Question

Need confirmation on my explanation of the solution. I think i understand but since the book doesn't clearly explain it i want to know if my reasoning is correct. (Question will follow shortly).

Answer 1

Here's the question:

a) Find the solution of the given initial value problem in explicit form
b) plot a graph of the solution
c) determine (at least approximately) the interval in which the solution is defined.

\[\frac{ \delta r( \theta ) }{ \delta \theta } =\frac{ r^2 (\theta ) }{ \theta }\]

r(1)=2

now the general solution is

\[\frac{ 2 }{ 1-2\ln( \theta ) } \in 0 < \theta < \sqrt{e}\]

now there's two things i would like to confirm. Let's call the Q and W

Q) The function above does not actually end after Sqrt(e), however it has no positive value after that point. Since the question originally asked for a r^2 i guess negative solutions don't count and that's why the solutions end at sqrt(e), is that correct?

W) when \[\int\limits_{\infty}^{-\infty} \frac{ 1 }{ \theta } \delta \theta \]

normally you'd get \[\ln \left| \theta \right| \]

however in this case \[\int\limits_{}^{} \frac{ 1 }{ r^2 } dr = \int\limits_{}^{} \frac{ 1 }{ \theta } d \theta \]

you get simply \[\ln \theta \]

is that because \[\frac{ -1 }{ r }=\ln \left| \theta \right| \]

would not make sense?

Answer 2

Actually the solution i posted is not the general solution but the initial value solution. but that doesn't really matter for the question.

Answer 3

Might I try to solve it myself (the differential equation) ?

Answer 4

Sure, go ahead.

Answer 5

In case it isn't clear from context it's actually r^2 (as a function of theta) on the top function, not r^2 multiplied by theta.

Answer 6

how do you get your equations large and colored btw? mine are always so small and hard to read sometimes.

Answer 7

\large \Large \LARGE \huge \Huge
these are varying degrees of font size, type them before the equations to adjust size.

\color{colour you want}{coloured text}

colors that I know work are white, black, yellow, blue, red, green, violet, orange... I don't really see why you'd need any more than those ^_^

now...

Answer 8

btw you made a mistake at your first equation

Answer 9

What's that?

Answer 10

\[\large \frac{ dr }{ r^2 } = \frac{ d \theta }{ \theta } \]

Answer 11

Oh yeah... crud.
Sorry about that.

Answer 12

See why you shouldn't consult me? I'm error prone XD

Answer 13

\[\Large -\frac1r= \ln|\theta|+C \]

Answer 14

wait did you just edit your post?

Answer 15

I didn't.
Should I have?

Answer 16

I just corrected it by posting under it.

Answer 17

Posts cannot be edited, btw, only deleted.

Answer 18

oh ok, i thought i went crazy when all of a sudden the wrong equation was replaced by a new one i guess.

Answer 19

anyway yes terenz that is what logically i would say as well, however it actually incorrect.

Answer 20

The ln does not have absolute signs in this case. Let me try if i can prove it somehow.

Answer 21

What do you mean it doesn't have absolute signs?

Answer 22

Why shouldn't it?

Answer 23

well

Answer 24

that's what i was trying to figure out as well

Answer 25

In fact i think for the given domain it doesn't even matter.

Answer 26

What do you mean it doesn't end at \(\large \sqrt e \)?

Answer 27

Well, if you draw the graph while ignoring the limits given, the graph just continues (after the asymptotic behavior) on the negative axis

Answer 28

Yes it does, so it's fair to say the function simply isn't defined at \(\large \theta = \sqrt e\)
Should it be?

Answer 29

http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427egh142rk7te

Answer 30

Yes it should.

Answer 31

It should? What is the apparent value of r when \(\large \theta = \sqrt e\)?

Answer 32

Oh in fact no, i misinterpret your question. It's undefined at that point because division by 0

Answer 33

http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ecjt00okj8e

Answer 34

Okay, so it's undefined at that point, is that a problem? :)

Answer 35

It's a bit hard to see from these graphs but both the ln with and without absolute marks actually completely overlap the initial value problem of the differential equation.

