Question

Hi everyone! Can anyone help explain how to define the "interval" for this diff eq? I have the entire answer, I just want to discuss how they arrived at the correct interval so I can understand better. Thanks! :O)

Answer 1

I understand that P(x) = cot(x) and that you need to find that interval which goes
from (0, pi), but I don't understand what they are talking about when they say "We also have the sec(x) term..."

Why do they include searching for an interval for sec(x) ?
I know the the right side of the differential equation has sec^2(x)csc(x), but why single out only the sec(x), what about the csc(x)? And for that matter, are they even talking about the right side of the equation at all or am I missing something?

Answer 2

@ganeshie8 @aaronq @amistre64 @jim_thompson5910 @phi @zepdrix

Answer 3

Maybe the reason they didn't make mention of the csc(x) is because it has the same discontinuities as cot(x).

So you're not getting any NEW information about your boundaries from the csc(x).

Answer 4

Okay everyone, I may have an answer but I need someone to confirm it for me...

I previously thought that you use P(x)=whatever as your interval but then looking at the "SOLUTION" to this diff eq, I thought it would make sense to pull your interval from whatever "y equals"...

For example, the solution to this question is y = sec(x) +C csc(x)

Well the interval of both tan(x) and sec(x) is: (-pi/2, pi/2)
and the interval of both csc(x) and cot(x) is: (0,pi)

Since the solution "y = sec(x) +C csc(x)" has both sec(x) and csc(x) in it, maybe the
interval for the solution is wherever those 2 functions overlap which would be between
(0, pi/2)

And since the solution to this question is: y = sec(x) +C csc(x) ; 0<x<pi/2
I'm starting to think that is the correct way to get the answer every time.

Can anyone tell me if this thinking will work for every linear diff eq where they ask you to supply the interval?

Answer 5

@zepdrix does all that I just wrote mean the same thing you just wrote?

Answer 6

I dunno >.< I'm thinking lol

Answer 7

lol :o)
I hope what I said is correct!

Answer 8

Before step 1, what was the original problem?
\[\Large\rm (\cos^2x \sin x) y'+(\cos^3x) y=1\]Something like that?

Answer 9

one sec

Answer 10

cos^2(x)sin(x) dy + (ycos^3(x) - 1) dx = 0

Answer 11

http://www.wolframalpha.com/input/?i=cos^2%28x%29sin%28x%29+dy+%2B+%28ycos^3%28x%29+-+1%29+dx+%3D+0

Answer 12

So you have this form:
\[\Large\rm y'+P(x)y=Q(x)\]I think you want to try to find the largest interval where P(x) and Q(x) are both continuous. That's what it seems like they're saying in Step 1 at least (They called it f(x) instead of Q(x) though).

Answer 13

the only that makes me scratch my head to that is when creating P(x), don't you kinda have the option to make it into different stuff depending on what you put over to the right hand side?

Answer 14

like there is not "one" perfect P(x) right?

Answer 15

Since you always want your y' term isolated, you should always end up with the same P(x).

I mean.. lemme try to think of an example...

\[\Large\rm y'+(x^2+1)y=2\]There is no "stuff" that we can send over to the other side because it's all multiplying the y. We can't move the +1 to the other side because of the y.

Answer 16

ok so going back to what you wrote>>>
So you have this form:
y′+P(x)y=Q(x)
I think you want to try to find the largest interval where P(x) and Q(x) are both continuous. That's what it seems like they're saying in Step 1 at least (They called it f(x) instead of Q(x) though).

it seems like we have 2 lines of thought then...
1) find the intervals of P(x) and Q(x) or
2) find the intervals of each term from y=whatever..."the solution"

In my book, the question actually states:
"State an interval on which the general solution is defined"

so which line of thought would answer that question? 1 or 2 ?

Answer 17

is the general solution y' +P(x)y = Q(x) or y = stuff ?

Answer 18

y=stuff :)

Answer 19

okay...so would you say my thought was correct then? find y = stuff and then find where all the intervals of all the stuff overlap?

