Homework 5 collaboration area
4.4 Problem 7. Can anyone provide some guidance on finding the Eigenvector for the imaginery eigenvalues in this problem? I am stuck trying to get the matrix to row reduce.
Re: 4.4 Problem 7: First you get the matrix that represents the differential equation: [[0 -2],[8 0]]. Solve the determinant for det(A - Lambda*I) = 0 = Lambda^2 + 16. Then solving for lambda, we obtain purely imaginary eigenvalues. When plugging in the eigenvalues, the trick to row reducing them (to get a bottom row of 0's) is to multiply by "i" and then divide out the constants (be careful to note that i^2 = -1). This should get you the eigenvectors.
4.4 Problem 11. Does anyone have advice as to how this problem should be approached? It isn't a system of equations, so how do we get the eigenvalues/vectors?
Reply: Say that y = y1, and y2 = y1'. Then, you will have a system.
So on this problem, the constant they give is "k".. Coming from an Engineering background I assumed this is a spring constant (i.e. always real). However, should we approach this problem in subcases (assuming "k" is not a spring constant)? In that case would we have to do case 1 = k is real, and case 2 = k is pure imaginary, and case 3 = complex? Or am I over-thinking it?
Reply from Bell: You may assume that k is real. In fact you may assume that it is a positive real number. (It is traditional in science and engineering to use k^2 to represent a strictly positive constant.)--Steve Bell 12:22, 27 September 2010 (UTC)
Reply: Based on the examples we have done in class, I don't think you are over thinking it. However, the book only lists one answer so I think going with the spring constant assumption should be fine (Prof. Bell will want to confirm).
My issue with this problem is I don't understand how to obtain hyperbolas and the given equation. I can get solve for y1 an y2 just fine, but from there I'm stuck. I would assume some kind of substitution would lead to the answer but Im not having any luck.
Reply from Bell: You'll want to eliminate the t variable from the two equations you get for y1 and y2. You might notice that if you add the two equations, you can eliminate, say e^(-kt), and solve for e^(kt) in terms of y1 and y2. Then you could maybe subtract the two equations and eliminate e^(kt) in a similar way. When you plug your expressions for e^(kt) and e^(-kt) back into one of the original equations, you'll get an equation involving only y1 and y2. It will be the equation for a family of hyperbolas.
Reply: Page 143 example 3 and 4 demonstrate how the book generates their solution.
4.5 Problem 1. What do we do about the y2 + y2^2? Specifically the square?
Reply: factor out the y2. Get y2(1+y2). Set the function equal to zero (to find the critical points).
So you should get two solutions(y2 = 0 and y2 = -1). To find the corresponding values of y1, set the second function ( y1 = 0, and solve). The two critical points should come out to be (0,0) and (0,-1).
The next thing you'll need to do is linearize the system at the critical points. The problem is asking for the eigenvalues of the linearized system at the critical points and what they imply about the stability of the solutions near the critical points.
More on P. 150 # 7: Towards the end of Lecture 11 we were given the general solution of a linear system with complex eigenvalues, but I'm a little confused as to why the book does not have c1 or c2 listed in the solution. Any thoughts? Thanks.
Reply: Aren't A and B also constants? Then using a c1 and c2 terms as well seems a bit redundant... -Just my thoughts
4.6 Problem 9: Can you explain the critical points here. I do not understand why we don't get a (0,0) critical point for the y1' term, and I am really stuck on the critical point for y2'.
Reply: The reason (0,0) isn't a critical point is because cos(0) ~= 0. After you turn the ODE into a system you should have x1' = x2 and x2' = -cos(x1). Then solving this system is just like previous problems.
General question: When solving a homogeneous system with complex eigenvalues, we obtain two solutions per eigenvalue. Im confused how we obtain a general solution. In lecture 12, we did an example at the beginning of class and came up with two complex eigenvalues: -1 +/- i. We determined the eigenvector for -1+i and wrote down the general solution. Why did we not analyze the second eigenvalue?
Reply from Bell: If you were to repeat the procedure for the other complex eigenvalue and eigenvector, you'd get exactly the same REAL general solution.
(Actually, the complex solution corresponding to the other eigenvalue is the complex conjugate of the solution you got from the first one. So the real part would be the same as the first real solution you got, and the imaginary part would give you MINUS the second solution you found. But that's ok. That minus sign can be absorbed into the arbitrary constant c_2, and it is clear that the general solution is the same. The moral of the story is: When you find the general solution, stop working.)
Set 4.6 #3: If we try $ \vec{y_p} =\vec{a} t+\vec{b} $ and substitute it in the given system, we'll get a system of 2 equations with 2 unknown vectors ($ \vec{a} $ and $ \vec{b} $). Each vector has 2 components, so that's 4 unknowns. But we only have 2 equations. How do we solve it? To get the same answer as the book gives, we need to set $ a_1 = 0 $ and $ b_2 = 0 $, but why?--Bpavlov 16:22, 26 September 2010 (UTC)
Reply from Bell: When you did the method of undetermined coefficients back in your sophomore year, you would collect the coefficients in front of each power of t, say like so
[...]t + [___] = 0
For that function to be the zero function, the stuff in front of t must be zero and the constant term stuff must be zero. You can do the same thing with vectors in place of scalers.
Let's say you want to find a particular solution to
x' = A x + ct + d
where c and d are constant column vectors and A is a matrix. You try a particular solution of the form
x = at + b
where a and b are constant column vectors. When you plug in and try to make it work, you get
x' = a = A (at + b) + ct + d.
Collect the parts multiplied by t and the parts not multiplied by t to get
[ Aa + c ]t + [ Ab - a + d] = 0
Both of the parts in square brackets must be zero for this equation to hold for all t. Setting the first thing in square brackets equal to zero gets you the a vector. Next, you can plug what you got for the a vector in to the second system you get by setting the second thing in square brackets equal to the zero vector. That will get you b. (Remember, c and d were given.) --Steve Bell 12:35, 27 September 2010 (UTC)
Reply: If you write the problem out you will notice that $ y_2' = -4*y_1 + 5 => b_1 = -4*a_1*t - 4*a_0 + 5 $ therefore $ a_1 $ must be 0. You can then use this to solve the rest of the variables in the two equations.--Whitsonl 14:38, 28 September 2010 (UTC)
Question: What is $ a_0 $ in your reply? Maybe this is why I can't understand it..I thought we only have $ a_1 $ and $ a_2 $. Thank you.--Bpavlov 22:04, 26 September 2010 (UTC)
Question: Oh, ok, I think I got it. You meant $ b_1=-4a_1t-4b_1+5 $. Now, since $ b_1 $ should be a constant (a number), we conclude that there should be no $ t $ in equation for $ b_1 $. Therefore $ a_1 $ must be zero. Then we can do a similar reasoning for $ b_2 $. Is this right?--Bpavlov 11:48, 27 September 2010 (UTC)
