREFed example matrix this would be:

Matrix Inverses:

Matrix Determinants:

Linear Combinations:

Linear Independance / Dependance:

Bases:

Second Order Linear Diff. Equations: