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standelds
science forum addict
Joined: 09 Sep 2005
Posts: 55
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 Posted: Sun May 21, 2006 12:13 am Post subject:
Initial Value Problem
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Solve the IVP:
L dI/dt + RI + q/c = E(t); q(0) = some initial value of q, and I(0) = some
initial value of I.
where L is the inductance in Henrys, R is the resistance in ohms, C is the
capacitiance in farads, E(t) is the EMF in volts, q(t) is the charge in
coulombs on the capacitor at time t, and I = dq/dt is the current in
amperes. Find the current at time t if the charge on the capacitor is
initally zero, L =10H, R = 20 ohms, C = (6260)^-1 F, and E(t) = 100V.
[HINT: Differentiate both sides of the original equation to obtain a
homogeneous linear second order diff eq for I(t). Then use the original
equation todetermine dI/dt at t=0]
Here's my work so far:
d/dt (10 dI/dt + 20 I + 6260 q )= d/dt (E(t))
10 I'' + 20 I' + 6260 I = 0
Auxiliary equation: 10 r^2 + 20 r + 6260 = 0
r = -1 +/- 500 i
a = -1; B = 500
I(t) = c1 e^(-t) cos(500t) + c2 e^(-t) sin(500t)
I'(t) = e^(-t) [ 500 c1 sin(500t) - c1 cos(500t) + 500 c2 i cos(500t) - c2 i
sin(500t)
q(0) = 0
So where do I go with this last bit from the problem statement? How do I
continue the problem?
Any help you can give will be most appreciated.
Dustin |
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William Elliot
science forum Guru
Joined: 24 Mar 2005
Posts: 1906
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| Posted: Sun May 21, 2006 9:49 am Post subject:
Re: Initial Value Problem
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From: standelds <standelds@hawaii.rr.com>
Newsgroups: alt.math.undergrad
Subject: Initial Value Problem
| Quote: | Solve the IVP:
L dI/dt + RI + q/c = E(t); q(0) = some initial value of q,
and I(0) = some initial value of I.
where L is the inductance in Henrys, R is the resistance in ohms, C
is the capacitiance in farads, E(t) is the EMF in volts, q(t) is the
charge in coulombs on the capacitor at time t, and I = dq/dt is the
current in amperes. Find the current at time t if the charge on the
capacitor is initally zero, L =10H, R = 20 ohms, C = (6260)^-1 F, and
E(t) = 100V. [HINT: Differentiate both sides of the original equation
to obtain a homogeneous linear second order diff eq for I(t). Then
use the original equation todetermine dI/dt at t=0]
Here's my work so far:
d/dt (10 dI/dt + 20 I + 6260 q )= d/dt (E(t))
10 I'' + 20 I' + 6260 I = 0
|
Oh??? E(t) is a constant E ?
| Quote: | Auxiliary equation: 10 r^2 + 20 r + 6260 = 0
r = -1 +/- 500 i
|
No. r^2 + 2r + 626 = 0
r = (-2 +- sqr(4 - 4*626))/2 = -1 +- sqr(1 - 626) = -1 +- 25i
??? Some in class jargon?
| Quote: | I(t) = c1 e^(-t) cos(500t) + c2 e^(-t) sin(500t)
I'(t) = e^(-t) [ 500 c1 sin(500t) - c1 cos(500t)
+ 500 c2 i cos(500t) - c2 i sin(500t)
|
No. There's no reason for i sqr -1
I'(t) = -I(t) + e^(-t)(-500 c1.sin 500 t + 500 c2.cos 500t)
BTW, correct for previous mistake, 25 instead of 500.
| Quote: | q(0) = 0
So where do I go with this last bit from the problem statement?
How do I continue the problem?
Any help you can give will be most appreciated.
|
What's next? Find c1 and c2? You need some values,
like for example I(0) and I'(0).
q(0) = 0 does nothing because upon solving
I(t) = dq(t)/dt
for q(t), you've introducted another constant which is
determined by q(0) = 0, in terms of c1 and c2.
