Thanks for you response. I feel as if I made this problem more complicated than it should.
For ∂f/∂x:
∂f/∂x = 6xy = 6x(3x4 - 2x +1)
and df/dx:
df/dx = ∂f/∂y * dy/dx
= d/dy(3x2y + 2yy') * (12x3 - 2)
= (3x2 + 2(d/dy(yy'))) * (12x3 - 2)
= (3x2 + 2(y * d/dy(y') + y')) * (12x3 -...
Homework Statement
Consider the function f(y,y',x) = 2yy' + 3x2y where y(x) = 3x4 - 2x +1. Compute ∂f/∂x and df/dx. Write both solutions of the variable x only.
Homework Equations
Euler Equation: ∂f/∂y - d/dx * ∂f/∂y' = 0
The Attempt at a Solution
Would I first just find...
After correcting my derivative I got,
x' = (-Aβ - Btβ + B)e-βt
Then using my initial conditions v = 0 at t = 0,
0 = (-Aβ + B)(1)
→ B = Aβ
so, x(t) = (A+Bt)e-βt where B = Aβ
and then I can use A = A/2 when t = 1s correct?
Thanks for you reply!
x' = (-Aβ - Btβ + β)e-βt
The velocity is 0 at t=0 so,
0 = (-Aβ + β)(1)
β=0 and A = 1
Would I then go on to use t = 1 sec at A0 = A0/2
where A0 = A+βt, the initial amplitude, or is A0 equal to the value I just solved for, A = 1?
Homework Statement
The shock absorber of a car has elastic constant k. When the car is empty, the design is such that the absorber is critically damped. At time t=0 the absorber is compressed by an amount A from its equilibrium position and released.
a) If after 1 second the absorber...
My apologies on the late reply guys. It was a late night and I ended up finishing the problem the following morning. Thanks for your help!
For my equation of motion I used x(t) = Ae-βtcos(ω1t) = Ae-βmTcos(ω1mT)
where ω1 = √(ω02-β2) (and yes my original formula was missing a square root.)...
Homework Statement
The amplitude of an underdamped oscillator decreases to 1/e of its initial value
after m complete oscillations. Find an approximate value for the ratio ω/ω0.Homework Equations
x''+2βx'+ω02x = 0 where β=b/2m and ω0=√(k/m)
x(t) = Ae-βtcos(ω1t-δ) where ω1 has been defined as...