 Staredit Network > Forums > Null > Topic: Free kisses from me Free kisses from me
Sep 11 2007, 12:35 am
By: MillenniumArmy

Sep 11 2007, 12:35 am MillenniumArmy Post #1   Free kisses from me if you are the first one to find the solution of the given First order differential equation of the given initial value problem:

y'+2y = te^(-2t)
at y(1) = 0

gogogo

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Sep 11 2007, 12:35 am blacklight28 Post #2   Is this just like your homework or something?

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Sep 11 2007, 12:40 am MillenniumArmy Post #3   Oh I already know how to do this. I just feel like giving someone an incentive to receive a nice kiss from me. And perhaps, even a personal shout out from me in my mini-game party map =-O

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Sep 11 2007, 12:41 am blacklight28 Post #4   Well I have no clue what it is but I want a personal shout out from you anyway for trying!! PLZ!!

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Sep 11 2007, 1:02 am Akar Post #5   whats the y' mean, is that just y?
damnit calculus! You asked 3 years too early >.>
Heres what I can do:
2y = 0
y = 0

y' + 2y = te ^ (-2t)
---------------------
2y
and thus no solution xD because y = 0

OR:

0' + 2(0) = te ^ (-2t)
0 = te ^ (-2t) < I'm assuming 0' is a function of multiplication
1/t = e ^ (-2t)
1/t ^ (-2t) = e < I'm guessing on this >.>
0 = t[ 1/t ^ (-2t)]
0 = 1 ^ (-2t)
I'm guessing I did that wrong becuase nothing in squares = 0.

Post has been edited 4 time(s), last time on Sep 11 2007, 1:22 am by Akar.

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Sep 11 2007, 1:49 am MillenniumArmy Post #6   Quote from Akar
whats the y' mean, is that just y?
damnit calculus! You asked 3 years too early >.>
Heres what I can do:
2y = 0
y = 0

y' + 2y = te ^ (-2t)
---------------------
2y
and thus no solution xD because y = 0

OR:

0' + 2(0) = te ^ (-2t)
0 = te ^ (-2t) < I'm assuming 0' is a function of multiplication
1/t = e ^ (-2t)
1/t ^ (-2t) = e < I'm guessing on this >.>
0 = t[ 1/t ^ (-2t)]
0 = 1 ^ (-2t)
I'm guessing I did that wrong becuase nothing in squares = 0.
It means what it means; y prime.

And ur answer is nowhere even close clue: to solve this, you have to be able to know how to do integrals ;P

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Sep 11 2007, 1:51 am Akar Post #7   never the less my first statement is correct, even if it isn't what your asking.

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Sep 11 2007, 2:22 am DT_Battlekruser Post #8   Yuck, I fail at isolating y and dy, t and dt.

Too tired to give it any really serious thought. (I assume y' = dy/dt)

EDIT ~ Looked a few things up online - I haven't had Differential Equations yet (or Calculus III). As far as I can tell, a simple two-semester knowledge of calculus isn't sufficient.

Post has been edited 1 time(s), last time on Sep 11 2007, 2:40 am by DT_Battlekruser.

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Sep 11 2007, 2:39 am MillenniumArmy Post #9   Quote from Akar
never the less my first statement is correct, even if it isn't what your asking.
No even your very first statement is incorrect because you're not using the given info correctly. You're probably thinking that "y(1) = 0" means "y = 0." y(1) = 0 means that when the variable t equals 1, then the variable y equals 0, but you shouldn't even use this information after integrating the whole equation.

Quote
Yuck, I fail at isolating y and dy, t and dt.
It's not about isolating any of those you just mentioned; for problems like this you're suppose to only isolate the y' (or in other words dy/dt) but in this case it's already isolated, so the next thing you're suppose to do is to multiply both sides by an unknown function (which you have to figure out how to do) that will make the left side of the equation integrateable (meaning after integration y' will not exist) None.

Sep 11 2007, 2:41 am DT_Battlekruser Post #10   Yeah, I looked it up, I haven't learned how to do that yet.

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Sep 11 2007, 2:42 am MillenniumArmy Post #11   Well technically all you need is calculus BC to solve this. You only need to use differentials and mainly integrals to solve this. The thing is you just gotta know how to apply them lol

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Sep 11 2007, 2:45 am DT_Battlekruser Post #12   Well, I'll wait for someone with more time to come solve it I'm sure I'll get it once I see the solution.

