Question

(ODE) Alright, working through a BVP, legit not sure how to deal with these, lol...

Answer 1

http://i.imgur.com/cIikLdS.png

Answer 2

I'm not quite sure how you deal with BVP's differently or what makes them fundamentally different from an IVP.

Answer 3

"There is an important mathematical difference between these kinds of problems: if we have a second order differential equation with given initial values whether or not a unique solution exists depends only on the equation, not the initial conditions, while whether or not there exists a solution to the same equation, with given boundary conditions, depends upon the boundary conditions as well."

http://mathhelpforum.com/differential-equations/180630-difference-between-ivp-bvp.html

Answer 4

Alright, so that doesn't really mean anything to me practically, but hey! lol. Food for thought.

Answer 5

"Difference between an Initial Value Problem(IVP) and a Boundary Value Problem(BVP):
- IVP: all the values needed to solve the particular problem are specified at a single point.
- BVP: all the values needed to solve the particular problem are specified at different points."

Just putting this here to think about. Still not sure how to approach this problem, gonna go read about it for a sec.

Answer 6

Okay, it looks like even if the conditions of that theorem 4.1.1 ( http://i.imgur.com/sLoesmC.png) are satisfied, there may be one solution, many (infinitely many, possibly) solutions, or none at all.

Answer 7

IVP : y(0) = .., y(0) = ..
BVP : y(0) = .., y(1) = ..

Answer 8

notice that in BVP the points given are at different times : x = 0 and x = 1

Answer 9

but the points given in IVP are at same times : x = 0 and x = 0

Answer 10

Yeah, I see that, I'm just wondering mechanically how to approach this.

Answer 11

you approach it exactly same as the IVP that you have solved earlier

Answer 12

get two equations and solve the constants

Answer 13

This has to be in some way fundamentally different from the first way we attempted to solve VP's, because it seems like you'll have to be simultaneously plugging in values for the same variable, but that makes no sense to me. Alright, I'll take a shot or something like that, lol.

Answer 14

Wait, I think I got it

Answer 15

\[c_{1}+c_{2}=y=0.\]\[c_{1}e+\frac{c_{2}}{e}=1\]

Answer 16

Alright, now just algebra.

Answer 17

yeah give it to wolfram :O

Answer 18

I can see how to solve this by substitution and stuff, I think I'm-lol

Answer 19

http://www.wolframalpha.com/input/?i=solve+c_1%2Bc_2%3D0%2Cc_1e%2Bc_2%2Fe%3D1

Answer 20

Okay, cool, makes sense. On to another problem type. Thank you for helping me through this stuff, it means a lot.

Answer 21

lol that kid is a psycho

Answer 22

kid? Wait, what kid, wot

Answer 23

(Lol what)

Answer 24

oh you're done with b and c ?

Answer 25

...............

Answer 26

See step 15. http://www.smbc-comics.com/comics/20130616.png

Answer 27

But yeah, taking a look at B, thank you for reminding me, in all seriousness.

Answer 28

(Alright, be right back, like, literally 3 minutes DON'T GO ANYWHERE :E)

Answer 29

sorry to remind you about more work lol but part b looks similar to part a and you can finish off part c somehow..

Answer 30

Alright, I'm back. Yup, gonna take a shot.

Answer 31

I'm not sure, but I think I might end up having to use the exponential definitions of cosh and sinh. Either way, setting up the general form now.

Answer 32

\[c_{3}\cosh(0)+c_{4}\sinh(0)=0;\]\[c_{3}\cosh(1)+c_{4}\sinh(1)=1.\]

Answer 33

Alright, erm.......lol, better go look for hyperbolic identities or properties, or otherwise bust out the exponential definitions of both.

Answer 34

Well, I'm not really familiar with the functions, but apparently

Answer 35

\[\cosh(0)=1, \ \sinh(0)=1.\]

Answer 36

\[c_{3}+c_{4}=0; c_{3}=-c_{4}.\]

Answer 37

\[-c_{4}\cosh(1)+c_{4}\sinh(1)=1; \ \ \ c_{4}(\sinh(1)-\cosh(1))=1.\]

Answer 38

Time to look for identities.

Answer 39

Aaaaaaand nothing of value was gained.

Answer 40

your first equation directly gives you \(c_3 = 0\)

Answer 41

(?)

Answer 42

\[c_{3}\cosh(0)+c_{4}\sinh(0)=0 \]

\[c_{3}(1)+c_{4}(0)=0 \]

\[c_3 = 0\]

right ?

Answer 43

I plugged into google both cosh(0) and sinh(0) and got them to be identically one, not zero, lemme double check on that.

Answer 44

Nah, you're right, understood.

Answer 45

If \[c_{3}=0, \]then \[c_{4}\sinh(1)=1, c_{4}=\frac{1}{\sinh(1)}\]

Answer 46

Which is a really nasty value numerically.

Answer 47

not really, use the definition of hyperbolic functions

Answer 48

Alright, lemme look 'em up, one sec

Answer 49

\[\sinh x = \dfrac{e^x-e^{-x}}{2}\]

Answer 50

\[\cosh(x)=\frac{e^{x}+e^{-x}}{2}, \ \ \ \sinh(x)=\frac{e^{x}-e^{-x}}{2}\]

Answer 51

Oh, and since it's reciprocal

Answer 52

We have like, \[\rm csch(1)=c_{4}\]

Answer 53

\[\sinh(1)=\frac{e^{1}-e^{-1}}{2}=\frac{e}{2}-\frac{1}{2e}\]

Answer 54

So the reciprocal of the entire right side is our answer expressed by the definition.

Answer 55

Looks like we made mistakes with both part A and part B? http://i.imgur.com/OFQN6L9.png

Answer 56

\[y = 0\cosh x +\frac{1}{\sinh(1)} \sinh (x)\]

Answer 57

you need to plug the constants back to the given solution ^

Answer 58

Yeah, nevermind, p1 was good, and you're right, that's always the thing I'm comically worst at remembering to do, the easiest and last part.

Answer 59

Alright, well, cool, problem solved. Going to move on to some more stuff.

Answer 60

Exam in 13 hours.

Answer 61

Ohk dont waste time on part c, its just algebra..