Question

DIFFERENTIAL EQUATIONS QUESTION:

Evaluate L{t^2 + 4t - 5}

can i do it like

L{t^2} + L{4t} - L{5}?

Answer 1

I guess the L here denotes the La Place Transform?

\[ \Large \mathcal{L}\lbrace t^2+4t-5\rbrace \]

Answer 2

yup

Answer 3

The La Place Transform is a linear operator.

\[ \Large \mathcal{L}\lbrace c_1f(t)+c_2g(t)\rbrace= \int_0^\infty e^{-st}(c_1f(t)+c_sg(t))dt \\= \Large c_1 \int_'^\infty e^{-st}f(t)+c_2\int_0^\infty e^{-st}g(t)dt \\ =\Large c_1 \mathcal{L}\lbrace f(t)\rbrace + c_2 \mathcal{L} \lbrace g(t)\rbrace \]

Answer 4

therefore, yes you can.

Answer 5

o.O

Answer 6

...so i can right?

Answer 7

L{t^2} = 2/s^3 right?

Answer 8

yes, it's a linear operator and right, if I remember it correctly, I usually use a cheat sheet *laughs*

Answer 9

Let me check.

Answer 10

well i use \[L{t^n} \implies \frac{n!}{s^{n+1}}\]
lol

Answer 11

\[ \Large t^n = \frac{n!}{s^{n+1}} \]

Answer 12

yes, you're correct.

Answer 13

http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf I should save that somewhere.

Answer 14

and L{4t} would be 4\s^2 right?

Answer 15

correct, you can pull the 4 out of the laplace transform and leave it as a scalar identity.

Answer 16

so \[\huge L\left[t^2 + 4n - 5 \right] \implies \frac{2}{s^3} + \frac{4}{s^2} - \frac 5s\]
right?
idk how to do those curly brackets in latex lol

Answer 17

Curly brackets are .\ lbrace and .\ rbrace, the Laplace Transform is written as .\ mathcal{L} if you want to be rigorous about it

Answer 18

And yes, you got it.

Answer 19

sweet

Answer 20

(-;