Question

How to count initial value for:

f' + 4y = 12e^(2t), y(0) = 5

Answer 1

you mean
y' + 4y = 12e^(2t)
?

Answer 2

yeah, sorry

Answer 3

so you have the general solution already?

Answer 4

well, thats all I know, and I want to know the initial value

Answer 5

y(0)=5
is what the initial value is here
in general, y(0)=whatever
is the initial value

Answer 6

okay, I want to know y(t)

Answer 7

or how to count it, I know the answer

Answer 8

that means you want to solve the IVP
this is a linear equation, so you can use an integrating factor
have you tried that?

Answer 9

yeah, i want to solve the Initial value problem, but i do not understand what you mean with "you can use an integrating factor"

Answer 10

do you know any techniques on how to solve linear DE's ?
I don't know where to start if you have never heard of an integrating factor
perhaps it's just a difference in terminology, I do not know your what native language is
does\[\large\mu(t)=e^{\int p(t)dt}\]look familiar as an integrating factor?

Answer 11

Well my native language is swedish but I study in Finnish :P

OK, I think that I should learn that formula; what does it mean?

Answer 12

this is really best explained through reading up on it
here is a good link
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx
I suggest you read it, along with perhaps other sections of that site, in order to understand this type of problem

Answer 13

I tried to search that site for help but did not find a fitting article, Thanks! Will read it.

Answer 14

if you read that page (which you could not possibly have done in this amount of time) you would see that it shows you exactly how to solve this kind of problem
it even derives how it works

look, I will work through this one with you, but read the page after!

Answer 15

one sec...

Answer 16

in general, say we have the form\[y'+p(t)y=q(t)\]that is called a linear DE
the way this is solved is by first multiplying the whole thing through by something called an integrating factor \(\mu(t)\) defined as\[\large\mu(t)=e^{\int p(t)dt}\]still with me?

Answer 17

I guess..

Answer 18

well let's try it:
your problem is\[y'+4y=12e^{2t}\]so what are \(p(t)\) and \(q(t)\) in this case?

Answer 19

p(t) is 4 and q(t) is 12e^(2t)?

Answer 20

exactly :D
now since I have given you a formula for the integrating factor \(\mu(t)\) you should be able to tell me what that is too
what is it?

Answer 21

µ(t) = e^(4t)?

Answer 22

excellent!
now multiply the whole equation by that integrating factor and show me what you get
(write both sides of the expression please)

Answer 23

do you mean: e^(4t) * y' + 4ye^(4t) = 12e^(2t) * e^(4t)

Answer 24

yes, which we can simplify to\[e^{4t}y'+4ye^{4t}=12e^{6t}\]right?
now here is here we need to be observant to understand the following step...

Answer 25

yeah

Answer 26

look carefully at the left-hand-side
notice that it can be written as a result of the product rule\[e^{4t}y'+4ye^{4t}=\frac d{dt}(e^{4t}y)\]so we can rewrite this whole expression as\[\frac d{dt}(e^{4t}y)=12e^{6t}\]this will \(always\) be the case when we do this integrating factor technique!

Answer 27

(in general we will always get the form\[\frac d{dt}(\mu(t)y)=q(t)\mu(t)\]from which we will now proceed)
still with me up to\[\frac d{dt}(e^{4t}y)=12e^{6t}\]?

Answer 28

this, i did not understand... thinking...

Answer 29

yes, this is the tricky bit I suppose...
-you were with me up to\[e^{4t}y'+4ye^{4t}=e^{6t}\](after multiplying by the integrating factor)
right?

Answer 30

12e^6t, yes

Answer 31

^right, my bad
-and just looking at the LHS, do you not agree that (by what may seem to be a pure coincidence) it is a fact that\[e^{4t}y'+4ye^{4t}=\frac d{dt}(e^{4t}y)\]?

Answer 32

nooo....?

Answer 33

product rule:\[(uv)'=u'v+v'u\]let \(u=e^{4t}\) and \(v=y\)
\[(e^{4t}y)'=(y)'e^{4t}+y(e^{4t})'=y'e^{4t}+4ye^{4t}\]hence\[y'e^{4t}+4ye^{4t}=(e^{4t}y)'\]

Answer 34

\[y(e^{4t})' \] be \[4tye^{4t}\] ?

Answer 35

chain rule:\[(e^{4t})'=(e^{4t})'(4t)'=4e^{4t}\]

Answer 36

sorry ofcourse :)

Answer 37

it happens :)

Answer 38

ok, I'm with you

Answer 39

so...
now you comprehend\[e^{4t}y'+4ye^{4t}=\frac d{dt}(e^{4t}y)\]which setting equal to the RHS again gives\[\frac d{dt}(e^{4t}y)=12e^{6t}\]now this is a separable DE, which you probably have done in calc I

Answer 40

I think it is best just to refresh your memory and demonstrate the next few steps, so....

Answer 41

\[\frac d{dt}(e^{4t}y)=12e^{6t}\]\[d(e^{4t}y)=12e^{6t}dt\]\[\int d(e^{4t}y)=12\int e^{6t}dt\]now note the LHS, being the integral of a differential, just becomes the integrand.
so we are left with\[e^{4t}y=12\int e^{6t}dt\]so once we can do the integral we are very close to solving for y!

why don't you try to do the integral and solve for y yourself?

Answer 42

e^(4t) * y = 12 * 6 e^(6t)

y = 72*e^(2t)

Answer 43

gotta be way more careful with that calculus\[\int e^{6t}dt\neq6e^{6t}\]and perhaps even more important, you forgot to add the constant +C from the indefinite integral
DE's is where that constant you learn to add in calc I really comes into play
try that integral again...

Answer 44

e^(4t) * y = 12 * (1/6) e^(6t) + C

y = 2e^(2t) + C*e^(-4t)

Answer 45

and that is the general solution!
nice job :D

Answer 46

c = 3

Answer 47

y(t) = 2e^(2t) + 3e^(-4t)

Answer 48

yes :)

Answer 49

Thank you so much for your time and patience!

Answer 50

yeah I had model solutions but they were crappy :)

Answer 51

you're welcome

try again to read the link I gave you, it has great explanations and example problems

note that there are many more constants of integration that poped up along the way, but they can be combined at the end if you know the situation you are dealing with well enough, so I ignored them

in general though, remembering to add the constant of integration for every indefinite integral is indispensable in differential equations