Recent content by pki15

Set y=x, and see what the limit is when x->0. Then try setting y=0, and see what the limit is as x->0. This is the idea for proving any limit in multiple variables does not exist, just go along different lines, if you get different answers, the limit does not exist.

Well you should be able to solve this equation as usual. (Basically ignoring the imaginary bit) y'' -iay' + by=0 r^2-iar+b=0 Now use the quadratic equation... r = \frac{ia \pm \sqrt{(-ia)^2-4b}}{2} = \frac{ia \pm \sqrt{a^2-4b}}{2} Let the two solutions be A,B. Then y = c_1 e^{At} + c_2...

You should get 3 equations: 12a + 2b = 0 (from 2nd derivative) 12a + 4b = -12 (from 1st derivative) 8a + 4b + c = -11 (from original equation) Just solve for a,b,c from that info... I think you have all the right ideas, just getting wrong values.

1st solve dy/dt = dx/dt. You should then be able to solve for y. Next you could plug y into dx/dt = -xy to get a seperable equation. Then solve for x & y. Be careful with your constants!

Begin with a formula for the distance(D) from (x,y) to the origin. Then you can plug in what you know about y. Finally differentiate (implicitly) the formula with respect to time. You should get an equation involving dD/dt, x, and dx/dt. You know x, and dx/dt. Find dD/dt.

Note that 1-2x-x^2 = -(x+1)^2 + 2. Now separate, and integrate. This is completing the square. Also you could multiply the bottom out and long divide to get a similar result.

You're correct, now multiply out and integrate! :smile:

Okay, suppose G is non-abelian. Let a,b \in G such that ab \not = ba. Then \phi (x) = axa^{-1} is the inner automorphism, and \phi(b) = aba^{-1} \not = b. So \phi is a nontrivial automorphism. Now if G is abelian, G = Z_{p_1^{n_1}} \oplus Z_{p_2^{n_2}} \oplus \dots \oplus Z_{p_k^{n_k}} by the...