---- |
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standelds
science forum addict
Joined: 09 Sep 2005
Posts: 55
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| Posted: Sun May 21, 2006 7:05 pm Post subject:
Re: Initial Value Problem
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"William Elliot" <marsh@hevanet.remove.com> wrote in message
news:Pine.BSI.4.58.0605210248350.7697@vista.hevanet.com...
| Quote: | From: standelds <standelds@hawaii.rr.com
Newsgroups: alt.math.undergrad
Subject: Initial Value Problem
Solve the IVP:
L dI/dt + RI + q/c = E(t); q(0) = some initial value of q,
and I(0) = some initial value of I.
where L is the inductance in Henrys, R is the resistance in ohms, C
is the capacitiance in farads, E(t) is the EMF in volts, q(t) is the
charge in coulombs on the capacitor at time t, and I = dq/dt is the
current in amperes. Find the current at time t if the charge on the
capacitor is initally zero, L =10H, R = 20 ohms, C = (6260)^-1 F, and
E(t) = 100V. [HINT: Differentiate both sides of the original equation
to obtain a homogeneous linear second order diff eq for I(t). Then
use the original equation todetermine dI/dt at t=0]
Here's my work so far:
d/dt (10 dI/dt + 20 I + 6260 q )= d/dt (E(t))
10 I'' + 20 I' + 6260 I = 0
Oh??? E(t) is a constant E ?
|
How else would you handle it? I made the assumption based on the
information given in the problem that E(t) = 100V.
| Quote: |
Auxiliary equation: 10 r^2 + 20 r + 6260 = 0
r = -1 +/- 500 i
No. r^2 + 2r + 626 = 0
r = (-2 +- sqr(4 - 4*626))/2 = -1 +- sqr(1 - 626) = -1 +- 25i
|
Yep, dropped the division by 20 in that half of the quadratic formula.
| Quote: |
a = -1; B = 500
??? Some in class jargon?
|
The formula for complex conjugate roots from the book:
for a diff eq ay'' + by' + cy = 0 with auxiliary eq ar^2 + br +c = 0 where r
has complex conjugate roots:
alpha +/- beta * i
two linearly independant solutions are:
e^(alpha*t) cos(beta * t) and e^(alpha*t) sin(beta*t)
then a general solution is:
y(t) = c1 e^(alpha*t) cos(beta * t) + c2 e^(alpha*t) sin(beta*t)
where c1 and c2 are arbitrary constants.
however I don't have alpha or beta characters on my keyboard. So I use a
and B since they are kind of similar.
| Quote: |
I(t) = c1 e^(-t) cos(500t) + c2 e^(-t) sin(500t)
I'(t) = e^(-t) [ 500 c1 sin(500t) - c1 cos(500t)
+ 500 c2 i cos(500t) - c2 i sin(500t)
No. There's no reason for i sqr -1
I'(t) = -I(t) + e^(-t)(-500 c1.sin 500 t + 500 c2.cos 500t)
BTW, correct for previous mistake, 25 instead of 500.
|
Yes. I see that now. I don't know how many times I reworked this problem
and didn't see that mistake.
| Quote: |
q(0) = 0
So where do I go with this last bit from the problem statement?
How do I continue the problem?
Any help you can give will be most appreciated.
What's next? Find c1 and c2? You need some values,
like for example I(0) and I'(0).
q(0) = 0 does nothing because upon solving
I(t) = dq(t)/dt
for q(t), you've introducted another constant which is
determined by q(0) = 0, in terms of c1 and c2.
----
Okay. however I have typed the problem above. Even copied the hint. How |
do you get the values? |
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standelds
science forum addict
Joined: 09 Sep 2005
Posts: 55
|
| Posted: Sun May 21, 2006 7:38 pm Post subject:
Re: Initial Value Problem
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|
"standelds" <standelds@hawaii.rr.com> wrote in message
news:AR2cg.5083$G95.225@tornado.socal.rr.com...
| Quote: |
"William Elliot" <marsh@hevanet.remove.com> wrote in message
news:Pine.BSI.4.58.0605210248350.7697@vista.hevanet.com...
