# hello i would like some help with laplace transforms.

#### stephan2124

##### New member
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

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
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)