Recent content by epsilonzero

Thanks, that helped me out a lot.

My bad. My last post was referring to a different problem. Here's the other problem I'm working on: R' = 0 J' = aR + bJ {{0, 0},{a,b}} L^2 - bL L1 = (b+sqrt(b^2))/2 = b L2 = (b-b)2/ = 0 e1 = {{b},{L-a}} = {{0},{b}} e2 = {{L-d},{c}} = {{-b},{a}} Now how do I go from those...

Another question related to this problem. For the eigenvectors I got (L=lambda) e1 = {{b},{L-a}} = {{0},{b}} e2 = {{L-d},{c}} = {{-b},{a}} Now how do I use those eigenvectors to sketch a phase portrait of the system? My book doesn't explain it well. Thanks.

Good call. I have no idea how I got (a+b) when it should be -ab.

Homework Statement R' = aJ J' = bR What happens to the graphs of R(t) and J(t)? The Attempt at a Solution I made the matrix {{0, a}{b, 0}} and then got the equation (L=lambda) L^2 - 0L + (a+b) after computing the trace and determinant of that matrix. I then solved for the...

I got t=(ln(-A/B))/(b-a) So I just say that's at most 1 root if either A or B is negative? For the second part I have x(0) = Ae^(a*0)+Be^(b*0) = A + B x'(0) = Aae^(a*0)+Bbe^(b*0) = Aa + Bb I know 0=Ae^(a*t)+Be^(b*t) so Ae^(at)=-Be^(bt) a,b<0 and t>0 Now I'm stuck at what to do next. I...

I don't see how A + Be^{(b-a)t} = 0 would tell me that there is at most one zero. Could you elaborate a little? Also any ideas on part b? Thanks

Homework Statement Consider x = Ae^{at} + Be^{bt} a) Show this can have at most one zero. b) Show that if x(t)=0 for t>0, then either x(0)>0 and x'(0)<0; or x(0)<0 and x'(0)>0 The Attempt at a Solution a) x = Ae^{at} + Be^{bt} 0 = Ae^{at} + Be^{bt} 0 = ln(Ae^{at} + Be^{bt}} 0 = ln(A)...

Ok, just got it. Thanks.

Homework Statement Consider ax''+bx'+cx=0 for b^2-4ac=0 with k= -b/(2a). Show x=e^(kt) is a solution. The Attempt at a Solution x=e^(kt) x'=(-b/2a)e^((-b/2a)t) x''=(-b^2/4a^2)e^((-b/2a)t) Then i plugged these into ax''+bx'+cx=0 and simplified to get e^((-b/2a)t)(-((b^2)/2)+c)=0...

I'm still at this step: x=(t(k-1)+C)^(1/(1-k)) and I want to know at what rate x(t) goes to 0 for different intervals of k. I'm not sure how to figure out what these intervals should be or how to calculate the rate. I think for rate there's an equation (x(t)-a)/(e^mt) as t->infinity where m...

what about this start out with the true statement: 0=0 using additive inverse: 0+(-0)=0 which is an additive identity for: -0=0

I just copied and pasted so yeah, it seems like there were a few problems. The x should have been an x'. Also I think the question is asking what rate does x(t)->0. What I've done so far is x'=-x^k (-x^-k)dx=-dt -Int(x^-k)dx=Int(1)dt 1/(-k+1)*x^(-k+1)=t+C Where do I go from here to...

Hey, I'm having trouble starting this assignment. If you could just tell me how to get started and the path I should be on then that would be great and I can do the rest. Here are the questions: 1) Solutions to x= - x^k for x0>0 satisfy x(t) goes to 0 as t goes to infinity. Characterize the...