Question

laplace ivp x" + 8x' + 15x = 0 , x(0) = 2, x'(0) = -3

Answer 1

L(x" + 8x' + 15x) = L(0)

Answer 2

L(0) = 0

Answer 3

show what u tried on left side...

Answer 4

so L(x" + 8x' + 15x) = 0

Answer 5

so L(x") + 8L(x') + 15L(x) = 0

Answer 6

yup. and you know
L(x'')

Answer 7

L(x") = s^2X(s) - sx(0) - x'(0)

Answer 8

go on....put initial vales....do L(x') simultaneously.

Answer 9

what is L(x')

Answer 10

L[x'] = s L(x) -x(0)

Answer 11

L(x") + 8L(x') + 15L(x) = 0
s^2X(s) - sx(0) - x'(0) + 8( s L(x) -x(0) ) + 15L(x) = 0
put the initial values
and isolate L(x)

Answer 12

i did that but i cant get the answer they have in the back of the book

Answer 13

show your work here.

Answer 14

X(s) = 2s + 13/(s^2 + 8s + 15) where X(s) = L(x)

Answer 15

s^2X(s) - 2s +3 + 8 s L(x) - 16 + 15L(x) = 0
thats correct.

Answer 16

use partial fraction, denominator can be factorized.

Answer 17

answer is 1/2(7e^(-3t) - 3e^(-5t))

Answer 18

okay so X(s) = 2s + 13/(s+5)(s+3)

Answer 19

X(s) = (2s + 13)/(s^2 + 8s + 15)
= A / (s+5) + B /(s+3)

Answer 20

find A,B

Answer 21

okay so A/(s+5) + B/(s+3) now i multiply A/(s+5) by what

Answer 22

u don't know partial fractions ?

Answer 23

2s+13 = A(s+3) + B(x+5)

Answer 24

ist been awhile

Answer 25

to get A , put x= -5

Answer 26

to get B , put x= -3

Answer 27

did u get how i got
2s+13 = A(s+3) + B(x+5)

Answer 28

2s + 13/(s+5)(s+3) = A/(s+5) + B/(s+3)

Answer 29

now i cant remember what to do

Answer 30

(2s + 13)/(s+5)(s+3) = A/(s+5) + B/(s+3)
= [A(s+3)+B(s+5)] / (s+5)(s+3)
so 2s+13 = A(s+3) + B(x+5)
did u get this ?

Answer 31

i dont understand

Answer 32

which part ?
did u get
(2s + 13)/(s+5)(s+3) = A/(s+5) + B/(s+3)
= [A(s+3)+B(s+5)] / (s+5)(s+3)
just cross-,ultiplied.

Answer 33

*multiplied

Answer 34

how you get A(s+3) + B(s+5)

Answer 35

ohhhh so you multiply both sides by (s+5)(s+3)?

Answer 36

i wanted to make the common denominator as (s+5)(s+3)
below A there was only s+5
so i multiplied and divided by s+3
A(s+3) / (s+5)(s+3)
ok ?

Answer 37

yeah.

Answer 38

it took me awhile but i got it haha

Answer 39

good, so can u find A,B?

Answer 40

to get A , put x= -5

Answer 41

once, u find A,B .
just take inverse L.T
you'll get the answer.

Answer 42

wait i have 2s + 13 = (A+B)s + 3A + 5B

Answer 43

keep it in this form only
2s+13 = A(s+3) + B(s+5)
put s= -5
you'll get A

Answer 44

WHY DOES s = -5?

Answer 45

to find the value of A
that equation is satisfied for any value of s
so i choose s=-5
so that we eliminate B directly.

Answer 46

so to find B i plug in s = -3

Answer 47

absolutely.

Answer 48

ive never done it like that before hmm

Answer 49

i got to go
u find A,B .
just take inverse L.T
you'll get the answer.

Answer 50

okay

Answer 51

i didnt get it but oh well

Answer 52

2s+13 = A(s+3) + B(s+5)
put s= -5
3=A(-2)
A = -3/2

2s+13 = A(s+3) + B(s+5)
put s= -3
7=B(2)
B= 7/2

(2s + 13)/(s+5)(s+3) = A/(s+5) + B/(s+3)
=(-3/2) / (s+5) + 7/2 (s+3)
taking inverse laplace
ivp = (-3/2) e^(-5t) + (7/2) e^(-3t)
which is the given answer.

Answer 53

@amaes