Question

Using laplace trasform solve :

dx/dt+3x=0

Answer 1

have u tried it ?
U know L[dx/dt] = ?

Answer 2

i know the ans is 2e^-3t by solving it normally ... but i don/t know the lplace of dx/dt?

Answer 3

i meant laplace lol

Answer 4

laplace of derivative is L[dx/dt] = sF(s)-F(0)

Answer 5

where F(s) is laplace of f(x)

Answer 6

L[df(x)/dt] = sF(s)-F(0)

Answer 7

i missed the intial value x(0)=2.. still confused?

Answer 8

i am doing some typos here, sorry,
L[df(x)/dt] = sF(s)-f(0)
here your f(x) is x .

Answer 9

f(0)=x(0)

Answer 10

right so is it sF(s)-f(0) +3/s^2

Answer 11

yes, put f(0) as 2

Answer 12

if f(0)=2 then what about sF(s)?

Answer 13

isolate F(s)
i think u know that to solve it, we need to take inverse laplace transform also, by first finding F(s) in terms of s

Answer 14

2/s+2) = 2e^-2t correct or not?

Answer 15

3/(s+2)
3e^{-2t}

Answer 16

if you solve the normal way you get 2e^-3t ? how come it is a diffrent ans?

Answer 17

dx/dt = -3x

(-1/3)(dx/x) = dt

dx/x = -3dt

ln x = -3t + C

x = e^(-3t+C)

x = Ce^(-3t) [The sloppiness I mentioned]

2 = Ce^0

C = 2

x(t) = 2e^(-3t)

Answer 18

dx/dt+3x=0

sF(s)- f(0)+3F(s)=0
sF(s)- 2+3F(s)=0
(s+3)F(s) = 2
F(s) = 2/(s+3)
f(x) = 2e^{-3t}
i think i should sleep.....

Answer 19

ask if u didn't get any of the step above.....

Answer 20

i under now ... lol im new to the topic thats why im so confused !

Answer 21

with practice, this topic becomes simpler ....

Answer 22

understand i mean rofl

Answer 23

how would i get about solving the laplace of d^2x/dt^2

Answer 24

Like is their a table for this ?

Answer 25

yes, there is table ...
without initial conditions,
L(d^2 f(x)/dt^2) = s^2 F(s)

Answer 26

oh i see... now!

Answer 27

any more doubts ?

Answer 28

how did you get 3F(s) for the last Question?

Answer 29

im just looking back now .. and noticed...

Answer 30

laplace of x(t) = F(s)

Answer 31

i thought 3x= 3/s^2 according to tables?

Answer 32

its not x, its x(t), right ?

Answer 33

ohh damn your right!..

Answer 34

any chance of working through another question?

Answer 35

any more questions ?

Answer 36

yes , im working on dx/dt+2x=2te^-2t ; x(0)=3

Answer 37

so what is laplace of dx/dt+2x ?

Answer 38

sF(s)- f(0)+2F(s)=> 2te^-2t how do i use splace for this?

Answer 39

asking laplace for this 2te^-2t, right ?

Answer 40

yes .. do i use tables

Answer 41

yes, from that what is L{t^n f(t)} = ?

Answer 42

2/s^3 and 1/s+2 multiplied together i guess?

Answer 43

no, laplace of product doesn't equal product of laplace
\(\large L(tf(x))=-d/ds(F(s))\)

Answer 44

so first find L(e^{-2t})

Answer 45

so how im i supposed to solve 2te^-2t , if tables can't be used?

Answer 46

use table to find L(e^{-2t}) first

Answer 47

L(e^{-2t})= 1/s+2

Answer 48

yes, so now L{t e^{-2t}} will be -d/ds(1/s+2)
got this ?

Answer 49

oh so you find the laplace of 2t now ?

Answer 50

where is -d/ds coming from is that a standard result?

Answer 51

no!
according to rule \(\large L(tf(x))=-d/ds(F(s))\)
you first find the laplace of the function that is mul.tiplied by t
then take its derivative(of laplace) w.r.t s

Answer 52

yes, its standard result

Answer 53

right, i have never been taught that.... first time seeing this! WOW! lol

Answer 54

so does that mean the laplace of 2 is solved separately?

Answer 55

2 is constant, it can be taken out of laplace, just like in derivatives or integration

Answer 56

can't you work it out like this L(2t)*L(e^-2t)

Answer 57

no, as i said , laplace of product is not product of laplace
\(L(f(x)g(x))\ne L(f(x))L(g(x))\)

Answer 58

right thats wierd beacuse i was taught that it is linearly propotional and can be solved this way.. my lecturer is quiet confusing i guess!

Answer 59

so now can u find L{t e^{-2t}}

Answer 60

sF(s)- 3+2F(s)=-d/ds(1/s+2) ... how can i proceed forward when i have -d/ds?

Answer 61

d/ds is just the first derivative w.r.t x
so what is d/ds(1/(s+2)) = ?

Answer 62

i don't get why you suppose to diffrentiate?

Answer 63

because the formula says so :P

Answer 64

-1/s^2 i think...

Answer 65

d/ds(1/(s+2)) = -1/(s+2)^2

Answer 66

yep i made mistake.. so i got (s+2) f(s)= 1/(s+2)^2+3 is that correct...

Answer 67

yes, isolate F(s)

Answer 68

f(s)= 1/(s+2)^2 +3/(s+3)
F(x)= t^2e^-2t +3e^-2t

Answer 69

(s+2) F(s)= 1/(s+2)^2+3
F(s) = 1/(s+2)^3 +3/(s+3)
isn't it ?

Answer 70

1/(s+2)^3 i think its 1/(s+2)^2

Answer 71

(s+2) f(s)= 1/(s+2)^2+3
dividing s+2 throughout the equation, what u get ?

Answer 72

yh your right! , but looking at tables their is no solution to 1/(s+2)^3

Answer 73

finally figured it out, it is 1/(s+2)^3....

Answer 74

1/s^3 ?

Answer 75

for 1/(s+2), it is ?

Answer 76

yes the power of t is added in to order to match 1/(s+a)^2

Answer 77

thats correct, so what u get for 1/(s+2)^3 ?

Answer 78

F(x)= t^2e^-2t +3e^-2t

Answer 79

final ans

Answer 80

first term will be divided by 2, isn't it ?

Answer 81

f(s)= 1/(s+2)^3 +3/(s+3)

F(x)= t^2e^-2t +3e^-2t

Answer 82

all that is correct, i was saying that it should be F(x)=(1/2) t^2e^-2t +3e^-2t

Answer 83

where is the 1/2 coming from?

Answer 84

what is laplace inverse of 1/s^3 ?

Answer 85

according to tables it seems right?

Answer 86

according to table laplace inverse of 1/s^3 = t^2/2 isn't it ?

Answer 87

t^2

Answer 88

L(t^n)=n!/(s^(n+1))

Answer 89

Im not sure ...

Answer 90

isn't that in the list ?

Answer 91

i did this f(t)=t^m , and according to f(s)=m!/s^m+1 M=2 ??? or m=1

Answer 92

m=2, because power is 2

Answer 93

well thats like saying 2x1/s^3 which is incorrect ... when it is 1/(s+2)^3

Answer 94

i meant laplace inverse of 1/x^3 = t^2/ 2

Answer 95

i think it is that... not sure 100%

Answer 96

i am pretty sure about that .
laplace inverse of 1/x^3 = t^2/ 2

Answer 97

if your sure then i guess it must be correct !

Answer 98

so final answer is F(x)=(1/2) t^2e^-2t +3e^-2t