Jake Shellman

(symplectic stuff)

In particular, could someone help me understand lemma 2?

Hey, I could really use some help understanding this post: math.stackexchange.com/questions/417365/…

Hey, I'm trying to show that X = Y a.s. => X =Y in law. Is this argument correct? Since X = Y a.s., put S = {w : X(w) != Y(w)}, and observe that P(S) = 0. and P(S^c) = 1. Hence, for a borel set B, P(X^{-1}(B)) = P(X^{-1}(B) \cap S) + P(X^{-1}(B) \cap S^c) = P(X^{-1}(B) \cap S^c) = P(Y^{-1}(B) \cap S^c) = P(Y^{-1}(B) \cap S^c) + P(Y^{-1}(B) \cap S) = P(Y^{-1}(B))

I still dont get it lol

But how do you mathematically solve for an eigenvector?

Wait dont you solve eigenvectors by doing (A-rI)x = 0, and x is the eigenvector for the eigenvalue r?

A - 1I = 0, 0, 0, 3

Yah...I know its diagonal...but I have to show it using the characteristic equation not the short cut

@AlecTeal I keep getting 0,0 when I use them though

@Alec

I already did....the eigenvals are 1, 4

?

Can someone help me find the eigenvectors for the 2x2 matrix ( (1 0) (0 4)) ?

but now what

s^2Y(s) + 2sY(s) + 5Y(s) = G(s)

@Mike we need to use the Laplace Transform

oops forgot to say that

using Laplace transform

pleaseseseses

Hey guys....can you help me solve: y'' + 2y' + 5y = g(t) ; y'(0) = y(0) = 0 (generally for any function, g(t).)

Hey guys....can you help me solve: y'' + 2y' + 5y = g(t) ; y'(0) = y(0) = 0 (generally for any function, g(t).)

Okay...can anyone help ?

I can't use reduction of order

@IanMateus I think I am supposed to use the Laplace Transform somehow....

?

Hey guys....can you help me solve: y'' + 2y' + 5y = g(t) ; y'(0) = y(0) = 0 (generally for any function, g(t).)

@usukidoll stability is determined by the value of y' for values of t less than or greater than the equilibria. So, if y' tends to be positive and negative (toward the equilibrium) it is stable, but if they point away it is unstable (in terms of the phase line)...if they both point in the same direction, it is termed semistable

@robjohn So then you could look at the parameter a and since $y' = y^2 + ay + 1$, you could say that if a = 2 or a = -2, there is only one equilibrium solution (plus or minus 1) and if a is between -2 and 2, then there are no equilibria...and if a is less than -2 or greater than 2, there are two equilibria

@robjohn

which are by definition equilibrium solutions right?

@usukidoll @robjohn so then, just use quadratic formula to find the equilibria, no?

@robjohn you don't need to solve the ODE to figure out equilibria right?

well, you don't need to solve the ODE to figure out equilibria right?

these are the equilibria right?

y = (-a +/- sqrt(a^2 - 4)/2

So....I think you just generally solve using quadratic equation right?

@robjohn nope

nope

Wouldn't you just set the autonomous equation equal to zero, then use quadratic formula to solve for y?

The question asks to determine equilibrium solutions based on the parameter a

y' = y^2 + ay + 1

sorry..:) here is the problem:

hmmm....can someone help me with this diff eq problem?

yup

thats sad

lol