Question

ODE question asking for a general solution and an interval on which the general solution is defined. Posted as soon as the OS thread interface starts working again.

Answer 1

Alright, http://i.imgur.com/Rp7Sdnc.png

Answer 2

That's the prompt (Not going to write it out in LaTeX anymore, lol, time is of the essence)

Answer 3

Okay, seems pretty straightforward. You know the kind of equation and how to approach it?

Answer 4

Thinking about it, half of the time I'm doing this, particularly because most of this stuff I've learned in the last four hours are so, my memory of application is horrific. Give me a second to try to remember it on my own.

Answer 5

kk, no problem :)

Answer 6

Alright, so this is a nonhomogeneous, linear ODE with non-constant coefficients.

First order linear ODE's when put into proper form are best solved with either separation of variables or using an integrating factor, IIRC.

Answer 7

Yep, ya figure its one of the two.

Answer 8

\[x \frac{dy}{dx}+(4x+1)y=x^2e^{4x}\]Wait, this isn't a Bernoulli DE, is it? I don't remember the general form of one but for some reason this is reminding me of a Bernoulli DE.

Answer 9

Nevermind, nevermind

Answer 10

Bernoulli would be a higher power of y on the other side of the equation.

Answer 11

Yeah. Alright, then, either separation of variables or using an integrating factor, one minute.

Answer 12

Going to first put it in the proper form to use an integrating factor by getting y prime to have a coefficient of one.

Answer 13

f(x) y' + g(x) y = h(x)

linear equation

Answer 14

\[\frac{dy}{dx}+\frac{4x+1}{x}y=xe^{4x}\]

Answer 15

Mhm. So definitely the form where you'd have an integrating factor.

Answer 16

Alright, now have to assume that the LHS is the derivative of the product of some mystery function mu, and y.

Answer 17

Oh, nevermind, I need to multiply the whole function by mu first in order to do that.

Answer 18

\[\mu \frac{dy}{dx}+\mu \frac{4x+1}{x}y=\mu xe^{4x}\]

Answer 19

\[\mu '=\mu P(x) = \mu \frac{4x+1}{x}\]

Answer 20

\[\frac{\mu '}{\mu}=\frac{4x+1}{x}\]

Answer 21

\[\int\limits_{}^{}\frac{\mu '}{\mu} d \mu=\int\limits_{}^{}\frac{4x+1}{x}dx\]

Answer 22

\[\ln(\mu)+c_{1}=4x+\ln(x)+c_{2}\]

Answer 23

I know I need to isolate mu by exponentiating both sides, but I start getting confused about how to properly handle these constants of integration, if at all; I know sometimes they are amalgamated together into a single constant.

Answer 24

\[\ln(\mu)+c_{1}=4x+\ln(x)+c_{2}\\
\ln(\mu)=4x+\ln(x)+c_{3}\\
\mu=e^{4x+\ln x+c_3}\\
\quad=Ce^{4x+\ln x}\]

Answer 25

you just need some \(\mu\) so set the constant to 0

Answer 26

Alright, yeah, now I remember how that works, thank you. One sec.

Answer 27

I dont think your professor really wants you to carry out this process in every single problem. you could simply use
\[IF = e^{\int p~dx}\]

Answer 28

He does.

Answer 29

\[\frac{dy}{dx}+p(x)y=q(x)\]\[\mu(x) =e^{\int p(x)\,dx}\]
\[\frac{d}{dx}(\mu(x) y)=\mu(x) q(x)\]

Answer 30

We're not allowed to memorize that formula, and we have to work it that way every single time.

Answer 31

Which is ridiculous and sucks, but it's the way he does it. The book also implores us repeatedly in more than one section, after it's given a definitive proof for how to get there, to do it from the ground up. Every single time.

Answer 32

Because that's soooo much more of a complicated way to think about it. Once you get the coefficient of y' to be 1 in an equation, all you need to do is take the integral of the function of x multiplying y and raise that as an exponent of e. Given an equation in the form: y' + P(x)y = Q(x), the integrating factor is equal to
\[\mu = e^{\int\limits_{}^{}P(x)dx }\] For you, P(x) = 4 + 1/x, so you get
\[\mu = e^{4x + lnx} = xe^{4x}\]

The only constant of integration you need is the last one. Now is when you worry about multiplying by mu. In the end, your integral will always end up being
\[y \mu = \int\limits_{}^{}\mu Q(x)dx\] Your Q(x) is \(xe^{4x}\), so
\[yxe^{4x} = \int\limits_{}^{}x^{2}e^{8x}dx\]

Wait, you have to do it the long way every time? Lame :(

Answer 33

thats a good thing to enforce on grade school kids but im sure that must be frustrating for grown up students lol

Answer 34

Yeah, it is, heh. Because these are freaking timed exams.

