Question

Find the solution of the given Initial Value Problem.

Answer 1

\(y''+2y'+2y=h(t)\)
\(y(0)=0\)
\(y'(0)=1\)

\[
h(t)=\{
\begin{array}{ccc}
1,&\pi\le t <2\pi
~~~~~~~~~~~0,&0\le t < \pi
\end{array}
\]

Answer 2

That is supposed to be a multi-line brace. I wasn't able to get it to display properly. It is supposed to be piecewise style...

Answer 3

hmm, heaviside seems to come to mind with piecewise stuff

Answer 4

what pat has you stymied?

Answer 5

*part ... different keyboard makes my typos even more abundant :/

Answer 6

I am looking at the book and it is showing parts where the h(t) is equal to functions of u sub some number of 't' and I am not sure where they are getting them.

Answer 7

any foureir stuff in the lesson by chance?

Answer 8

Nope, that isn't for another 4 chapters.

Answer 9

ut is usually something to do with finding a ... forget what they call it .... function.

Answer 10

I have the answer to the question, it is just a matter of how the heck do I get there :/

Answer 11

can you attach a picture of the problem and solution?

Answer 12

BVP, boundary value problem seems to be something to this

Answer 13

I am working on getting the pics on my PC now. Just a moment. Also, it is just an answer, not a worked out solution.

Answer 14

Here we are, problem 2 is my assigned problem.

Answer 15

Just so you know, I only need to do part (a) as well.

Answer 16

it looks like they are using u(t) as an arbitrary "constant" in the solution

Answer 17

I think so, but then in other examples, they have very specific subscripts which then carry over as part of how to solve the problem.
I has me rather confused all around.

Answer 18

can you provide a very specific example? :)

Answer 19

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

heaviside seems applicable

Answer 20

the subscripts just indicate where the pieces are redefined .... they represent an interval marking

Answer 21

Ok, do you understand this well enough that you could help me through this problem?

I am reading through those notes and I can kinda see how they are using the subscripts, now I just need to know how to apply this to this problem.

Answer 22

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

example 4 looks like a good run thru,

without reading it, im assuming they convert the h(t) defined by the piecewise into its heaviside equivalent ..... have to read further to test that hypothesis :)

Answer 23

Ok, so going by their example I get something that looks like this,

\(\large{h(t)=1+(-1)u_{2\pi}(t)}\)

Answer 24

Or would it be,

\(\large{h(t)=1+(-1)u_{3\pi}(t)}\)

Answer 25

im having to do a review of the heaviside ...

the u parts seems to identify a switch that turns on when its subscript is reached.

h(t) = 0 , 0 <= t < pi

f(t) = 0 , to start with
..................................
h(t) = 1 , pi <= t < 2pi

f(t) = 0 + (1)\(u_{pi}(t)\)
...................................
h(t) = 0, 2pi <= t

since the initial, and the first switch are on; we have a value of 1, we need a value of -1 when the switch is turned on to equal 0 again

f(t) = 0 + (1)\(u_{pi}(t)\) - (1)\(u_{2pi}(t)\)

Answer 26

Okay, I think I am following...

Answer 27

In the example, it said something about a "shift", how would that be done with this problem?

Answer 28

in most cases, the piecewise function is not defined in terms of constant values, but as function. the shifting seems to be a way to stabalize functions .... if im reading it right

Answer 29

"most switches will not turn on and take constant values. Most switches will turn on and vary continually with the value of t."

Answer 30

your problem gives us constants so im assuming that its a simpler setup ...

Answer 31

I am reading this stuff, I am not seeing what they are doing to them to get this. It almost seems random, to me anyway.

Answer 32

the heaviside page explains alot of what they are brushing over since it came before this and is assumed to be prior experience :)

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

Answer 33

\[L\{u_c(t)\}=\frac{e^{-cs}}{s}\]

Answer 34

\[L\{u_{pi}-u_{2pi}\}=\frac{e^{-pi~s}}{s}-\frac{e^{-2pi~s}}{s}\]

Answer 35

we also need to take the laplace of the equation

\[L\{y''+2y'+2y\}\]
you recall how to take the laplace of that?

Answer 36

\(\large{ s^2Y(s)-y(0)-y'(0)+2(sY(s)-y(0))+2(Y(s))=\frac{e^{-\pi s}}{s}-\frac{e^{-2\pi s}}{s}}\)

This look right?

Answer 37

\[(s^2Y(s)-sy(0)-y'(0))+2(sY(s)-y(0))+2Y(s)\]
filling in the initial conditions give us:
\[s^2Y(s)-1+2sY(s)+2Y(s)\]
factoring Y(s)
\[(s^2+2s+2)Y(s)-1\]
and solving for Y(s), gives us ...
\[Y(s)=\frac{e^{-pi~s}-e^{-2pi~s}+s}{s(s^2+2s+2)}\]

Answer 38

yes, yes it does :)

Answer 39

i think i ran over a 2 someplace in mine

Answer 40

nah, y=0 so it didnt distribute worth anything ....

Answer 41

Okay.... I have....

\(\displaystyle Y(s)(s^2+2s+2)-1=h(t)\)

I think this is similar to what you got...

Where would I go from here?

Answer 42

add 1 and divide the s^2 poly

Answer 43

\(\displaystyle Y(s)=\frac{e^{-\pi x}}{s(s^2+2x+2)}-\frac{e^{-2\pi s}}{s(s^2+2x+2)}+\frac{1}{(s^2+2x+2)}\)

This is where I am at....

Answer 44

good, now we can essentially ignore the e parts and work out the laplace inverses of:\[F(s)=\frac1{s(s^2+2s+2)}\]and\[G(s)=\frac1{s^2+2s+2}\]

Answer 45

On F(s) I believe we can even break that into two different parts as well, can't we?

Answer 46

a by parts decomp would be fine

Answer 47

notice that G(s) gives us the e^(-t) sin(t) part of the solution

Answer 48

Yep, I see that.

Answer 49

the partial decomp should provide the rest of the solution they have when we apply it to the upi and -u2pi inverses

Answer 50

without a lot of experience with heaviside and Laplace, i think that worked out pretty nicely :)

Answer 51

Yeah! Thanks very much, makes more sense to me now anyway :P

Answer 52

youre welcome :)