Sorry to be so dense, but I get lost at this point.
I think I am then suppose to do
ut=X(x)[-C(∏/2 + 2m∏)sin(∏/2 + 2m∏)t + D(∏/2 + 2m∏)cos(∏/2 + 2m∏)t)
ut(x,0) = D(∏/2 + 2m∏)cos(∏/2 + 2m∏) = x(1-x) ----> D ≠ 0
t = 0 f(x) = ∑ Dsin (∏/2 + 2m∏)t
u(x,t) = ∑ D sin ((∏/2 + 2m∏)t sin...
Solve the problem.
utt = uxx 0 < x < 1, t > 0
u(x,0) = x, ut(x,0) = x(1-x), u(0,t) = 0, u(1,t) = 1
The Attempt at a Solution
Here is what I have so far but I'm not sure if I am on the right path or not.
u(x,t) = X(x)T(t)...
I need to visualize the wave equation with the following initial conditions:
u(x,0) = -4 + x 4<= x <= 5
6 - x 5 <= x <= 6
du/dt(x,0) = 0
subject to the following boundary conditions:
u|x=0 = 0
I'm not sure I understand the...
Okay now I have the following:
F = ma
F = EA(x)U(x)
= d/dx[EA(x)d/dx)] = m(x) d^2u/dt^2
u(x,t) = U(x)T(t)
d/dx[EA(x) d/dx[U(x)T(t)]] = d^2/dt^2 [m(x)T(t)]
t(t) d/dx [EA(x) d/dx [U(x)T(t)] = 1/T(t) d^2/dt^2 T(t) = -w^2
d/dx[EA(x) d/dx U(x)] + w^2m(x)U(x) = 0
I need to get to
Homework Statement Derive the differential equation governing the longitudinal vibration of a thin cone which has uniform density p, show that it is
1/x/SUP] d/dx(x du/dx) = (1/c) d u/d[SUP]t
Hint: The tensile force sigma = E du/dx where E is the Young's modulus (a constant), u is the...
okay I have been studying my book all weekend and this is what I had.
Laplace Equation (1/r) d/dr(rdu/dr) + (1/r^2)d^2u/ds = 0 (s represents theta)
(1/Rr)d/dr(rdR/dr) + (1/(r^2S))(d^2S/ds^2) = 0
(1/S)d^2S/ds^2 = -w^2 and (r/R)d/dr(rdR/dr)= w^2
Therefore S = A cos ws +...
Show that the solution u(r,theta) of Laplace's equation (nabla^2)*u=0 in the semi-circular region r<a, 0<theta<pi, which vanishes on theta=0 and takes the constant value A on theta=pi and on the curved boundary r=a, is
u(r,theta)=(A/pi)[theta + 2*summation ((r/a)^n*((sin...
Obtain all solutions of the equation partial ^2 u/partial x^2 - partial u/partial y = u of the form u(x,y)=(A cos alpha x + B sin alphax)f(y) where A, B and alpha are constants. Find a solution of the equation for which u=0 when x=0; u=0 when x = pi, u=x when y=1...
Sorry to be a pain but I think I am finally getting the idea and want to make sure this is true. ux = f'(2x+y2)*2 + g'(2x-y2)*2
uy = f'(2x+y2)*2y + g'(2x-y2)*-2y
uxx = f''(2x+y2)*0 + g''(2x-y2)*0
uyy = f''(2x+y2)*2 + g''(2x-y2)*-2
then I just plug these results into the equation y2uxx +...
I am still confused, I do not see where the chain rule comes into play, f(x,y) = 2x + y2. So why isn't fx equal to 2? I do not see how you are coming up with (2x + y2)*2. I desperately need to understand this whole concept! Thanks!
Show that u=f(2x+y^2)+g(2x-y^2) satisfies the equation y^2 d^2u/dx^2 + (1/y) du/dy - d^2u/dy^2=0 where f and g are arbitrary (twice differentiable) functions.
The Attempt at a Solution
I came up with fxx=0 fyy=2 gxx=0 gyy= 2. But didn't...
Please accept my apologies. I had been studying all day and had hit a wall. I stepped away and when I came back, it all made sense. Your assistance was greatly appreciated! Thank you for all your help!