One of the weird/fun  things about having been a visible face of science on TV is that I sometimes get science and math questions from people I never met.. some are simple .. others are pretty sophisticated..   some are jsut plain wierd..

Last night I got a note from a guy I; ve never met who needed help choosing between two laptops..   Ha !

My favorite came today.. I was busy ‘working’ during a conference call.. when the following window popped up in facebook .. It was from a guy named Charlie in Sydney Australia. whom I’ve never met.  He has a differential equations test tomorrow.. and he needed help on a problem They were studying the Laplace transform .. and all is friends were already asleep..

Here’s the problem..

 My friend wayne and I stared at it a it. got out the Laplace tables and came up with an answer .. (i think..)

I transformed

y” + 16y = f(t)
into

s^2y(s) – sy'(0) + y(0) + 16 y(s) = 0

which simplified to (when using boundry conditions of y'(0) =0 and y(0) = 1 , for the case t> pi

y(s) (s2+16) = 1

so

y(s) = 1/(s^2+16)

which un-transforms (i think) to

1/4 sin(4t)

for the case 0< t < pi, I inverse transformed cos(4t)

i got s^2y(s) -1 + 16y(s) = 1/(s^2 + 16)

which I think simplifies to 4sin(4t) (t/2+1)

i think we need to add constants to make the boundary conditions work out..

so maybe

1/4 sin(4t) + 1 : t> pi
4sin(4t) (t/2+1) +1 : 0 < t < pi

But that doesn’t  look right..
What did  I do wrong ?

I’ll go sleep on it

good luck on your test Charlie.. !

nite all , nite sam
-me