Question

Question Below

Answer 1

i dont see it...

Answer 2

As a spring is heated, its spring constant decreases. Suppose the spring is heated so that the spring constat at time t is given by k(t)=6-t N/m. If the unforced spring-mass system has mass m=2kg and a damping constant of \(\gamma=1N-sec/m\), with initial conditions x(0)=3m and x'(0)=0m/sec, first wright the initial value problem (second degree homogenous ODE) problem which governs displacement x(t).
Then use a series-solution method to find the first four non-zero terms n a power series expansion about t=0 for the displacement; i.e. find a series-solution to this problem and write the first four non-zero terms of that series.

Answer 3

This making any sense to you?

Answer 4

no not at all

Answer 5

first write*
sorry typo.

Answer 6

amistre, please tell me you know how to do this.

Answer 7

do we have the diffyQ written? that might be the only issue id have with it at the moment

Answer 8

using a series for a solution is simple enough, but i dont think i could come up with the diffyQ to begin with :/

Answer 9

I am trying to figure it out. I have no clue why this problem was made so hard :/

Answer 10

hmm, we seem to have smartscores back for a moment .. time to do some mod work

Answer 11

It is unforced, so it will be something equal to zero...

Answer 12

Okay, I think I got this. The general form is:

\(\Large{mu''(t)+\gamma u'(t)+ku(t)=F(t)}\)

So I think it would make sense that the equation is,

\(\Large{2u''(t)+u'(t)+(6-t)u(t)=0}\)

Answer 13

@amistre64
Do you think you could help me from here?

Answer 14

let \[u=\sum_0a_n~(6-x)^n\]
\[u'=-\sum_1a_n~n(6-x)^{n-1}\]
\[u''=\sum_2a_n~n(n-1)(6-x)^{n-2}\]

Answer 15

well, 6-t if you want to get picky lol

Answer 16

Okay, and what do you do from there?

Answer 17

\[{2u''(t)+u'(t)+(6-t)u(t)=0}\]

plugging in we get
\[{2\sum_2a_n~n(n-1)(6-t)^{n-2}-\sum_1a_n~n(6-t)^{n-1}+(6-t)\sum_0a_n~(6-t)^n=0}\]

simplifying
\[{\sum_22a_n~n(n-1)(6-t)^{n-2}+\sum_1-a_n~n(6-t)^{n-1}+\sum_0a_n~(6-t)^{n+1}=0}\]

now what i usually do is get the exponents worked to lowest by adding or subtracting to an index

Answer 18

What are the numbers on the bottom of the sigmas? And is top value supposed to be infinity?

Answer 19

the bottom index value is the starting value of "n" and yes, to infinity

\[{\sum_22a_n~n(n-1)(6-t)^{n-2}+\sum_{1+1}-a_{n-1}~(n-1)(6-t)^{n-1-1}\\ \hspace{5em}+\sum_{0+3}a_{n-3}~(6-t)^{n-3+1}=0}\]

\[{\sum_22a_n~n(n-1)(6-t)^{n-2}+\sum_{2}-a_{n-1}~(n-1)(6-t)^{n-2}\\ \hspace{5em}+\sum_{3}a_{n-3}~(6-t)^{n-2}=0}\]

now that we have all the same exponents, we can align the index values by pulling out the first few terms

Answer 20

the numbers might be tiny for you to see now that i think of it

Answer 21

right click, settings, zoom if need be

Answer 22

Why was it +3 on the last one?

Answer 23

think of it as shifting a function left or right.

to shift f(n) to the right by 3 units; we subtract .. f(n-3)

we do not want to change any values tho, so we need to "start" the index at +3 to account for the shift

Answer 24

okay, but what I mean, you made the other one so that it was two... why was that one three?

Answer 25

let f(n) = n+1 ; for n=0,1,2,3,4,...

we want it to be n-2, f(n+3) = n-3+1

Answer 26

we are trying to align all the exponent parts so that they can be factored out

Answer 27

Ah, okay

Answer 28

since the lowest exponent part is n-2, we want to shift all the others to an exponent of n-2

Answer 29

Yep, I see that now :)

Answer 30

once we get all the n parts shifted; we need to consider making them all start at the same place so that we can group them into one summation:

\[2a_2~2(2-1)(6-t)^{2-2}+{\sum_32a_n~n(n-1)(6-t)^{n-2}\\
-a_{2-1}~(2-1)(6-t)^{2-2}+\sum_{3}-a_{n-1}~(n-1)(6-t)^{n-2}\\ \hspace{10em}+\sum_{3}a_{n-3}~(6-t)^{n-2}=0}\] ^^
all lined up

Answer 31

Okay, I think I follow.

