Question

can you solve this for me?
inverse laplace transform:
f(s) = 10/s(s+2)(s+3)^2

Answer 1

do you know partial fractions ?

Answer 2

A-Level Pure Math 3

Answer 3

@hartnn I know a little bit

Answer 4

@edwinls9 what do you mean?

Answer 5

\(\Large \dfrac{10}{s(s+2)(s+3)^2}=\dfrac{A}{s}+\dfrac{B}{s+2}+\dfrac{C}{s+3}+\dfrac{D}{(s+3)^2}\)

can you find A,B,C,D ??

Answer 6

10 = A (s+2)(s+3)^2 + Bs (s+3)^2 + Cs(s+2)(s+3)+Ds(s+2)
That is true for every value of s

put s=0,
10 = A(2)(3)^2 =....
Find A

put s = -2
10 = B(-2)(1)^2 =....
Find B

put s= -3
10 = D(-3)(-1)=....
Find D

put s= -1
10 = A(1)(2)^2 + B(-1)(2)^2 +C(-1)(1)(2)+D(-1)(1)
Find C

@fariza

Answer 7

did you try finding A,B,C, D ??
what values you got ?

Answer 8

I'm not finding it yet cuz I don't know partial fractions :/

Answer 9

http://www.mathsisfun.com/algebra/partial-fractions.html

Answer 10

@undeadknight26 I'm studying this note and trying to understand .

Answer 11

ok, are you clear till here,
\(\Large \dfrac{10}{s(s+2)(s+3)^2}=\dfrac{A}{s}+\dfrac{B}{s+2}+\dfrac{C}{s+3}+\dfrac{D}{(s+3)^2}\)
?

Answer 12

yes @hartnn

Answer 13

now do you know how to make a common denominator on right side ?

Answer 14

no

Answer 15

ok, lets take example of only 2 fraction,

say,
a/b + c/d
how would you simplify that ? nake a common denominator

Answer 16

ad+cb/bd
true?

Answer 17

yes
so, how about 3 fractions ?

a/b +c/d + e/f = ... ?

Answer 18

common denominator will be bdf
right ?
what about numerator ?

Answer 19

I don't know

Answer 20

okk, lets go other way round,
\(\Large \dfrac{10}{s(s+2)(s+3)^2}=\dfrac{A}{s}+\dfrac{B}{s+2}+\dfrac{C}{s+3}+\dfrac{D}{(s+3)^2}\)

now multiply EVERY term by
s (s+2) (s+3)^2

what do u get ?

Answer 21

do I have to expand it?

Answer 22

just multiply each term by s(s+2)(s+3)^2
do not expand

Answer 23

what answer do u get?

Answer 24

\(\Large \dfrac{10}{s(s+2)(s+3)^2} \times s(s+2)(s+3)^2 \\ \Large =\dfrac{A}{s} \times s(s+2)(s+3)^2+\dfrac{B}{s+2}\times s(s+2)(s+3)^2 \\ \Large +\dfrac{C}{s+3}\times s(s+2)(s+3)^2+\dfrac{D}{(s+3)^2}\times s(s+2)(s+3)^2\)


simplify each term, cancel whatever gets cancelled from numerator and denominator.

example, on left side, only 10 remains

Answer 25

let me help you get started,

\(\Large \dfrac{10}{ \cancel {s(s+2)(s+3)^2}} \times \cancel {s(s+2)(s+3)^2 }\\ \Large =\dfrac{A}{\cancel s} \times \cancel s(s+2)(s+3)^2+\dfrac{B}{\cancel {s+2}}\times s\cancel {(s+2)}(s+3)^2 \\ \Large +\dfrac{C}{\cancel {s+3}}\times s(s+2)(s+3)^{\cancel 2}+\dfrac{D}{\cancel {(s+3)^2}}\times s(s+2)\cancel{(s+3)^2}\)

Answer 26

so what the next step? I'm trying to understand this

Answer 27

did you get how those terms are getting cancelled ?

Answer 28

next step :

\(10 = A (s+2)(s+3)^2 + Bs (s+3)^2 + Cs(s+2)(s+3)+Ds(s+2) \)

Answer 29

ok then?

Answer 30

you got that ?
because after that, things are easy.


that is true for all values of s
so, put s=0 in that and tell me what u get?

Answer 31

why s=0?

Answer 32

because we can choose any value of s
if we choose s = 0
then B,C,D will vanish
and we will be able to find A :)

Answer 33

the 's' of 'Bs,Cs and Ds must be 0 too?

Answer 34

yes

Answer 35

could you get A ?

Answer 36

18?

Answer 37

10 = A (18)
you got that, isn't it ?

Answer 38

Yaa I got it . so the next step is?

Answer 39

10 = A (s+2)(s+3)^2 + Bs (s+3)^2 + Cs(s+2)(s+3)+Ds(s+2)
That is true for every value of s

put s=0,
10 = A(2)(3)^2 =....
Find A

put s = -2
10 = B(-2)(1)^2 =....
Find B

put s= -3
10 = D(-3)(-1)=....
Find D

put s= -1
10 = A(1)(2)^2 + B(-1)(2)^2 +C(-1)(1)(2)+D(-1)(1)
Find C

Answer 40

find B,C,D

Answer 41

why the value of s is different?

Answer 42

s can take any value
wee need to choose
we choose value of s such that every other constant gets eliminated

Answer 43

okay :]

Answer 44

Usually it's a requirement to have Calc II on board before Differential Equations..

Answer 45

@FutureMathProfessor means?