Question

Good morning OpenStudy! I have a few questions about autonomous differential equations, such as dx/dt=f(x). I understand how to get the critical points given f(x)=0. I can generate a slope field. I'm not sure how to get the explicit solution for x(t). I'll attach a screenshot of a problem like the one I'm working on.

Answer 1

Even as I'm posting this, I think I know what to do now. I think working with the integrating factor, rho(x), I'll get the answer I'll need. I'm still a little clear as to how I'm supposed to infer the exact solution from the slope field, though. Thoughts?

Answer 2

to get the explicit solution, use separation of variables

Answer 3

either approach actually works, you can also use integrating factor

Answer 4

but its faster to do separation of variables
dx/ (x-4) = dt
integrate both sides

Answer 5

Ok. I'll try separation first. Just a sec.

Answer 6

Ok, so here's what I have to this point.

Answer 7

good so far

Answer 8

may i ask what program you are using to draw that (thats pretty cool )

Answer 9

Photoshop. :)

Answer 10

why is the background black

Answer 11

It's just how I set it. I could make it just about anything.

Answer 12

cool :)

Answer 13

i have microsoft paint,

Answer 14

so youre using adobe photoshop?

Answer 15

now raise both sides to the power of e

Answer 16

Yup. I've been using it since 2.0, but only over the last few years as a math tool. I draw with it too.

Ok, so the solution my book shows is x(t)=4+(x_0-4)e^t. I'm not sure how to get there.

Answer 17

it should be
ln | x - 4| = t + c , because we are finding the general antiderivative

Answer 18

Hmm...the problem says that after finding f(x) and analyzing its sign, to find the explicit solution, either using a slope field or the exact solution.

Answer 19

I got to the same answer as you for the general solution, though.

Answer 20

i might be missing some information
you wrote f(x) = 4 ?

Answer 21

Yeah, so we're starting from \[\frac{\text{dx}}{\text{dt}}=x-4\] where f(x)=0=(x-4)

Answer 22

ok so we have
dx/dt = f(x)
where f(x) = x-4

it doesnt make sense to say f(x) = 0. If that were true then you have
dx/dt = 0

Answer 23

To solve the differential equation explicitly you need to be given a point x(t*) = c
If t = 0, then you might see it as
x(0) = xo

Answer 24

Actually, it's autonomous, so f(x) isn't dependent...I'm not sure I completely understand that. Maybe it'll help if I give the verbatim instructions.

Answer 25

autonomus differential equation just means you can get the differential equation in the form :
dy/dt = f(y)
or
dx/dt = f(x)

Answer 26

one moment, reading this :)
http://tutorial.math.lamar.edu/Classes/DE/EquilibriumSolutions.aspx

Answer 27

Ah, I should have gone there as well. :)

Answer 28

This is helping out:

https://www.youtube.com/watch?v=KgpUyKUC85Q

Answer 29

To solve it explicitly, i think they mean find the solution x(t) in terms of t.
I don't think it means find a particular solution since we were not given an initial condition. For example if you remember the function y = x^2 , y is an explicit function of x. But in the expression x^2 + y^2 = 1, y is a function of x implicitly. To make it explicitly you have to solve for y.

dx/dt = x - 4
dx/(x-4) = dt
ln|x - 4 | = t + c
x-4 = e^(t+c)
x -4 = C*e^t
x = C*e^t + 4

Therefore the explicit solution is

x(t) = Ce^t + 4

Answer 30

are you using mathematical to generate that slope field, or matlab.
I have maple.
x = 4 is an unstable equilibrium solution

Answer 31

Hmm.. They have it as:\[x(t)=4+\left(x_0-4\right) e^t\]

Answer 32

I'm using Mathematica. Here's the slope field I got.

Answer 33

The 4+ piece makes sense to me. We're looking for the behavior of plots slightly more or less than 4.

