A times (-1,-1,1) i (incorrectly) did:
1x-1 + 0x-1 + 2x-1
0x-1 + 1x-1 + 2x-1
2x 1 + 2x 1 + 3x 1
giving
-1-2
-1-2
2+2+3
resulting in (-3,-3,7)
which is incorrect.
i know how to do it now, thanks ehild and lanedance.
shouldve remembered this from years ago...
ok i was multiplying column of matrix by each row of the vector, when i should be multiplying the rows of the matrix by the vector and adding terms.
my bad.
eigenval/vect question, double check my answers, dont make sense.............
Homework Statement
find eigenvalues and eigenvectors of matrix A =
102
012
223
Homework Equations
A -λI etc blah
The Attempt at a Solution
I'm getting λ = 1, -1, 5.
For λ = 1, v = (-1,1,0)t (t...
ok found some stuff in my notes.
i think the du/dx (0,y) = 0 line tells you that something is insulated.
i've done some question in my book (obviously lecturer has done it and i was merely copying it, lol), and i have written down:
uxx + uyy = 0
u(0,y) = 0
u(a,y) = 0
u(x,0) = x (maintain...
ehh my head's not working, i should go to bed soon.
question 4. no clue.
H and L are just arbitrary points on the y and x axis respectively, i presume..
what do each of those initial constraints mean, please?
and how do i know what is insulated?
ughh i really have no idea what to do now.
also i've wasted too much time on one question.
gonna try to do the PDE one.
you said
i'll give this a shot and post up working out as i go.
this is my working out:
http://i.imgur.com/1hsQS.jpg
i sort of figured out how to do this a few mins ago lol. it doesnt seem too hard.
it's sort of like... multiplying the first number in the matrix A by it's position in the matrix (x1 * x1) which is basically the coordinates of the value...
im not used to tex but i'll keep that in mind.
yes my f is in that form.
so i have
a_0 = \frac{1}{2}
b_n = \frac{1-(-1)^{n}}{n*Pi}
would i just let c_0 and d_n equal a_0 and b_n respectively?
http://i.imgur.com/0MiDc.jpg
my result as a sum of infinite series at the top, the rest is getting the first 4 non-zero terms.
and im not following you from your second sentence onwards...
yeah that too!
ok particular solution, this one i'd have to guess depending on what f(x) is (which i've never learnt how to do lol). method of undetermined coefficients, i think it is.
my guess, something related to a trig function because we have a couple of sine's in f(x)?
so according to...
ok to avoid confusion..
question: http://i.imgur.com/FfowZ.jpg
3. a.
b.
c.
so now do i say
(A*cos\sqrt{10}*x + B*sin\sqrt{10}*x) = 2 + \frac{2}{Pi}*sin(x) + \frac{2}{Pi}*sin(3x)
?
edit: ugh the LHS is just y, so i'd have to make y" + 10y equal to f(x)
i think the period would be root 10 or something?
amplitiude has something to do with the A and B i think...
edit: oh, well im not sure how to graph that. i can do them separately lol (2, 2/pi sinx, 2/pi sin3x) but not together..
yess i was just going to say, it's in the other thread lol.
how's my solutions from that thread?
and are you talking about sketching y = (A*cos√10*x + B*sin√10*x)? because i wouldnt know how to do that..
oh, that second quote, i meant if the root is in the form λ = a ± bi. but a=0 in this case, so e^0x is just e^0, which is 1.
working off this:
no alpha in our roots, only beta.
ok so ive worked out a0, an, and bn again, and im getting
a0 = 1/2,
an = 0,
bn = \frac{1 - (-1)n}{n*pi}
erm it isnt showing up right for me..
bn = [1 - (-1)n] / (n*pi)
is this right?
and then i sub all into f(x)?
ok so after consulting maths notes from 2 years ago, the roots are lambda = +- root (-10)
so y = (A*cos\sqrt{10}*x + B*sin\sqrt{10}*x)
yesh?
supposed to be e ^ alpha*x(Acosblah + Bcosblah) but alpha = 0, so the e at the beginning becomes 1
complimentary solution.. do you mean the λ2 + 10λ = 0, λ = 0 and -10 line?
or yhomogenous= A + Be-10x?
both look okay, unless theres something really stupid im doing...
for the first one (c) would i just first do the left hand side.
turn that into λ2 + 10λ = 0, solve it, so then λ = 0 and -10
so then
yh = Ae0x + Be-10x
= A + Be-10x
and so now f(x) = A + Be-10x and work this out as a fourier question, but we have A and B and i dont know what to do wtih them...
ohh the bolded bit i've seen before but i never actually understood what the lecturer was doing.
i think i'll understand this better if i do the question again but without halving the coefficients lol.
i basically dont know how to do pde's, so i'm learning it from scratch today for my test which is tomorrow (which is in 24hrs from now, for those who dont live in australia), and notes/the internet arent nearly as good as explaining things as people are.
so how would i go by starting these...
firstly, thank you so much for this detailed response!
much much much appreciated and i couldn't've asked for a better response :)
yeah for some reason i thought if i split up the integral i have to half the coefficient -_-
but thats not required.
ughhh silly mistake, i did another...