From: standelds <standelds@hawaii.rr.com
Newsgroups: alt.math.undergrad
Subject: Initial Value Problem
Solve the IVP:
L dI/dt + RI + q/c = E(t); q(0) = some initial value of q,
and I(0) = some initial value of I.
where L is the inductance in Henrys, R is the resistance in ohms, C
is the capacitiance in farads, E(t) is the EMF in volts, q(t) is the
charge in coulombs on the capacitor at time t, and I = dq/dt is the
current in amperes. Find the current at time t if the charge on the
capacitor is initally zero, L =10H, R = 20 ohms, C = (6260)^-1 F, and
E(t) = 100V. [HINT: Differentiate both sides of the original equation
to obtain a homogeneous linear second order diff eq for I(t). Then
use the original equation todetermine dI/dt at t=0]
Here's my work so far:
d/dt (10 dI/dt + 20 I + 6260 q )= d/dt (E(t))
10 I'' + 20 I' + 6260 I = 0
Oh??? E(t) is a constant E ?
How else would you handle it? I made the assumption based on the
information given in the problem that E(t) = 100V.
Auxiliary equation: 10 r^2 + 20 r + 6260 = 0
r = -1 +/- 500 i
No. r^2 + 2r + 626 = 0
r = (-2 +- sqr(4 - 4*626))/2 = -1 +- sqr(1 - 626) = -1 +- 25i
Yep, dropped the division by 20 in that half of the quadratic formula.
a = -1; B = 500
??? Some in class jargon?
The formula for complex conjugate roots from the book:
for a diff eq ay'' + by' + cy = 0 with auxiliary eq ar^2 + br +c = 0 where
r has complex conjugate roots:
alpha +/- beta * i
two linearly independant solutions are:
e^(alpha*t) cos(beta * t) and e^(alpha*t) sin(beta*t)
then a general solution is:
y(t) = c1 e^(alpha*t) cos(beta * t) + c2 e^(alpha*t) sin(beta*t)
where c1 and c2 are arbitrary constants.
however I don't have alpha or beta characters on my keyboard. So I use a
and B since they are kind of similar.
I(t) = c1 e^(-t) cos(500t) + c2 e^(-t) sin(500t)
I'(t) = e^(-t) [ 500 c1 sin(500t) - c1 cos(500t)
+ 500 c2 i cos(500t) - c2 i sin(500t)
No. There's no reason for i sqr -1
I'(t) = -I(t) + e^(-t)(-500 c1.sin 500 t + 500 c2.cos 500t)
BTW, correct for previous mistake, 25 instead of 500.
Yes. I see that now. I don't know how many times I reworked this problem
and didn't see that mistake.
|
I spoke too soon. After the correction to from 500 --> 25 I get:
I(t) = c1 e^-t cos(25t) + c2 e^-t sin(25t)
I'(t) = c1[-e^-t cos(25t) - 25 e^-t sin(25t)] + c2[-e^-t sin(25t) + 25 e^-t
cos(25t)]
= -c1 e^-t cos(25t) - c2 e^-t sin(25t) - 25 e^-t sin(25t) + 25 e^-t
cos(25t)
= -I(t) - 25 e^-t [sin(25t) - cos(25t)]
or did I make another mistake?
| Quote: |
q(0) = 0
So where do I go with this last bit from the problem statement?
How do I continue the problem?
Any help you can give will be most appreciated.
What's next? Find c1 and c2? You need some values,
like for example I(0) and I'(0).
q(0) = 0 does nothing because upon solving
I(t) = dq(t)/dt
for q(t), you've introducted another constant which is
determined by q(0) = 0, in terms of c1 and c2.
