Question

Need help...Use the Laplace transform to solve differential equation:
dx/dt + 7x = 5cos(2t)
Initial conditions: x(0) = 4, x'(0) = -4
Assume forcing functions are 0 prior to t = 0-

Answer 1

I did the first step taking the laplace of both sides.
SX(s) - x(0-) + 7X(x) = 5s/(s^2 +4)

Answer 2

x(0-) = 4
SX(s) - 4 + 7X(s) = 5s(s^2 + 4)
(s + 7)X(s) = 4s^2 + 5s+ 16 / (s^2 +4)
X(s) = (4s^2 + 5s + 16)/(s + 7)(s^2 + 4)

Answer 3

So I guess my question is taking the inverse laplace transform and that has to do with partial fractions

Answer 4

hey myininaya satellite helped me out with the partial fraction

Answer 5

\[\frac{35s}{53(s^2+4)}+\frac{20}{53(s^2+4)}+\frac{177}{53(s+7)}\]

Answer 6

thanks... I got the inverse laplace transform:
35/53cos(2t) + 10/53sin(2t) + 177/53e^(-7t)

Answer 7

i wish i new jack about these transforms

Answer 8

guess i need a book

Answer 9

I don't teach this stuff enough to have all the transforms memorized...I always look them up

Answer 10

yeah it really helps it turns differential equations into linear algebra makes it easy to solve. in this case I'm just using a table to substitue

Answer 11

if it is like those stultifying formulas for "techniques of integration" or ode's i will forget about it

Answer 12

http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx

Answer 13

but it seems to get asked fairly frequently so it must be pretty common

Answer 14

so for instant laplace of (sin3t) = 3/(s^2 + 9)

Answer 15

laplace of dx/dt = sX(s) - x(0)

Answer 16

helps you solve ODE algebraically

Answer 17

yeah someone tried to explain it to me. said it turns ode's into polynomials

Answer 18

yeah

Answer 19

so like x' + 5x = t

Answer 20

with initial conditions x(0) = 0 x'(0) = 0

Answer 21

laplace [x'] = sX(s)) - x(0)
laplace [5x] = 5X(s)
laplace [t] = 1/s^2

Answer 22

So you have:
sX(s) - 0 + 5X(s) = 1/s^2

Answer 23

(s+5)X(s) = 1/s^2
X(s) = 1/[(s^2)(s+5)]

Answer 24

So to find the solution which is x(t) you take inverse laplace of the expression on the right which involves partial fractions.

Answer 25

great, more partial fractions!

Answer 26

1/5s^2 + 1/25(s+5) - 1/25s

Answer 27

haha...yeah..So inverse laplace will give us
x(t) = t/5 + 1/25e^(-5t) - 1/25u(t)

Answer 28

apparently you get
\[x(t)=\frac{1}{25}(5t=e^{-5t}-1)\]

Answer 29

As you do more practice you'll get the hang of it.

Answer 30

yep

Answer 31

\[x(t)=\frac{1}{25}(5t+e^{-5t}-1)\]

Answer 32

U(t) just represents the step function it is 1 when t > 0 and 0 when t <0

Answer 33

is it fun or donkey work?

Answer 34

lol it gets tiring when the laplace transforms get complicated with partial fractions...but that's when matlab comes in lol we are also learning to use that software.

Answer 35

good luck
btw
http://www.wolframalpha.com/input/?i=solve+x%27+%2B+5x+%3D+t%2C+x%280%29%3D0%2C+x%27%280%29%3D0

Answer 36

thanks for your help too

Answer 37

i think your initial conditions are off, it should be x(0) = 4, x'(0) = -23
http://www.wolframalpha.com/input/?i=solve+x%27+%2B+7x+%3D+5cos%282t%29%2C+x%280%29%3D4%2C+x%27%280%29%3D-23

i solved it but i didn't need to use LaPlace transform

Answer 38

well you only need the first initial condition if you're solving with laplace because laplace of dx/dt only involves initial condition x(0) but i'm checking the work over again

Answer 39

oh ok, yeah thats true, the second condition isn't neccessary, i just noticed our solution x(t) didn't match up with x'(0) = -4

Answer 40

Okay I went through solving using method of undetermined coefficients and for that the result for the particular solution is the same 35/53cos2t +10/53sin2t, and when I plug in initial condition of x(0 = 4 I get the same thing c1 is 177/53 so the other solution the homogenous is 177/53e^(-7t). in fact in both cases because we are dealing with a first oder DFQ the initial condition of x'(0) is not important.

Answer 41

it's just giving in the problem.