Question

Diff EQ IVP: Non-elementary integration

Answer 1

\[\frac{ d ^{2}y }{ dx ^{2} }=e ^{-x ^{2}}\]
\[ y'(3) = 1, y(3) = 5 \]

Answer 2

Non-elementary integration obviously :/

Answer 3

Or is this one of those problems where you have to do some general solution or....something like a taylor series approximation? O.o

Answer 4

HELP ME WITHMY PROBLEM http://openstudy.com/study#/updates/52201a9ae4b0750826e1352a

Answer 5

@Loser66

Answer 6

yo are the one that did differentials like 2 days ago!!

Answer 7

ive forgotten all about diff equation methods!!

Answer 8

2nd order non homogenous equation so

Answer 9

you could use that formula

Answer 10

the variation of paremeter formula

Answer 11

solve y''=0
gets your general equation and plug into variation of paremeter formula

Answer 12

Yeah, problem is Im not allowed to do that, haha. I mean, there ARE methods, but in regards to where we've advanced in the class we don't have that option.

Answer 13

http://www.youtube.com/watch?v=fKP-qFY-L9g

Answer 14

oh i see okay

Answer 15

i know then!! assume solution

Answer 16

no i dont remember!!

Answer 17

what are the other ways asgain?

Answer 18

there undetermined coeff, cauchy way, the e^mx way, and the all working variation of parmenets and wronskin way

Answer 19

Yeah, I thought power series, too.....but then I thought that was cheating, haha.
Well, this is the beginning of the semester, so we're not far at all. I mean, its all review for me until probably halfway through the semester, so we're barely into integrating factors. Problem is this is hw and he never went over this.

Answer 20

which method u learn till that will help

Answer 21

None, haha. I mean, this question is even in a section of the textbook where theres no crazy techniques shown. This is a chapter 2 question, haha. Thats why I was thinkin gmaybe power series because what else could we possibly know as a way to solve it when we're only in chapter 2? x_x

Answer 22

Yes it is.

Answer 23

Then ol' prof be trollin me xD

Answer 24

Yeah, this was hw given to us after the 3rd lecture, he just didnt go over it yet.

Answer 25

Ah x_x Then I think that would confirm theres some sort of old method, power series or somesort of just....general form that isnt a full solution, dunno.

Answer 26

Lol, well Im trying to think of everything Ive done before that would make any sense.

Answer 27

No worries ^_^ Sorry this is such a funky question.

Answer 28

Some piece of crap one, lol. "A First Course in Differential Equations With Applications" Author zill.

Answer 29

err....Theres two, ill just put both:

!SBN-13: 978-1-111-82705-2
ISBN-10: 1-111-82705-2

Answer 30

You found it? :o

Answer 31

Can you show me where by chance?

Answer 32

Howd you get that? xD

Answer 33

You just bought it? O.o

Answer 34

any progress on this yet?

Answer 35

why dont you solve with power series i want to see that solution

Answer 36

Yeah. Looks like loser has a solution manual and I have an example x_x

Answer 37

kk

Answer 38

Ex:

\[\frac{ dy }{ dx }=x ^{x ^{2}}; y(0) = 3\]

Answer 39

I wanted to show this before I just took the solution >.<

Answer 40

You know that in this level, people don 't give you step by step. They jump from this part to another part. My prof, in class, jumps as if we, students, are his Ph.D classmates. ha!!!

Answer 41

@SithsAndGiggles help him, please. ha!! you are lucky.

Answer 42

\[\int\limits_{0}^{x}\frac{ dy }{ dt }=\int\limits_{0}^{x}t ^{t ^{2}} \]

y(t) limits x to 3:
\[y(x) - y(0) \]
\[y(x) = 3 + \int\limits_{0}^{x}t ^{t ^{2}} \]

Make any sense?

Answer 43

That was the example I found on it.

Answer 44

you have SithAndAngle here, I have nothing to do. hehehe.. I need sleep.

Answer 45

Lol, night then xD

Answer 46

Ooops, typo on the limits in the example obviously x_x

Answer 47

The integral parts are correct, just what I actually typed is off, I meant 0 to x, lol.

Answer 48

For this kind of problem, I remember learning this formula: Given some initial condition \(y(x_0)=y_0\), the solution to the following DE is
\[y'=f(t)~~\Rightarrow~~y=\int_{x_0}^xf(t)~dt+y_0 \]
So, extending this to the second order case, you'd have
\[\frac{d^2y}{dt^2}=e^{-x^2}\]
\[\frac{dy}{dt}=\int_{x_0}^xe^{-x^2}~dx+y'_0\]
Here, \(y'(x_0)=y'_0\), so \(x_0=3\) and \(y'_0=1\).

Similarly,
\[y(t)=\int_{x_0}^x\left(\int_{x_0}^xe^{-x^2}~dx+y'_0\right)~dx+y_0\]
And here, \(x_0=3\) again, and \(y_0=5\), as per initial conditions.

Answer 49

Oh wow O.o So just kind of a general solution method. Not supposed to truly get a full answer. Yeah, Ill write that down, that's awesome! ^_^ Thanks.

Answer 50

You're welcome