Recent content by mistereko

Brilliant, thanks again.

Just looked at it with fresh eyes. y(x) = 1/x * cos(2ln(x)) - 1/x * sin(2ln(x)) + x

Final solution is e^(-pi/4)*e^(-t)sin(2t) +e^(-t)cos2t + e^t

Your dead right, it's been a very long night.

No need, I just realized a huge mistake I made and figure it out. Thanks anyway :)

2 = c1*e^0*sin(0) + c2*e^0*cos(0) + e^0 that works out nicely to C2 = 1 For C1 it's a bit messier. I let 2Sinh(pi/4) = (e^(pi/2) -1)/ e^(pi/4) Therefore (e^(pi/2) -1)/ e^(pi/4) = C1*exp(-pi/4)*sin(pi/2) + exp(-pi/4)*cos(pi/2) + exp(-pi/4) multiplied by exp(pi/4) and found c1 to...

can anyone verify that the final solution is y(t) = (cos(pi/2) + 2 + exp(pi/2)) e-t sin(2t) + e-t cos(2t) + e^t please?

My general solution is y(t) = C1 e-t cos(2t) + c2 e-t sin(-2t) + e^t, but I don't know how to chnage the boundary values from x to t. Having a slow moment.

The first one you wrote. Cheers.

It's definitely a linear ODE right?

Homework Statement I've got y'' - ω2y = sin(ωx) + sinh(ωx) where y(a) = A, y(b) = B Homework Equations The Attempt at a Solution Yc = C1 Sinh(ωx) + C2 Cosh(ωx) and I got my Yp to be -1/2*sin(ωx) + 1/2*sinh(ωx) I'm not sure about getting the Y general. Any pointers...

Homework Statement How do I get y'' - yω2 = y'(sinωx + sinhωx) + y(cosωx*ω + coshωx*ω) equal to y'' - yω2 = sinωx + sinhωx I'm baffled. Homework Equations The Attempt at a Solution

Anyone?

Homework Statement S[y] = \int21dx ln(1 + xny'), y(1) = 1, y(2) = 21-n where n > 1 is a constant integer, and y is a continuously differentiable function for 1 ≤ x ≤ 2. Let h be a continuously differentiable function for 1 ≤ x ≤ 2 and ε a constant. Let ∆ = S[y + εh] − S[y]. Show that...

That gives me Aa^4exp(ax) - Aa^4exp(ax) = exp(ax) :(