Differential Equations, Part 3: Elimination of Arbitrary Constants with a Single Variable in Two Factors
SAME PRINCIPLES BUT MORE COMPLEXITY:
The same techniques used in hub#12.11( linked) will be used in this hub, but as you can see it is considerably more complicated when the single variable is in a second factor. This additional complexity arises from the fact that we must use the product rule( see L2) of differentiation to differentiate the original function. In hub#12.11 the first derivative gave us two terms but in this hub it gives us 4 terms. In hub#12.11 the second derivative gave us two terms again, but in this hub it gives us 8 terms. We still have the simultaneous equations to solve but it is now very cumbersome to do so with all these terms.
TWO CONSTANTS TO ELIMINATE WILL REQUIRE 1ST AND 2ND DERIVATIVES:
This was explained at hub#12.11. Using the product rule at L2, the first derivative is taken at L3 and put in order at L4. The 2nd derivative of Y(located at L1) is taken by differentiating Y′( located at L4), and this is done at L5 and L6 by using the product rule throughout. Y′′ (the 2nd derivative of Y) is organized at L7, and its factors are assigned the symbols a1, a2, etc at L7. When the math gets so cumbersome, I find it much easier to condense the work further with symbols as these. L8 is just L7 rewritten with the new symbols. These symbols( a1, a2, a3, and a4) are also common to Y and Y′ as listed at L9.
THIS PROBLEM WILL BE SOLVED TWO WAYS---THE HARD WAY AND THE EASY WAY:
4. First the hard way; it is being included because I thought the reasoning behind it will benefit students and others who are interested. This reasoning hinges on L10. The easy way begins at paragraph 12.
5. If x1 is distributed across the entire equation of y'' then it becomes a factor to every term in that equation; i.e. x1y'' = -5x1a1--5x1a2--12x1a3+12x1a4. The same statement goes for x2, and x3. Let us consider the first column in the brackets to be column 1. So column 1 would be our homogeneous equation( see L11 also): x1y'' + x2y' + x3y = 0. Homogeneous can have different meanings in math but in this case it means an equation that equals zero. Linear algebra( LA) is a powerful branch of mathematics, and in LA a set of simultaneous equations can be solved with matrices. We did this in hub#21.1( linked). We can think of the rows( or columns) in a matrix as being vectors with the elements of the matrix representing their vector's coordinates. At L10 we will consider the rows to be vectors; therefore, column 2 with the a1 terms can be viewed as the i( x-axis) coordinate; column 3 with the a2 terms as the j( y-axis) coordinate; column 4 with the a3 terms is k( the z-axis) coordinate, and column 5 with the a4 terms is the 4th spacial dimension axis, and gives the coordinate in the 4th dimension. We live in only 3 dimensions, but our equations can go into as many spacial dimensions as we want. If this is confusing you may remember when two simultaneous equations were solved with two line equations, and where the two lines intersect was the solution to the two simultaneous equations. This is the same reasoning here except we are in four spacial dimensions, not just two, and we have three 4D vectors at L10.
L10 CONTINUED; USE VECTOR COORDINATES TO SOLVE EQUATIONS:
6. So we have 3 vectors in 4D space, and where all 3 vectors intersect will be the solution to our system of equations at L10. If we add the first column as at L11 we have our homogeneous equation( it equals zero). Remember when we have 3 equations in 4 unknown then we have infinite solutions to the system of equations. But L11 adds up to zero; therefore, the three vectors at L10 must add to zero also. The only way this is possible is if each of the 4 coordinates are equal to zero. This is why the vertical arrows all point to zeros at L10. There is no way vectors can add to the zero vector unless all of their respective components add to zero also. So -5x1a1 + 2x2a1 + x1a1 must equal zero. The same argumentation goes for the other three columns.
