Recent content by scienceguy288

I have solved the problem. Thanks for the help.

Nevermind...I got C. Still running into some trouble finding A and B, but will continue to work on it. If someone can give me a shove in the right direction, that would make my life that much easier...

It turns out that this is in fact the case. Still stuck, though...

I don't know. Are e1 and e2 standard symbols for the standard basis vectors? Otherwise I think they are just referring to the general vectors {e1, e2}, rather than any specific vector. Perhaps I have to find those vectors first? That is how I have been approaching the problem thusfar, but as...

I can't figure out how to take the first bite out of this one. Homework Statement Let T1: R^2 --> R^2 and T2: R^2 --> R^2 have the indicated properties. Find matrices A, B, and C such that: T2T1x=Ax, T1T2x=Bx, (T1+T2)x=Cx Homework Equations T1e1=(1,3), T1e2=(2,2), T2e1=(-1,1)...

Thanks for helping me, I finally got the right answer. Your advise was truly indispensable!

Alright, so we get p=t+/-Sqrt[tc+t^2], integrating that in terms of t to get x=(t (t + Sqrt[2 c + t^2]) + 2 c Log[2 (t + Sqrt[2 c + t^2])])/2 and x=(t (t - Sqrt[2 c + t^2]) - 2 c Log[2 (t + Sqrt[2 c + t^2])])/2

(p^2)/2-pt=constant?

Well, Integrating both sides will give us p/2(p-2t)+c=(p^2)/2-pt (p^2)/2-2tp/2+c=(p^2)/2-pt ((p^2)/2-pt)+c=(p^2)/2-pt 0=0

Okay, so the left will become p/2(p-2t)+c. The right is the integral of a derivative, so it will become simply (p^2)/2-pt. After simplification this just results in 0=0, though?

Right, so [(y^2)/2)]'=pp'. I still don't see how that is helpful...Sorry for my thickheadedness, but I am just not seeing it...

Nope... I see that [pt]'=p't+p, but I don't see how that can help solve this...

Still stumped. I don't recognize that as anything I have seen before as far as I know...

I am stuck and after several attempts, have made little progress. Homework Statement Solve: (x'-t)x''-x'=0 The Attempt at a Solution I know that the dependent variable x is missing. Therefore, I substitute x'=p and x''=p', giving me, pp'-p't-p=0. However, here I get stuck. If I try to...