Question

for x" +9x=1, x(0)=x'(0)=0; I get X(s)= 1/(s(s^+9)) How do I solve this to get x=...?

Answer 1

You've found X(s) so far, you need to find the laplace inverse of this to get the solution of the original DE. You need to use partial fractions here as follows:
\[{1 \over s(s^2+9)}={A \over s}+{Bs+C \over s^2+9} \implies A(s^2+9)+(Bs+c)s=1\]
\[\implies s^2(A+B)+cs+9A=1 \implies c=0, A=\frac{1}{9}, \text{ and } B=\frac{-1}{9}.\]
So, \(\large X(s)=\frac{1}{9}(\frac{1}{s}-\frac{s}{s^2+9}) \implies x(t)=\frac{1}{9}(1-\cos(3t)).\)

Answer 2

Anwar has done it for you.

If you want to see another worked example, watch this, beginning around minute 36:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-20-derivative-formulas/

If you're still sketchy on the theory, watch from the beginning.