Answer 20

Mmm math! Brains.. everywhere :(
sec, lemme gather my thoughts lol

Answer 21

if the general solution actually is y = stuff
and the questions states: "State an interval on which the general solution is defined"
doesn't that leave the only choice to get the intervals from is from y = whatever ?
Isn't that logical? :o/

Answer 22

Here is what I'm thinking...
If you simply look for an continuous interval directly from your solution, you can't be certain that interval will satisfy the Differential Equation.

I think that's why they're looking for the interval sooner than that:
They're looking for an interval which is true both for the General Solution and still works for the Differential Equation.

This might be nonsense though^
So I was trying to see if it made sense before writing it, but oh well lol.

Answer 23

we need more input from people then...i will shout out at someone else...if you know anyone that can help, please @them
@satellite73 @AccessDenied @aaronq

Answer 24

hey @zepdrix , I think i'm correct and here is why:

using: y' +P(x)y = Q(x) form where (dy/dx) + tan(x)y=csc(x)
here we have a tan(x) and a csc(x) where tan(x) has an interval of (-pi/2,pi/2) and
csc(x) has an interval of (0,pi)

however the solution to this question I just made up is y = stuff with sin and cos
and sin and cos have intervals of (-infinity, +infinity)

according to wolfram, the solution has an interval of (-infinity, +infinity)
which would make me think that we ONLY have to concern ourselves with the intervals from y = whatever

check out wolfram>>>

http://www.wolframalpha.com/input/?i=%28dy%2Fdx%29+%2B+tan%28x%29y%3Dcsc%28x%29

doesn't this sound correct?

Answer 25

How did you get the interval from wolfram? I don't see it listed.
I see a log(cosx) and log(sinx) in the solution though.
Which certainly don't allow from -infinity to infinity, yes? :o

Answer 26

Here are my immediate thoughts; I just came from another class, so I'll refine it a bit after winding down. lol

In general, the domain of the derivative of a function is a subset of the domain of the original function. So taking the interval information from the general solution rather than the initial conditions is sort of like simplifying: (x+1)^2/(x+1) = x+1. The left side clearly cannot have x=-1 because the denominator will be zero. The right side, you can't really tell that the left side happened.

Btw, the solution you presented contains natural logarithms of sine and cosine, and the natural log function can only take positive values. The sine/cosine must not be negative or zero, then.

Answer 27

drads...
okay, so my idiot book doesn't tell me how to deal with the intervals...
do we look at the left and right sides of y' +P(x)y = Q(x) or the left and right sides of y=whatever ?
a combination of both?
does anyone have an answer to this craziness? :o)~

Answer 28

where can we find the procedure to deal with intervals?
i can't even find it on youtube

Answer 29

if I take a pic of the questions and answers from my book, will that allow us to come up with a general methodology to always find the correct intervals?

Answer 30

What I mean by that analogy is, more or less, that you could have fabricated some information in obtaining the general solution y that wasn't initially there. So I think you really want to take the information from your initial equation. Or maybe I'm just crazy and all.

I'll do some more research though so I don't feel like I'm making a crazy guess. D:

Answer 31

i'm going to post a pic of q and a's from my book...
if you are correct, then all the intervals from the answers must come from the initial equations...gimme a sec to post it

Answer 32

Here are some simply `separable DE's` worked out on Paul's Notes:
http://tutorial.math.lamar.edu/Classes/DE/Separable.aspx

It looks like he's getting the valid intervals directly from the solutions like you were suggesting. (But these are with given initial data).

Now I'm all confused :d lol

Answer 33

simple* not simply

Answer 34

Maybe they both just work, since he also took from the initial equation once it was set up as a linear DE:
http://tutorial.math.lamar.edu/Classes/DE/IoV.aspx

Answer 35

does this help at all?

Answer 36

Pauls notes say:

Notice as well that the interval of validity will depend only partially on the initial condition. The interval must contain to, but the value of yo, has no effect on the interval of validity.

what is ysubzero and t sub zero?

Answer 37

t_0 and y_0 are values for initial conditions. They're usually given in a particular type of problem called initial value problems. Basically they're just extra pieces of information about the general solution that let us determine the integration constant.