----
Okay. however I have typed the problem above. Even copied the hint. How
do you get the values?
|
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standelds
science forum addict
Joined: 09 Sep 2005
Posts: 55
|
| Posted: Sun May 21, 2006 9:41 pm Post subject:
Re: Initial Value Problem
|
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|
"standelds" <standelds@hawaii.rr.com> wrote in message
news:Nk3cg.3769$9W5.1926@tornado.socal.rr.com...
| Quote: |
"standelds" <standelds@hawaii.rr.com> wrote in message
news:AR2cg.5083$G95.225@tornado.socal.rr.com...
"William Elliot" <marsh@hevanet.remove.com> wrote in message
news:Pine.BSI.4.58.0605210248350.7697@vista.hevanet.com...
From: standelds <standelds@hawaii.rr.com
Newsgroups: alt.math.undergrad
Subject: Initial Value Problem
Solve the IVP:
L dI/dt + RI + q/c = E(t); q(0) = some initial value of q,
and I(0) = some initial value of I.
where L is the inductance in Henrys, R is the resistance in ohms, C
is the capacitiance in farads, E(t) is the EMF in volts, q(t) is the
charge in coulombs on the capacitor at time t, and I = dq/dt is the
current in amperes. Find the current at time t if the charge on the
capacitor is initally zero, L =10H, R = 20 ohms, C = (6260)^-1 F, and
E(t) = 100V. [HINT: Differentiate both sides of the original equation
to obtain a homogeneous linear second order diff eq for I(t). Then
use the original equation todetermine dI/dt at t=0]
Here's my work so far:
d/dt (10 dI/dt + 20 I + 6260 q )= d/dt (E(t))
10 I'' + 20 I' + 6260 I = 0
Oh??? E(t) is a constant E ?
How else would you handle it? I made the assumption based on the
information given in the problem that E(t) = 100V.
Auxiliary equation: 10 r^2 + 20 r + 6260 = 0
r = -1 +/- 500 i
No. r^2 + 2r + 626 = 0
r = (-2 +- sqr(4 - 4*626))/2 = -1 +- sqr(1 - 626) = -1 +- 25i
Yep, dropped the division by 20 in that half of the quadratic formula.
a = -1; B = 500
??? Some in class jargon?
The formula for complex conjugate roots from the book:
for a diff eq ay'' + by' + cy = 0 with auxiliary eq ar^2 + br +c = 0
where r has complex conjugate roots:
alpha +/- beta * i
two linearly independant solutions are:
e^(alpha*t) cos(beta * t) and e^(alpha*t) sin(beta*t)
then a general solution is:
y(t) = c1 e^(alpha*t) cos(beta * t) + c2 e^(alpha*t) sin(beta*t)
where c1 and c2 are arbitrary constants.
however I don't have alpha or beta characters on my keyboard. So I use a
and B since they are kind of similar.
I(t) = c1 e^(-t) cos(500t) + c2 e^(-t) sin(500t)
I'(t) = e^(-t) [ 500 c1 sin(500t) - c1 cos(500t)
+ 500 c2 i cos(500t) - c2 i sin(500t)
No. There's no reason for i sqr -1
I'(t) = -I(t) + e^(-t)(-500 c1.sin 500 t + 500 c2.cos 500t)
BTW, correct for previous mistake, 25 instead of 500.
Yes. I see that now. I don't know how many times I reworked this
problem and didn't see that mistake.
I spoke too soon. After the correction to from 500 --> 25 I get:
I(t) = c1 e^-t cos(25t) + c2 e^-t sin(25t)
I'(t) = c1[-e^-t cos(25t) - 25 e^-t sin(25t)] + c2[-e^-t sin(25t) + 25
e^-t cos(25t)]
= -c1 e^-t cos(25t) - c2 e^-t sin(25t) - 25 e^-t sin(25t) + 25 e^-t
cos(25t)
= -I(t) - 25 e^-t [sin(25t) - cos(25t)]
|
typo correction: = -I(t) - 25 e^-t c2 [sin(25t) - cos(25t)]
| Quote: |
or did I make another mistake?
q(0) = 0
So where do I go with this last bit from the problem statement?
How do I continue the problem?
