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.

Options

Pages: 1

Back to forum
Please log in to reply to this topic or to report it.

Members in this topic: None.

[12:03 am]

Pr0nogo -- Suicidal InsanitySuicidal Insanity shouted: But dat files probably would be a massive pain to support, so unlikely I would ever do that

would be nice to at least have the option to edit player map settings for ids past the default limit, but I guess we have ways around that already[11:01 pm]

Corbo -- Suicidal InsanitySuicidal Insanity shouted: Pr0nogo And custom color cycling !

and custom parallax support tbh[10:29 pm]

Suicidal Insanity -- But dat files probably would be a massive pain to support, so unlikely I would ever do that[10:28 pm]

Suicidal Insanity -- Pr0nogoPr0nogo shouted: he's already added expanded tileset functionality which requires an external plugin

And custom color cycling ![07:33 pm]

Pr0nogo -- meant to quote poiuy_qwertpoiuy_qwert shouted: There are some limit expander plugins floating around, at least one that allows expanded dat files