Question

I am doing these course Laplase Transformation and stuck at these question of recitation :
these the link of recitation 2nd question-http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/recitations/MIT18_03S10_rec_18.pdf
And these is the solution link - http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/recitations/MIT18_03S10_rec_18_sol.pdf

I don't understand why f(t)u(t) notation is used and what are the delta(t) & u(t). where they come from?

Answer 1

u(t) is a unit step function. The function can be represented below (H(x) can be replaced with u(t) in this case):

http://www.roymech.co.uk/images/fourier_form_82.gif

The f(t)u(t) notation just means that the function exp(-t)cos(3t) is defined for t > 0. Anything that is t < 0 is equal to 0.

delta(t) is known as a Dirac delta function, or a unit impulse. It is used to represent the discontinuous shift between f'( t < 0 ) = 0 and f'( t > 0 ) = -exp(-t)(sin(3t) + cos(3t)).

Therefore:
f'(t) = delta(t) - u(t)exp(-t)(sin(3t) + cos(3t))
discontinuty representation function only existing after t = 0

Let me know if this helps.

Answer 2

\[f(t)=e ^{-t}\cos{3t} \]
find out f' and then its laplace transform
\[L(e^{-t}\cos3t)=(s+1)/((s+1)^{2} + 3^2)\]\[L(f'(t))=-(s+10)/((s+1)^2+3^2)\]
now
\[L(f'(t))=sF(s)-f(0)\] t derivative rule
f(0)=1 by plugging in 0 in f(t)

now solve sF(s)-1 you will get L(f'(t))
njoy