Any help you can give will be most appreciated.
What's next? Find c1 and c2? You need some values,
like for example I(0) and I'(0).
q(0) = 0 does nothing because upon solving
I(t) = dq(t)/dt
for q(t), you've introducted another constant which is
determined by q(0) = 0, in terms of c1 and c2.
----
Okay. however I have typed the problem above. Even copied the hint.
How do you get the values?
|
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William Elliot
science forum Guru
Joined: 24 Mar 2005
Posts: 1906
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| Posted: Mon May 22, 2006 2:04 am Post subject:
Re: Initial Value Problem
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|
On Sun, 21 May 2006, standelds wrote:
| Quote: | "standelds" <standelds@hawaii.rr.com> wrote in message
"standelds" <standelds@hawaii.rr.com> wrote in message
"William Elliot" <marsh@hevanet.remove.com> wrote in message
From: standelds <standelds@hawaii.rr.com
Solve the IVP:
L dI/dt + RI + q/c = E(t); q(0) = some initial value of q,
and I(0) = some initial value of I.
where L is the inductance in Henrys, R is the resistance in ohms, C
is the capacitiance in farads, E(t) is the EMF in volts, q(t) is
the charge in coulombs on the capacitor at time t, and I = dq/dt is
the current in amperes. Find the current at time t if the charge
on the capacitor is initally zero, L =10H, R = 20 ohms, C =
(6260)^-1 F, and E(t) = 100V. [HINT: Differentiate both sides of
the original equation to obtain a homogeneous linear second order
diff eq for I(t). Then use the original equation to determine dI/dt
at t=0]
Here's my work so far:
d/dt (10 dI/dt + 20 I + 6260 q )= d/dt (E(t))
10 I'' + 20 I' + 6260 I = 0
Auxiliary equation: 10 r^2 + 20 r + 6260 = 0
r = -1 +/- 500 i
No. r^2 + 2r + 626 = 0
r = (-2 +- sqr(4 - 4*626))/2 = -1 +- sqr(1 - 626) = -1 +- 25i
I(t) = c1 e^-t cos(25t) + c2 e^-t sin(25t)
|
= e^-t (c1 cos 25t + c2 sin 25t)
I'(t) = -e^-t (c1 cos 25t + c2 sin 25t)
+ e^-t (-25c1 sin 25t + 25c2 cos 25t)
= -I(t) + e^-t (-25c1 sin 25t + 25c2 cos 25t)
| Quote: | I'(t) = c1[-e^-t cos(25t) - 25 e^-t sin(25t)] + c2[-e^-t sin(25t) + 25
e^-t cos(25t)]
= -c1 e^-t cos(25t) - c2 e^-t sin(25t) - 25 e^-t sin(25t) + 25 e^-t
cos(25t)
= -I(t) - 25 e^-t [sin(25t) - cos(25t)]
typo correction: = -I(t) - 25 e^-t c2 [sin(25t) - cos(25t)]
No, as above. You need to go slowly and methodically checking your work |
as you go and again upon finish.
| Quote: | q(0) = 0
So where do I go with this last bit from the problem statement?
How do I continue the problem?
What's next? Find c1 and c2? You need some values,
like for example I(0) and I'(0).
q(0) = 0 does nothing because upon solving
I(t) = dq(t)/dt
for q(t), you've introduced another constant which is
determined by q(0) = 0, in terms of c1 and c2.
How do you get the values?
|
From problem statement
q(0) = some initial value of q,
and I(0) = some initial value of I.
there is inadequate information.
Yet from above you said q(0) = 0, tho that has no bearing on the solution
as explained above. The best you can do is to assume some initial values
for I(t) and I'(t), or some initial and final value of I(t).
Let's say I(0) and I'(0) are known. Thus use
I(t) = c1 e^-t cos(25t) + c2 e^-t sin(25t)
I'(t) = -I(t) + e^-t (-25c1 sin 25t + 25c2 cos 25t)
to find c1 and c2 in terms of I(0) and I'(0). |
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