y" + 0.5y' + y = 1-cos(t); y(0) = y'(0) = 0
I used method of undetermined coefficients to get particular solution:
Y(t) = -2sin(t) + 1
To get homogeneous solution, I solved characteristic equation to get complex roots:
r_1,2 = -1/4 +- i*sqrt(15)/4
so homogeneous solution is:
y...
Thanks, I see what the notation means, and your way of visualizing the index pairs of the terms being summed was very helpful too, as I never thought to think about it like that.
Hi, how do I interpret the last sum:
http://planetmath.org/encyclopedia/LagrangesIdentity.html
Sum (...)
1<=k < j <= n
Is it the double sum:
Sum( Sum( (a_k*b_j - a_j*b_k)^2 from k = 1 to n) from j = 2 to n ) ?
Well I guess I get stuck on how to solve the ODE:
(\exp{(-i\omega)} - 2 + \exp{(i\omega)})\overline{u} = \overline{u}_t
So (\exp{(-i\omega)} - 2 + \exp{(i\omega)}) can be treated as a constant with respect to t. But when I solve I get exponents raised to exponents...The answer is suppose to...
Find the solution (in integral form) of the equation:
u(x+1,t) - 2u(x,t) + u(x-1,t) = u_t
u(x,0) = f(x)
Hint: Use the shift formula
F[f(ax-b)] = \frac{\exp{i\omega b/a}}{|a|} \overline{f}(\omega/a)
So I took the Fourier transform of each term using the shift formula:
\exp{(-i\omega)}...