Answer 32

to clean it up some
\[4a_2+{\sum_32a_n~n(n-1)(6-t)^{n-2}\\
-a_{1}+\sum_{3}-a_{n-1}~(n-1)(6-t)^{n-2}\\ \hspace{3em}+\sum_{3}a_{n-3}~(6-t)^{n-2}=0}\]


\[4a_2-a_{n-1}+\sum_{3}~[2a_n~n(n-1)-a_{n-1}(n-1)+a_{n-3}]~(6-t)^{n-2}=0\]

Answer 33

second term is mistyped lol -a1

Answer 34

mistyped... what was it supposed to be?

Answer 35

-a1 .. i typed it as a{n-1} when trying to edit up the latex coding

Answer 36

the rest of this is doing some algebra to define an in terms of a{n-1} and a{n-3}

Answer 37

Ohhh, at the very bottom of the post. Okay.
I am entirely incompetent in this. I hate it.

Answer 38

\[4a_2-a_{1}+\sum_{3}~[2a_n~n(n-1)-a_{n-1}(n-1)+a_{n-3}]~(6-t)^{n-2}=0\]

the summation is equal to 0 when the coefficients are all zero

\[2a_n~n(n-1)-a_{n-1}(n-1)+a_{n-3}=0\]

work the algebra to solve for an

\[2a_n~n(n-1)=a_{n-1}(n-1)-a_{n-3}\]
\[a_n=\frac{a_{n-1}(n-1)-a_{n-3}}{2n(n-1)}~:~n\ge3\]

Answer 39

4a2 - a1 = 0 when a2 = a1/4, or when a2=a1=0

Answer 40

keep in mind that we are looking for the setup:\[y=\sum_0a_n(6-t)^n\]or written another way:\[y=a_0+a_1(6-t)+a_2(6-t)^2+a_3(6-t)^3+...\]

Answer 41

using the formula for an:

\[a_n=\frac{a_{n-1}(n-1)-a_{n-3}}{2n(n-1)}~:~n\ge3\]
lets start defining the coefficients
\[a_0=a_0\]
\[a_1=a_1\]
\[a_2=\frac14a_2\]
\[a_3=\frac{a_{3-1}(3-1)-a_{3-3}}{2(3)(3-1)}\]
\[a_4=\frac{a_{4-1}(4-1)-a_{4-3}}{2(4)(4-1)}\]
\[a_5=\frac{a_{5-1}(5-1)-a_{5-3}}{2(5)(5-1)}\]
\[a_6=\frac{a_{6-1}(6-1)-a_{6-3}}{2(6)(6-1)}\]

you will also find that we can substitute in the parts we need to define this in terms of 2 unknown constants a0 and a1

Answer 42

nother typo lol .... a2 = 1/4 a1 from before

Answer 43

Do we need to solve for those a_0 and a_1? or are these answers correct?
And first 4 non-zero terms, are the first couple zero?

Answer 44

we are given initial conditions which will give us a way to define specific values with

Answer 45

a0 and a1 are the arbitrary constants that can be solved for with the initial conditions

Answer 46

ah, okay.
I am still reading through this trying to make sense of it lol.

Answer 47

its just alot of work, but the work itself is pretty simple to do

Answer 48

I am not entirely sure how to solve with the initial conditions.... I am looking at this, and I don't even know where you would plug this crap in.

Answer 49

if we start to try to fill it in so that an is in terms of a0 and a1, for the sake of less typing, lets say a and b
\[a_2=\frac{1}{4}b\]
gonna have to work a3 alittle bit more by the formula
\[a_3=\frac{2a_{2}-a_{0}}{2(3)(2)}\]\[a_3=\frac{1!~(\frac12b-a)}{2~3!}\]
same with a4
\[a_4=\frac{a_{3}(3)-a_{1}}{2(4)(3)}\]
\[a_4=\frac{2!~(3a_{3}-b)}{2~4!}\]
\[a_4=\frac{2!~(3(\frac{(\frac12b-a)}{2~3!})-b)}{2~4!}\]

and so on and so forth

Answer 50

and to clarify a=a0 b=a1

Answer 51

alot of work to try to code in ... but you would see that we get a fairly dependable pattern in terms of a and b .... with any luck

and if that pattern comes up alot and is useful in solving other equations, they give it a name like beta, or legendre, or something to seperate it from the overall mishmash

Answer 52

yes .... a_0 and a_1 are rather difficult for me to keep trying to type up over and over again is all

Answer 53

okay... makes sense so far.

Answer 54

there are simpler diffy qs to start getting the basics of it downpat with ... but i have to focus on getting some classwork done in the next 30 minutes. post some thought you have and ill see if i can address them later

Answer 55

okay. Thanks so much!