Answer 34

That is the explicit solution if you are given the initial condition
x(0) = x_o

dx/dt = x - 4
dx/(x-4) = dt
ln|x - 4 | = t + c
x-4 = e^(t+c)
x -4 = C*e^t

substitute x(0)=x_0
x_o - 4 = C*e^(0)

x_o -4 = C

so the solution is


x(t) = 4 + (x_o -4)*e^t

Answer 35

i guess thats a fair assumption, since you are often given such an initial condition
x(0)=x_0
or
y(0)= y_0

Answer 36

but our solution was equivalent
x(t) = 4 + (x_o -4) * e^t
is equivalent to
x(t) = 4 + C*e^t
since ' x_o -4 ' is a constant

Answer 37

Ah, finally a crack in the armor! So we use x_0 as our initial condition to find C. Excellent!

Genius award goes to .... perl. :)

Answer 38

right
The initial condition is often given as
x(0)=x_o
or
y(0)= y_o

Answer 39

Man, there wasn't a thing in the book that would have pointed me in that direction. Thanks for working me through that, I really appreciate it.

Answer 40

x_o is the value x(t) when t = 0 , this is a shorthand i guess :)

Answer 41

Cool, now I'm going to try one of the problems from the actual assignment and see how I fare. It ought to be a lot easier now. (Whew!)

Answer 42

:)

Answer 43

what is the name of the book you are using

Answer 44

It's Linear Algebra & Differential Equations, Edwards and Penny, but it's one of those custom jobs from Pearson for the University of Utah. I kind of wish they'd gone with the open-source book by Strang, would have saved me about $200.

Answer 45

yeah and i bet last year editions can be bought for 20 dollars

Answer 46

Right? I can't stand the textbook cartels. They charge you caviar prices and feed you from the garbage. As a non-traditional student, I'm probably a lot less complacent about it. At this point, I want to finish the physics degree so I can subvert the corruption by helping flood the market with free help. It's part of why I'm a huge fan of this site.

Answer 47

:) i like that philosophy

Answer 48

end the textbook cartels

Answer 49

A-MEN. :)

Answer 50

im gonna plug in your function into mathematica , one moment

Answer 51

the code you gave me

Answer 52

or could you copy and paste the code in here , the vector and stream

Answer 53

Cool. I'll probably superimpose the solution on top of the field, but here are two ways to do it:

ClearAll["`*"]
VectorPlot[{t, (x - 4)}, {t, 0, 10}, {x, 0, 10}]
StreamPlot[{t, (x - 4)}, {t, 0, 10}, {x, 0, 10}]

Answer 54

thanks, did you teach yourself mathematica? i would like to learn it, (i have mathematica 10)

Answer 55

Yeah, I've been struggling to learn it. My physics teacher told us we should learn it because it is "the industry standard". Of course whenever I've asked him simple questions about it, he's blanked out and solved things for me on paper instead. I might be the only one in class that has any grasp of it.

Answer 56

If I hadn't been a programmer for the last 15 years...well, I'd be screwed.

Answer 57

oh ...
yeah mathematica's coding looks pretty strange , i have some experience with c++ , i can write simple programs

Answer 58

Yeah, my bg is in PHP, C#, Python. If you have any programming experience at all, Mathematica will make you feel like you're on drugs. I like the markup, though. It can make it a lot easier to visually inspect an algorithm.

Answer 59

what is bg ?

Answer 60

Background. Sorry, my inner web developer is surfacing.

Answer 61

:)

Answer 62

also the mathematica output is strange, do you see how the vectors get bigger from left to right

Answer 63

That makes sense, because as t increases, x increases exponentially.

Answer 64

yes but we want the slope field , the derivative or slope at each point

Answer 65

check this out
http://stackoverflow.com/questions/8905726/how-do-i-plot-a-slope-field-using-mathematica

Answer 66

Ooh, that's nice. I looked at a few examples on StackExchange, but that's a lot better.

Answer 67

yeah i wonder what the difference between that and the code you sent me

Answer 68

I want to try it out, but I'm worried I'll never finish this assignment as is. :(

Answer 69

I'm going to need to check out for a few.

Answer 70

ok , ill probably be on later, nice meeting ya

Answer 71

or leave me a pm

Answer 72

Nice meeting you, too. Thanks again. Btw, I just solved the first one, and it worked. (Hooray!)