Recent content by fusi0n

Thanks. E = (1/2)y'^2 -cos(y) works. The y'^2 allows for an additional y' term that can cancel later!

Homework Statement Consider y'' = - sin(y) find a conserved quantity for this equation Homework Equations This looks an awful lot like a simplified version of a nonlinear pendulum equation The Attempt at a Solution For a conserved quantity I guessed: E = -cos(y) + y' because...

never mind, mathgician, i out the way to find m'(t). your input helped a lot. i will post my solution to the problem if anyone is interested.

this is an inhomogeneous differential equation. the method of variation of parameters is for homogeneous. in the wikipedia article it is said to "determine the homogeneous solution using a method of your choice" as for as the other method is concerned: how can I use this method to derive a...

Homework Statement Suppose that u(t) is a solution to y'' + p(t)y' + q(t)y = 0 Suppose a second solution has the form y(t) = m(t)u(t) where m(t) is an unknown function of t. Derive a first order linear differential equation for m'(t). Suppose y(t) = e^(2t). Use the method...

thank you everyone; I have proven that E is conserved.

Question: How do I differentiate E with respect to t when y and v are dependent upon t? here is my attempt... E = (1/2)(v^2) - (1/2)(y^2) - (1/4)(y^4) where v = (dy/dt) => dE/dt = (dv/dt) - (dy/dt) - (dy/dt) = y'' - 2y'

Homework Statement Given: y'' - y - (y^3) = 0 (equation 1) E = (1/2)(v^2) - (1/2)(y^2) - (1/4)(y^4) (equation 2) v = y' i. Show that E is a conserved quanitity ii. Find all the solutions with E = 0 2. The attempt at a solution I'm not sure how to show a...

should've known that... thanks

I was given an equation: P(f)=(sqrt(2/(Pi*N)))*(exp(-(nf^2)/2)) What math exercise is exp referring to? Thanks a lot for any help.

anybody?

1 cm=10^7 nm. I thought that 1 m^3 =10^27 nm^3... Am I right?

How do I convert grams per centimeter cubed to grams per nanometer cubed? Thank you! :confused:

Why will you see no difference?

If you hold a mirror at arm's length and look at your reflection, what will happen as you begin to run and a speed close to that of light (v=.99c). Will you still be able to see yourself? Will your image look any different?