Answer 35

The final solution to a first order DE, should have one constant in it. This means you dont have to worry about constants in the integrating factor, because you'll get a constant of integration later anyway

Answer 36

Alright, let me try to pick up where we were, got lost in the convo.

Answer 37

\[\mu = Ce^{\ln(x)+4x}\]

Answer 38

Right, give C your favorite value

Answer 39

\[\int\limits_{}^{}\frac{\mu '}{\mu} d x=\int\limits_{}^{}\frac{4x+1}{x}dx\\
\ln(\mu)=4x+\ln(x)\\
\mu = e^{4x+\ln(x)}\]

Answer 40

Lul "your favorite value"

Answer 41

Yeah, now just remembering how to finish the problem out

Answer 42

Favorite value, hehe
\[\mu = (banana)e^{\ln(x) + 4x}\]

Answer 43

finding the domain might be tricky

Answer 44

Alright, so plugging into the general form of the equation modified to have mu:

Answer 45

yes if banana is not a function of x :P

Answer 46

lol

Answer 47

Banana is an unknown constant D:

Answer 48

\[e^{\ln(x)+4x}\frac{dy}{dx}+e^{\ln(x)+4x}\frac{4x+1}{x}y=e^{\ln(x)+4x}xe^{4x}\]

Answer 49

(Jesus what is this I can't even)

Answer 50

This is such a gross looking problem :C https://33.media.tumblr.com/f8bdf5093c849397f4aca47fbfcf7b9c/tumblr_ndlex59ktC1rdfsmyo1_500.gif

Answer 51

Yeah, I'll always add cheesy humor to my math. If I need imaginary solutions, I consider chair an imaginary solution.

And thats the mess you get when you have to do it all th way through, step-by-step.

Answer 52

WELL

Answer 53

I think what I'm going to do if we get something like this on the exam is do like 90% of the workings up until this point....and then just sort of try to jump to the final result, near the end. lol.

Answer 54

Use the reverse product rule\[\mu(x)\frac{dy}{dx}+\mu(x)p(x)y=\mu(x)q(x)\]
\[\frac{d}{dx}\Big(\mu(x) y\Big)=\mu(x) q(x)\]

Answer 55

Alright, so, I got that, I just again am bad at remembering how to put these solutions in their proper forms once I have them. So what we have is
\[e^{\ln(x)+4x}y=e^{\ln(x)+4x}xe^{4x}\]

Answer 56

(Double checking that)

Answer 57

*derivative* symbol is missing on left side

Answer 58

Nah, I dun goofed, I better be careful, and thanking you for mentioning that, I see what I need to do, I just didn't realize it for a moment, integrate both sides.

Answer 59

\[\int\limits_{}^{}\bigg[\frac{d}{dx}\bigg(e^{\ln(x)+4x}y\bigg)dx\bigg]=\int\limits_{}^{}e^{\ln(x)+4x}xe^{4x}\]

Answer 60

... http://nailsobsession.files.wordpress.com/2012/01/mother-god-meme.jpg

Answer 61

\[\int\limits_{}^{}e^{\ln(x)+4x}xe^{4x}=\int\limits_{}^{}e^{\ln(x)}e^{4x}xe^{4x}=\int\limits_{}^{}x^2e^{8x}\]

Answer 62

Integration by parts? Really? This problem would take *forever* on a test, lord.

Answer 63

it should take less than 10 seconds to come this far if you stop worrying about ur professor

Answer 64

Lul

Answer 65

Yeah, this whole process is fast really. And even the by parts is fast. You know the tabular method for by parts, correct?

Answer 66

Yeah, well, at least I know the technique, I think I'll drop it here for the moment, if I conceptually get how to do it, I can leave the algebra for the test. For now, I'm running out of time on the more crucial thing, the concepts.

T-minus 19 hours, 30 minutes.

No, I mean, I've heard of it, but I'm not really familiar with it and don't use it.

Let me put it this way: I posted this problem 40 minutes ago and *still* haven't solved it, granted I wasn't sure/didn't entirely remember how to 40 minutes ago, lol.

Answer 67

>puts hands under butt, waits for tabular method fancy demonstration

Answer 68

Lol, you sure it wont be cramming extra info? :P

Answer 69

Yeah, you're right, I'll leave it as is, new question, new concept coming up.

Answer 70

Sounds good.

Answer 71

(Also, interval on which solution is defined, I'm going to find an easier/quicker one to do that with.)

HEY, EVERYBODY LOOKING AT THIS PROBLEM:

Answer 72

Do you think this was the right approach, or should I have attempted to solve this DE in another manner/using a different technique?

Answer 73

Yeah, method of integrating factors. Just would be nice if you didnt have to go through all the in between mess.

Answer 74

Alright, cool. Hi-ho, silver, next problem. Thanks, folks.

Answer 75

can you post the interval in which the solutions are defined if psble

Answer 76

^I'm not sure, one minute, focusing on the other problem