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

at y(1) = 0

gogogo

None.

Free kisses from me

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

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

at y(1) = 0

gogogo

None.

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

None.

Well I have no clue what it is but I want a personal shout out from you anyway for trying!! PLZ!!

None.

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.

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.

None.

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.

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.

And ur answer is nowhere even close

clue: to solve this, you have to be able to know how to do integrals ;P

None.

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.

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.

None.

Quote from Akar

never the less my first statement is correct, even if it isn't what your asking.

Quote

Yuck, I fail at isolating y and dy, t and dt.

None.

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

None.

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.

I'm sure I'll get it once I see the solution.

None.

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)))

===================

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)))

===================

None.

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?

Very crafty

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

None.

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.

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?

Very crafty

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

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.

None.

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)))

===================

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.

None.

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[10:18 pm]

Suicidal Insanity -- On game start you could detect who is at which start location, store that in switches / dcs, and use that to permanently force alliances. Lots of combinations though[09:43 pm]

MTiger156 -- That 2v2v2v2 mapping problem does sound like an interesting challenge. Its worth putting some experimentation into. I could make an example map for combining locked alliances (based on lobby forces) + random start location. Though an EUD solution would be the most efficient, I will avoid that for the sake of simplicity.[08:52 pm]

NudeRaider -- pallfypallfy shouted: thanks, but that sounds complicated

well, that's the gist of sc mapping. Either accept the limits sc sets for you and work within those parameters, or get creative and find complicated solutions to achieve what you want. The latter can be most satisfying. Sometimes comprised are inevitable though. Also see my respnse in the thread.[07:18 am]

NudeRaider -- FaRTy1billionFaRTy1billion shouted: also ya, I showed her my chair wasn't scary and now whenever I'm not in the chair she is

hard to believe that they are predators with almost no enemies in the wild[07:17 am]

NudeRaider -- Suicidal InsanitySuicidal Insanity shouted: http://photos.arlinghaus.org/Animals/Rheine/2019.03.19-ESK_4355-QHD-%7B38EF16AE-6255-4C3B-8034-F92BC3689132%7D.jpg <- "I want a name "

obviously francis
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