Recent content by jimagnus

oh right, I didn't think to break it up like that. so you end up with 1/s-[(s+1/2)/((s+1/2)2+3/4)+1/2 / ((s+1/2)2+3/4), which becomes 1-e-t/2cos3t/4-√3/3 * e-t/2sin√3/2 thanks!

If I complete the square, I get the denominator is (s+1/2)2+3/4, and the numerator is -(s+1), and I don't know what inverse laplace transform this would be.

IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform Homework Statement Solve x"+x'+x=1, given x(0)=x'(0)=0 Homework Equations The Attempt at a Solution Plugged in transforms: s2*Y(s)-s*y(0)-y'(0)+s*Y(s)-y(0)+Y(s)=1/s Plugged in initial value points, simplified...

For a, I used Xn=n*lamba*Z/a , Xn being a zero. The first zero on my graph was .02, so I plugged in Xn=.02 and n=1.