Question

use laplace transformation to solve the given initial value problem y''-2y'+2y=e^-t

Answer 1

@dan815

Answer 2

whats the initial value ?

Answer 3

u take laplace transform of the left and right

Answer 4

and u get to solve the f(t) given the inverse laplace transform

Answer 5

L(f')=sL(F)-f(0)
L(f'')=sL(f')-f'(0)
=s*(sL(F)-f(0))-f'(o)

Answer 6

ok i derived another equ

Answer 7

L(y'')=s^2L(y) -sy(0)-y'(0)

Answer 8

L{y^n}=s^nL(y)-s^(n-1)y(0)-s^(n-2)*y'(0).... and so on

Answer 9

initial values are y(0)=0 y'(0)=1 sorry for the delay

Answer 10

if n is the nth differential of y

Answer 11

its ok dont let it happen again

Answer 12

y''-2y'+2y=e^-t
s^2Y-sy(0)-y'(0)-2sY-y(0)=1/(s+1)
(s^2-2s)*Y-s*0-1-0=1/(s+1)

Answer 13

Y=(1/(s+1) +1) * 1/(s^2-2s)

Answer 14

solve or Y

Answer 15

and then inverse laplace t

Answer 16

make Y look nicer so u can apply transform

Answer 17

notice the capital Y and small y , Y is the L{y}

Answer 18

for start i wrote s^2Y(s)-sY(0)-Y'(0)-2[sY(s)-y(0)]+2Y(s)=1/s+1

Answer 19

oh theres a 2 ys oops :)

Answer 20

okay so all that in

Answer 21

yeap you missed that ;p

Answer 22

i don't know you to continue

Answer 23

(s^2-2s+2)Y-1=1/(s+1)

Answer 24

agree

Answer 25

okay so

Answer 26

Y=?

Answer 27

u will neeed to do partial practions

Answer 28

that bracket is not nice too umm

Answer 29

u will hav to make its a perfect square form then

Answer 30

and apply the shift formula

Answer 31

yes that not good that's why i could not finish it

Answer 32

(s^2-2s+2)Y=1/(s+1) +1
((s-1)^2+2-1)Y=1/(s+1) +1

Answer 33

now its a nice form

Answer 34

yes now it's better

Answer 35

where s-1 is the variable basically

Answer 36

or in other words e^at shift factor

Answer 37

now do partial fraction decomposistion

Answer 38

((s-1)^2+2-1)Y=1/(s+1) +1
Y=(1/(s+1) +1)*1/((s-1)^2+1)

Answer 39

rewrite
1/((s+1)((s-1)^2+1)))=A/(s+1)+(Bx+C)/(s-1)^2+1)

Answer 40

and u will have a 1/((s-1)^2+1) byitself too

Answer 41

that u already knw just see how the Bx+c part works out

Answer 42

ok thank you very much

Answer 43

so i will have 1=A(s-1)^2+(BX+C(s+1)+(s-1)^2?