MHB Hello i would like some help with laplace transforms.

[stephan2124]

hello if someone could please tell me if i am incorrect and where , and how to type it into a math program so it can understand it many thanks stephan2124

L -3e^{9t}+9 sin(9t)
L-3e^{9t}+L 9 sin (9t)
-3 Le^{9t}+9 L sin(9t)
-3 (1/s-9) +9 (9/(s^2+9^2))
-3 (1/s-9) +9 (9/(s^2+81))

into a math program
-3*(1/(s-9)+9*(9/s exp +81)

 

[HallsofIvy]

We have no way of knowing what your math program will accept as an answer!

Yes, the Laplace transform of -3e^{9t}+ 9 sin(9t) is -3 times the Laplace transform of e^{9t} plus 9 times the Laplace transform if sin(9t). Yes, the Laplace transform of e^{9t} is \frac{1}{s- 9} and the Laplace transform of sin(9t) is \frac{1}{s^2+ 9^2}= \frac{1}{s^2+ 81} so the Laplace transform of

-3e^{9t}+ 9 sin(9t) is \frac{-3}{s- 9}+ \frac{9}{s^2+ 81}.

Now, if your math program does not accept that, perhaps it wants the two fractions added: \frac{-3(s^2+ 81)}{(s- 9)(s^2+ 81)}+ \frac{9(s- 9)}{(s- 9)(s^2+ 81)}= \frac{-3s^2- 243}{(s- 9)(s^2+ 81)}+ \frac{9s- 81}{(s- 9)(s^2+ 81)}= \frac{-3s^2+ 9s- 324}{(s- 9)(s^2+ 81)}= \frac{-3s^2+ 9s- 324}{s^3- 9s^2+ 81s- 729}= -\frac{3s^2- 9s+ 324}{s^3- 9s^2+ 81s- 729}.

Any one of those is a correct answer. ​

 

[Prove It]

the Laplace transform of sin(9t) is \frac{1}{s^2+ 9^2}= \frac{1}{s^2+ 81}

Actually it's $\displaystyle \frac{9}{s^2 + 81} $.

so the Laplace transform of

-3e^{9t}+ 9 sin(9t) is \frac{-3}{s- 9}+ \frac{9}{s^2+ 81}.​

Actually it's $\displaystyle -\frac{3}{s - 9} + \frac{81}{s^2 + 81}$.To enter it in Weblearn you will need to write:

-3/(s - 9) + 81/(s^2 + 81)​