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Sep 11 2007, 2:45 am cheeze Post #13   y + y^2 = integral of that

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Sep 11 2007, 2:54 am Akar Post #14   damn foiled again.
a = b
a^2 = ab
a^2 -b^2 = ab - b^2
(a-b)(a+b) = b(a-b)
a+b = b
2b = b
2 = 1

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Sep 11 2007, 9:57 pm Dr. Shotgun Post #15   Ask me next year.

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Sep 12 2007, 8:15 am MillenniumArmy Post #16   Boo. No kisses from me then.

But anyways, here's the solution:

---------------------------------
y'+2y = te^(-2t)
at y(1) = 0
---------------------------------

y'+2y = te^(-2t)
becomes:

m(t)y'+m(t)2y = m(t)te^(-2t)

now m(t) is an unknown function, so we have to solve it so that the left side of the equation (the y'+2y) becomes integrateable and have the y' be eliminated so we do this:

m(t) = e^( S2dt) = e^(2t)
so now we have:

(e^(2t))y'+(e^(2t))2y = (e^(2t))te^(-2t)

now before you integrate both sides, you have to eliminate the y'. And to do this, you just do the product rule for differentiation backwards (this is not the same thing as integration). So you end up with:

(d/dt)ye^(2t) = t (the e^(2t) and e^(-2t) on the right side cancel each other out)

Now when you integrate both sides, you get:
ye^(2t) = (t^2)/2 + c

Isolate the y now and you get:
y = ((t^2)/2 + c)/(e^(2t))

------------------------------------

now at the beginning, we were given y(1) = 0. Basically this means that when t = 1, y is 0, so let's plug those into the new equation:
0 = ((1^2)/2 + c)/(e^2t))
solve for c algebraically, and you end up with:
c = -1/2
now replace c with -1/2 and you have your answer:

===================
y = (t^2 - 1)/(2(e^(2t)))
===================

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Sep 12 2007, 10:17 pm DT_Battlekruser Post #17   It makes sense, but I was definitely never taught that.

Very crafty How did you find m(t), or was it just looking and seeing it intuitively?

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Sep 12 2007, 11:14 pm Symmetry Post #18 Dungeon Master  Quote from Akar
(a-b)(a+b) = b(a-b)

aa+ab-ab-bb = ba-bb

0=0? I think you expanded it wrong, bud.    Sep 13 2007, 9:04 pm MillenniumArmy Post #19   Quote from DT_Battlekruser
It makes sense, but I was definitely never taught that.

Very crafty How did you find m(t), or was it just looking and seeing it intuitively?
The equation in my very first post was in the form of:

dy/dt + ay = g(t)

so a was 2 in that case.

Well this is actually something new that was taught to us in class, but pretty much the trick to this is that to find m(t), you set m(t) = e^(integral of a)

This only works for simple equations like this, where it's already in the perfect form.

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Sep 14 2007, 1:34 pm Paravin. Post #20   Quote from MillenniumArmy
Boo. No kisses from me then.

But anyways, here's the solution:

---------------------------------
y'+2y = te^(-2t)
at y(1) = 0
---------------------------------

y'+2y = te^(-2t)
becomes:

m(t)y'+m(t)2y = m(t)te^(-2t)

now m(t) is an unknown function, so we have to solve it so that the left side of the equation (the y'+2y) becomes integrateable and have the y' be eliminated so we do this:

m(t) = e^( S2dt) = e^(2t)
so now we have:

(e^(2t))y'+(e^(2t))2y = (e^(2t))te^(-2t)

now before you integrate both sides, you have to eliminate the y'. And to do this, you just do the product rule for differentiation backwards (this is not the same thing as integration). So you end up with:

(d/dt)ye^(2t) = t (the e^(2t) and e^(-2t) on the right side cancel each other out)

Now when you integrate both sides, you get:
ye^(2t) = (t^2)/2 + c

Isolate the y now and you get:
y = ((t^2)/2 + c)/(e^(2t))

------------------------------------

now at the beginning, we were given y(1) = 0. Basically this means that when t = 1, y is 0, so let's plug those into the new equation:
0 = ((1^2)/2 + c)/(e^2t))
solve for c algebraically, and you end up with:
c = -1/2
now replace c with -1/2 and you have your answer:

===================
y = (t^2 - 1)/(2(e^(2t)))
===================

I think my brain just exploded.

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