@Mark44
Hmm okay, so the first complex solution I get is ##{i}\choose{1.5}## ##e^{(0.5-1.5i)t}## and the other complex solution I get is ##{i}\choose{-1.5}## ##e^{(0.5+1.5i)t}##.
The general solution to this system can be given by
##{x}\choose{y}## = ##c_1 e^{\lambda_1} v_1 + c_2...
@Mark44 I had thought that just using one eigenvalue would yield the full general solution.
When you say
does that mean I need to work out the x(t) and y(t) for with the other eigenvalue/eigenvector pair, and add the two x(t) s for the general x(t) solution and add the two y(t) s for the...
So far, I have implemented the code below, using Java Priority Queues and Maps. I tried to identify each customer by the time they came in (ranging from 360 and onwards) and their grades. However, this is my first time using priority queues and maps, and I am not very sure if I'm doing things...
From solving the characteristic equations, I got that ##\lambda = 0.5 \pm 1.5i##. Since using either value yields the same answer, let ##\lambda = 0.5 - 1.5i##. Then from solving the system for the eigenvector, I get that the eigenvector is ##{i}\choose{1.5}##. Hence the complex solution is...
@Orodruin
Well, yes, which actually makes b true as well.
Hmm, so. So far I know that
a. -3 is an eigenvalue of A.
b. [4, 2] is an eigenvector of A.
are right.
c. The trajectory of this solution in the phase plane with axes x, y is part of a straight line.
should be right because I...
@Orodruin
I had thought that this was a repeated eigenvalue, since the second constant c2 = 0.
I had thought that the way solutions are formatted is
[x, y] = c_1 e^{##\lambda_1 t##} *eigenvector_1 + c_2 e^{##\lambda_2 t##}*eigenvector_2.
From this, I thought that c_2 = 0 (which led me to the...
Summary:: Suppose that [x, y] = e^{-3t} [-2, -1] is a solution to the system $x' = Ax$, where A is a matrix with constant entries. Which of the following must be true?
a. -3 is an eigenvalue of A.
b. [4, 2] is an eigenvector of A.
c. The trajectory of this solution in the phase plane with axes...
So in the class TJTwitter, I have the following defined:
private Twitter twitter;
private PrintStream consolePrint;
private List<Status> statuses;
private List<String> terms;
private String popularWord;
private int frequencyMax;
and in the same class, I have a few methods that are meant to...
@gneill
oh indeed, yes you are right—it was the sign convention. Then is it usually the case that power produced (which I believe is power coming out) is positive, and the power absorbed (power going in) is negative?
going with the previous idea that i_A = 1 and i_B = 0, I get v_1 = iA*R_1 = 2V and i_2 = 3-1 = 2A and hence v_2 = i_2*R_2 = 2V. And they turn out to be right.
But using the same method, my answers to Q2—2V and 0V for v_1 and v_2—turn out to be wrong... so it does seem like my method itself is...
@BvU
Okay—I take it then that maybe I shouldn't concern myself with R's, such as R_A and R_B?
From the graphs N_A and N_B, at I_S = 1 and V_S = 1, i_A = 1 and i_B = 0... That is however wrong since if I have i_B = 0, I would not have a valid V/I.
How else might I approach this?
We are given that ##i_D = 8\cdot 10^{-12} (e^{v_D/20m} - 1)##
Hence ##i_D' = e^{50 v_D}/2500000000## and ##i_D'' = e^{50 v_D}/50000000##
Then I have that ##\delta i_D \approx\frac{ e^{50 v_D}}{2500000000} \cdot \delta v_D = \delta v_D / 5## Cancelling ##\delta v_D## from boh sides and solving...