Answer 38

I seriously CANNOT BELIEVE that my book doesn't teach me how to do this!!!

Answer 39

the only thing i know about initial conditions is that they usually are given to look like this:

y(-2)=4 where x = -2 and y=4...so where in this initial condition is ysubzero?

Answer 40

SinginDaDiffEQBlues :c

Answer 41

\(\Large\rm x_o=-2\)
\(\Large\rm y_o=4\)

Answer 42

okay...subzero just means initial condition...it's y"knot" not subzero...i get that at least lol

Answer 43

Ah good call :) It's not like a "starting point", it's just initial information.
y"naut"

Answer 44

so going back to Pauls notes...
"The interval must contain t"naut", but the value of y"naut", has no effect on the interval of validity."
But look at my book below...it doesn't give initial conditions but wants the intervals! grrrr

Answer 45

Pauls notes say: "The interval must contain t"naut", but the value of y"naut", has no effect on the interval of validity."

my book gives neither! so how am I supposed to answer the questions?

Answer 46

I'm sorry guys but I'm freaking out and need coffee!
I know we aren't done with this problem yet but I just want to thank both of you in advance sooooo much for helping me! :o)

Answer 47

For #9, checking the original equation you would only find that \( x \ne 0 \) when putting it in the form of \(y ' + P(x) y = Q (x)\). The solution has that natural logarithm, which eliminates any x < 0 as well as the x=0.

Answer 48

I will be right back

Answer 49

so what is your summary of question #9 then? which do we use to find the intervals,
y' +P(x)y = Q(x) or y = whatever ?

Answer 50

brb

Answer 51

For #13, it doesn't appear you can conclude 0 < x < infinity from either the linear DE or general solution; only that \( x \ne 0\).
In fact the only place that seems to come up is at the integrating factor integral of 1/x dx = ln x, where x>0 because we didn't use an absolute value of x.

I'm gonna have to say, you gotta just check everything. That would cover all the bases!

Answer 52

Okay, so just so I understand you perfectly...

You are saying we have to check everything meaning to check:
1.) y' +P(x)y = Q(x)
2.) y = whatever
and 3.) the integrating factor

all these? is that correct?

Answer 53

That sounds good. I was thinking in general, check any assumptions you make. But those should be the major ones. Maybe there's a more refined strategy, but I think in general if you're keeping track of every possibility over the course of the problem, you'll hit all the marks you need to.

Answer 54

how is it that the book wants people to answer the question without telling them how to answer the question? lol

my teacher never said anything either!

Maybe I am the crazy one but I am always finding questions like this where everyone somehow does it, but doesn't know how or why...i must be crazy! how am i the only one that asks these things?

Answer 55

They ought to at least tell you where to check! At least the process of finding discontinuities at any point isn't tremendously difficult on its own, just a bit tedious

I remember I flat out skipped anything involving intervals while I was studying DE because I simply didn't care about them in solving those equations. :p

Answer 56

and your book never asked you for the intervals of the solutions?

Answer 57

something I should ask...in my book, the question says specifically:
"State an interval on which the general solution is defined"

does the word "an" basically say they don't care about "all" the intervals? just any old interval will do? am i reading that correctly?

Answer 58

As long as everything is defined, yes. Like with your trig solutions, those are periodic. So unlike the IVP problems where you already have information about the problem, you can choose the interval as long as everything works properly.

Also checking my book for "interval", it only ever came up with IVP, and either I ignored those parts of the problem or they were trivially easy. I studied DE on my own time though, so I had that liberty of ignoring boring parts. lol

Answer 59

you didn't get credit for this in college?

Answer 60

Nah, I'm still in Calc 2 right now for college. I just couldn't really test out of anything after Calc 1, even though I've studied far past that.

Answer 61

ewww...calc 2 is el stinko! hence my name! ;O)

Answer 62

okay well thanks for everything!
have a great weekend :o)

Answer 63

I guess the only thing I have yet to learn in Calc 2 is series convergence tests, which I also was too bored of to work on, although I have a basic idea of it. Otherwise I've got a few years of easy Math sailing!

Glad to help! And you too!