Question

método de Laplace y''-6y '9 y = ty (0) = 0 y' (0) = 1

Answer 1

work the method, where are you stuck?

Answer 2

also, fix your notation ...

Answer 3

wait can u even speak english @marquito94

Answer 4

Laplace works in other countries so i doubt english is needed

Answer 5

@amistre64 true

Answer 6

@amistre64 but it is necessary when the Asker needs more explanation or makes question on the problem. For example, if I don't understand your question "Where are you stuck", how can I answer that question?

Answer 7

gogle translate to the rescue!

Answer 8

but he is is not speaking english he is speaking spanish i speak spanish

Answer 9

@hartnn a lot of fun when using google translate

Answer 10

Yeah, google translate is off for so many things. I know this because I know other languages too.

Answer 11

metodo seems more italian to me than spanish

Answer 12

either way, the notation is off, something is missing in the equation.

Answer 13

between 6y' and 9y
whether it is + or -

Answer 14

\( \Large y''-6y ' \pm 9 y = t \\ \Large y (0) = 0 , ~ y' (0) = 1\)

Answer 15

\[L\{f''(t)\}=s^2L\{f'(t)\}-f'(0)-f(0)\]
sounds familiar

Answer 16

might be missing an s in there somplace lol

Answer 17

\(\Large s^2 Y - s\times 0 -1 - 6sY +6(0) \pm 9Y = \dfrac{1}{s^2} \)

Answer 18

yeah, its
\(\Large s^2Y(s) -sy(0) -y'(0)\)

Answer 19

:) thats the one ...

Answer 20

\(\Large Y (s^2 -6y \pm 9) = \dfrac{1}{s^2} +1\)

Answer 21

\(\Large y =L^{-1}[\dfrac{1+s^2}{(s^2+1)(s^2-6y \pm 9 )} ] \)

i suspect it is +9

that would make it (s-3)^2

Answer 22

\(\Large y =L^{-1}[\dfrac{1+s^2}{(s^2+1)(s^2-6y + 9 )} ]\)
\(\Large y =L^{-1}[\dfrac{1+s^2}{(s^2+1)(s+3)^2} ]\)

partial fractions

Answer 23

****\(\Large y =L^{-1}[\dfrac{1+s^2}{(s^2+1)(s-3)^2} ]\)

Answer 24

after partial fractions, its just looking up formulas in the laplace table