WRITE THE VECTOR COMPONENTS AS A SYSTEM OF EQUATIONS:
7. We do this at L12, and by solving for x1, x2, and x3 at L11 we have the solution to our problem, which is eliminating the arbitrary constants and coming up with a general differential equation( linear, ordinary, homogeneous). The first thing to notice about L12 is row 1 is identical to row 2; therefore, we can drop out one of them. They are two identical lines occupying the same space. Then we write the system of equations as a matrix with just the constant coefficients on the right side of L12. These equations equal "A" at L12, and "A" is a column vector equal to zero, with elements equal to zero. I did not bother to augment it to the coefficient matrix because any row manipulations we would do would not change it from zero. The symbols used at L13 have been explained at hub12.1( see link at par. 5). The matrix at L13 is not in echelon form, but it does not have to be. Row 2 has enough information to solve for the variables( x1, x2, and x3). Row 2 is the matrix representation of 4x1 + x2 = 0. So if we plug a value of 1 for x1, we have 4 + x2 = 0. We solve for x2, which is x2 = -4. We plug these values into row 1 and solve for x3 and we get -5--8 + x3 = 0; therefore, x3 = 13. This work was done at L15 and L16. The values for x1, x2 and x3 are plugged into L11 and we then get the final answer at L17.
CHECK YOUR ANSWER AT L17:
8. We use the differential operator as described at hub#12.16(linked) to see if we get the solution given at L1. Hub#12.16 explains the work at L18 through L25, and hub#12.12( linked) explains in detail how to handle these complex roots. So these two hubs( 12.16 and 12.12) brings us to L25 through L27. At L27 we do indeed have the same original function as at L1; however, there may be some confusion at this point; i.e. the complex " i " on the 2nd term of L25. Let's take a look at it.
DOES COS(X) + SIN(X) = COS(X) + i SIN(X)?
9. No not really. If you plug π/4 radians in for x on the left side of the equation, your Ti-89 calculator will give an answer of "approximately"( set your ti-89 in this mode, as well as radian mode) 1.4142, but if you plug π/4 for x into the right side of the equation your ti-89 will give you 0.7071 + i 0.7071. The reason why is the left side is a scalar and the right side is a vector. The presence of "i" tells us the value of sin(x) is the value of the y-coordinate. The right side is a unit vector pointing 45 degrees into the 1st quadrant of the xy plane. The left side can also be considered a vector but it is on the x-axis( the real number axis), and it is ,therefore, a real number, but the right side represents an imaginary number, and with the real part we have a complex number.
SO THEN WHY DOES L25 EQUAL L1?
10. Because they do. How is that for an explanation? I have three textbooks on DEs and they all say the same thing; i.e. plug in "A" for (C1 + C2), and plug in "B" for i(C1-C2) and then call A and B real numbers because they are arbitrary constants. Yap, that is essentially what they said! When I first read that I cocked my head like a dog, wrinkled my nose and said to myself, "You can't just do that!" It turns out you can just do that. C1 and C2, or A and B are arbitrary constants and in this case they can even be imaginary or complex, or you can ignore that they are imaginary or complex; it will not change the outcome. From the theorem at paragraph 11 we can plug in that imaginary term at the right of L25, or even the entire complex number at L25, and either way they will be a solution to the DE at L1. We can even put the " i " in front of C1 at the cosine term at L30, and then go through with the arithmetic from L30 to L36 and it will not change the solution. We will still get the same DE that we got at L36.
Y = (C1y1 + C2y2)----> C1( 0 ) + C2( 0 ) = 0
11. We are still on the above question, Why does L25 equal L1? Professor Wylie gave the derivation for the theorem at paragraph 12, and the derivation ended with C1(0) + C2(0) = 0. We can conclude from this that the arbitrary constant, under these conditions, can be any number including complex. What that means to us in this hub is that L25 does indeed equal L1 because the arbitrary constants are not relevant to the general solution.
AN IMPORTANT THEOREM OF DIFFERENTIAL EQUATIONS:
12. In C. Ray Wylie's quality textbook, Differential Equations, on page 4 he states: "One particularly important property of linear differential equations is given by the following theorem.
If y1 and y2 are two solutions of a homogeneous, linear differential equation, then for all values of the constants c1 and c2 the linear combination y = c1y1 + c2y2 is also a solution of the homogeneous equation."
And in the corollary following Theorem 1, he states,
"If y1, y2, . . . , yk are k solutions of a homogeneous, linear differential equation, then for all values of the constants C1, C2, . . . , Ck the linear combination, y = C1y1 + C2y2 + . . . + Ckyk is also a solution of the homogeneous equation."