Answer 36

No it's not a problem, i just wanted to confirm if my reasoning is true. The graph of the solution ends after sqrt{e} is that because r^2 doesnt allow negative solutions?

Answer 37

No, it doesn't have to end with \(\sqrt e\)

r^2 concerns the derivatice, and if that's the case, it only means the function r is increasing at all times (which it is, check the graph)

Answer 38

derivative^

Answer 39

but the initial value solution of the equation actually does stop after sqrt(e) why exactly is that?

Answer 40

Where does it say that it stops at sqrt(e)?

Answer 41

The graph shows so

Answer 42

It's not that it stops, it's more like it has an asymptote at \(\theta = \sqrt e\)

But it is also defined for values \(\theta > \sqrt 3\)

Answer 43

As the graph also clearly indicates.

Answer 44

Which one?

Answer 45

Your wolfram graph.

Answer 46

which of the ones? the one is from the solution, the other one is from the differential equation. As you can see the differential equation does not have negatives, while the solution does (if you don't exclude values greater than sqrt(e)

Answer 47

Here's all those 3 graphs plotted on my calculator http://imageshack.us/photo/my-images/35/7cw.JPG/

Answer 48

The differential equation has no negatives for all \(\theta > 0\) yes?

Answer 49

It doesn't even exsist for \[0 > \theta > \sqrt{e} \]

Answer 50

We can ignore the negative values of \(\theta\) But for \(\theta > \sqrt e\), it does in fact exist. The wolfram graph shows it....

Answer 51

Why is that so hard to believe?

Answer 52

Then why does the solution stop after that point?

Answer 53

Who said the solution stopped after that point?

Answer 54

well for one the textbook, for two my calculator. also, i think when you refer to wolfram, you are not refering to the graph of the differential.

Answer 55

The textbook explicitly says it is not defined for values greater than \(\sqrt e\) ?

Answer 56

Or is it just the graph?

Answer 57

http://www.wolframalpha.com/input/?i=dr+%2F+d+theta+%3D+r%5E2+%2F+theta+and+r%281%29%3D2

Answer 58

the textbook states it as i said, it's only defined between 0 and sqrt(e) i just wanted to confirm if my logic is not flawed.

I didn't want to know if the solution is wrong or not, i wanted to know the REASON behind the solution is correct or not.

Answer 59

Perhaps it was simply restricted to that domain?

Answer 60

it's the interval at which the solution is defined. that's what the textbook says.

Answer 61

Then I'm at a loss.
I still see no reason why it can't be defined for \(\theta > \sqrt e\) however.

Answer 62

because at that point you divide by zero and after that point the graph will continue in the negative. Which is NOT what the differential equation graph does.

Answer 63

You do not divide by zero when \(\theta > \sqrt e\) only when \(\theta = \sqrt e\)

Answer 64

yes that's what i meant, at that point (sqrt(e)) you divide by zero and after that point the solutions will all be negative (asymptotic towards 0).

Answer 65

And what's wrong when the solutions are negative?

Answer 66

The differential equation only goes UP from that point, and at Sqrt(e) the slope is infinite, so it will never pass that point. Hence, they will never reach negative values. Nor anything to the right of that value.

Answer 67

If however we would have picked another initial value then negatives would have been possible.

Answer 68

Okay... assuming that, what's wrong now? You seem to have everything under control...

Answer 69

I guess so, although it doesn't feel like i understand. Anyway thanks for your help. Talking about it did make it somewhat clearer i guess.

Answer 70

Though, at sqrt(e), the slope is infinite, still, at the right-side of the sqrt(e)
So what's wrong with having the slope decrease from then on?

Answer 71

I will leave the question open for a bit in case people want to add to it to clear it up a bit but it has no hurry, i'll move on to other problems from the book now.

Answer 72

We have an initial value problem, the graph that goes trough the initial value will not ever reach that point. So we can disregard all the points that will not be reached by the initial value?