Recent content by xVladx

Thanks very much for you help! :)

Hmm mistyped my Latex, I ment: y(t) = \left\{\begin{array}{cc}\frac{1}{16}-\frac{\cos(4t)}{16}&t<{\pi}\\0 & t\geq{\pi}\end{array}\right]

OK guess I am stuck then.

Would my answer then be y(t) = \left\{\begin{array}{cc}\frac{1}{16}-\frac{1}{64\sin(4t)}&t<{\pi}\\0 & t\geq{\pi}\end{array}\right]

Should I have gotten Laplace (1-u(t-\pi)) = \LARGE\frac{1}{s}-\frac{e^{-\pi s}}{s}

Nevermind this I found out about the Latex thing. Just going to restate the problem to give it a try \ddot{y} + 16y = f(t) = \left\{\begin{array}{cc}1&t<{\pi}\\0 & t\geq{\pi}\end{array}\right] with y(0) = 0 and \dot{y}(0) = 0

Also I've seen in other posts that some people can post the equation writing similar to the way it is done in MS word, how do you do that? would be much neater then

Take Laplace of both sides: s^2Y(s) -sy(0)-y'(0) +16Y= 1/s - e^-pi*s y(0) = y'(0) = 0 Y(s^2+16)=1/s - e^-pi*s Y = [1/s(s^2+16)] - e^-pi*s/(s^2+16) Using partial fractions [1/s(s^2+16)] = 1/16s - 1/16(s^2+4^2) Y(s) = (1/16s)-(1/(16(s^2+4^2)))-(e^(-pi*s)/(s^2+4^2))

Homework Statement Hi all came across this problem whilst doing some revision and i can't work out the answer Solve the following equation with laplace transformation Homework Equations y''+16y = f(t) = { 1 t < pi ] with y(0) = 0 and y'(0) = 0 _____________{ 0 t >= pi ]...