13. This theorem is employed at L30 through L37. Notice at L10 in the third row we have y = a1 + a2 . Both of the "a" terms are themselves solutions to the DE at L11. Our objective is to find the values of the "X" coefficients at L11. Remember that the "a" terms are symbolic of the terms at L7.
L30 THROUGH L37 EXPLAINED:
14. The theorem at par. 12 tells us that y1 is a solution to the DE. This hub is about elimination arbitrary constants, and we know that there are two of them, C1 and C2; therefore, we are still going to differentiate y1 twice at L31( 1st derivative), and at L32( 2nd derivative). At the 2nd line of L31 that is supposed to say "1st term is 2a1", and "2nd term is 3a3." Next at L33 we do just as we did above in this hub and just as we did in hub#12.11( see link at par. 2). As usual, all we need are the coefficients of a1 and a3, which we use to set up the matrix at L33. Hub 12.1( see link at par. 5) explains the row reduction at L33, L34 and L35, and at L36 is our final answer, which is the same as at L17. How is that for easier?
IS THERE A REASON FOR NATIONAL STARVATION AND DISEASE?
15. Have you ever asked these questions? Is there a reason for suffering? Does God plague people, and if so why? Does God execute terror, consumption of eyes and sorrow, and if so why? Does God send drought and starvation, and if so why? These questions were answered at paragraphs 7, 8, 9, 10 and 11 within hub#12.22( link is toward the end of this hub). Many books have been written concerning God's mercy, tolerance, understanding, compassion, long-suffering, patience and love. God not only is loving, but Scripture states that God is love( 1 John 4:8), and sending His Son to die in our place( 1 John 4:9 and 10) is one example, of thousands given in Scripture, of God executing and displaying His love. But what about all the Scripture given in hub#12.22. Can we just ignore those verses of God's Word and blithely go along believing God will bless us regardless if we disobey and sin; thereby ignoring God's commandments and warnings? Further answers to the questions in this paragraph can be found in Deuteronomy. My Dake's Annotated Reference Bible has headings for portions of each chapter. In chapter 28 prior to verse one is "Twenty-one blessings of obedience," and prior to verse 15 is "Fifteen curses upon children and material prosperity." Prior to verse 21 is "Thirty curses of sickness, crop failure, war, captivity, business failure, and poverty," and prior to verse 30 is "Twenty-six new and old curses of defeat, captivity, sickness, persecution and insanity." God gives the Israelites His reasons for these curses at verses 45, 46 and 47. Prior to verse 48 is "Twenty-one curses of slavery, death, cannibalism, and extreme poverty and death," and prior to verse 58 is "God again demands obedience: predicts twenty more curses."
GOD, AMERICA AND OUR PROSPERITY:
16. Some people do not believe in God or what His Word teaches us. They have their own standards and morality as "homosexuality is just an alternative lifestyle," and "abortion is a woman's choice," but this is the point of paragraph 15 above, and those mentioned in hub#12.22. When the general citizenry believe their collective subjective morality is the standard rather than God's objective Standard of morality, then He has the right and obligation to execute justice upon that nation. If He did not have a structure in place to handle this problem, then why can't nations and individuals do as they please as do animals in the wild? How much crime do you think we would have if no structure of laws and punishment were not on the books? If there was no deterrent at all, can anyone truly believe there would not be total anarchy? Why should God allow this in His universe? America was once a magnificent country, but we are now slipping down a precipitous slope of moral depravity. We will not get away with it more than anyone else.
Hub#12.22: Inverse matrix of the FS eigenvector matrix
- Finding the Inverse Matrix of the Fibonacci Sequence Eigenvector Matrix:
A message from God begins at par. 7: Poverty, sickness and catastrophe. Is there a reason for suffering? At par.8 Why are there plagues? Par.9 has warnings and promises. Par.10: terror and sorrow--why? Par.11 drought and starvation--why?
Finding the Determinant and Inverse of a 4x4 matrix
Paragraph 14 describes in detail why we cannot take credit for any of our beneficial accomplishments.
- Lagrange Multipliers; extrema, costs, profits
Paragraph 16 of this hub mentions the blessing of God, not only for our minds, but also for our prosperity. If we continue to ignore God's commandments, then there will be no more blessings regardless of how smart we are.
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