Recent content by mkay123321

The question is x^2dy/dx + y^2=0 , y(1)=3 I re-arrange the equation to get -1/y^2dy=1/x^2dx Seperated them, then I integrate both sides to get 1/y=-1/x + c Now I don't get how they got the answer y=3x/(4x-3), as when I try use the condition I get a different answer, could anyone help? I...

I did ae^-x - (-ae^-x+b) - 6(ae^-x+bx+c) = e^-x + 12x Then I got e^-x(-4a-1) - b(1+6c) = x(12+6b) After that I did -4a = 1 so a = -1/4 6c=-1 so c = -1/6 6b=-12 so b = -2 Could you tell me where I am going wrong?

I just tried it again. Got the same answer.

So the question is y" - y' - 6y = e^-x + 12x, y(0)=1,y'(0)=-2 First I found the general solution which came out to be, Ae^3x + Be^-2x I then Substituted y=ae^-x + bx + c y'=-ae^-x + b y"=ae^-x Then I just compared the coefficients to get a=-1/4, B=-2 and C=-1/6 So I am getting y =...

Oh yes I see that it is actually easier to check them, thanks a lot for that man! Much appreciated. I just didn't understand the x-y-z=0 properly but now you've clearly explained it to me.

Yes I was getting confused with the 1's and 0's everywhere in the notes. So for L=6 will it be easier to use Gaussian Elimination or simultaneous? I am trying to use G.E but then the matrix goes into fractions and makes it harder. What I got so far is [1 -4 1].

[4 3 1]?

Ha, [2 1 1], okay so it could be [4 2 2] as well and [6 3 3] etc?

[2 -1 -1]?

So [1 -1 -1] ? And you said there were 2 corresponding eigenvectors for L=2, how would I find the other?

Okay, So for L=2 I am getting [1 -1 -1:-4 4 4:1 -1 -1] and when I try using G.E I get [1 -1 -1:0 0 0:0 0 0]. From there x - y -z=0 x=y+z, that's why I thought it was [-1 1 1], I've never done this before so I am not sure if what I am doing is right. I am trying to learn off youtube but I am...

Sorry the matrix is [3 -1 -1:-4 6 4:1 -1 1] and one of the eigenvalue is 6 and the other is 2

Homework Statement So the 3x3 matrix involved is [3 -1 -1:-4 6 4:-1 1 1], The eigenvalues are L=6 and L=2. Homework Equations (A-LI)e=0 The Attempt at a Solution I stuck the eigenvalues into the matrix and got (-1 1 1)(not sure if its right) for L=2 but when I use L=6 in I can't...

Ahh I see now, thanks alot.

I get (2-L)((3-L)(2-L)-2) - 2((2-L)-1) + 2-(3-L) 2-L(4 - 3L - 2L - L^2) + 3L - 3 I tried doing all sorts of stuff to this, just can't get it.