Question

can someone refresh me on how to do this

find the general solution using laplace transform

y' = e^t ; y(0) = 2

Answer 1

firstly apply laplace transform to both side of equation

Answer 2

here's what i got..
\[y' = e^t\]
\[\implies sy(s) - y'(0) = \frac{1}{s-a}\]
\[\implies sy(s) - 2 = \frac{1}{s-a}\]
\[\implies sy(s) = \frac{1 + 2(s-a)}{s-a}\]
\[\implies y(s) = \frac{1 + 2s - 2a}{s(s-a)}\]
idk if it's right...and even if it is idk how to proceed from there

Answer 3

i think i did something wrong...

Answer 4

? \[y(s)=\frac{1}{s(s-a)}+\frac{2}{s}\]

Answer 5

\[L(y')=sy(s)-y(0)\]

Answer 6

uhh yeah...lemme do this again

Answer 7

and u have a=1

Answer 8

you combined but its much easier to keep them seperate

Answer 9

what do you mean i have a=1

Answer 10

just use partial n the first and the second is more simple

Answer 11

he means a=1 for \[e^{at}=e^{1t}\]

Answer 12

ohh lol yeah...i forgot

Answer 13

\[\frac{1}{s(s-1)}=\frac{A}{s}+\frac{B}{s-1}\]

use coverup method

Answer 14

coverup method?

Answer 15

i am completely clueless about what you're doing now....although im getting the feeling you're getting this more complicated than it is...

Answer 16

cover up method is just another way of partial fractions if you like to expand thats fine also

\[1=(s-1)a+sB

when s=0

-1=A

when s=1

B=1

Answer 17

did you tke the inverse laplace in the last step?

Answer 18

yes

Answer 19

i erased so that you could gt the answer yourself though

Answer 20

you can check it at the end for the initial to make sure it's write

Answer 21

right*

Answer 22

ohhh so you have to take tehe inverse laaplace in the end to get the final answer? always?

Answer 23

yes because you have

\[F(s)\] and you want

\[f(t)\]

Answer 24

ohhh

Answer 25

i didnt know tha..thanks

Answer 26

\[L({f(t)})=F(s)\]
\[L^{-1}(F(s))=f(t)\]

Answer 27

have you taken linear... these are a lot like kernels the chapter about them... my book actually classifies them as kernel (K(s))

Answer 28

i dont believe so