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| Author |
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standelds
science forum addict
Joined: 09 Sep 2005
Posts: 55
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 Posted: Mon May 22, 2006 4:33 am Post subject:
diff eq.
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Can someone please take the time to check my work:
Find a general solution to the DE using the method of variation of
parameters:
y" + 2y' + y = e^-t
homogeneous: y" + 2y' +y = 0
auxiliary equation: r^2 + 2r + 1 = 0
( r + 1)^2 = 0
r = -1 (repeated root)
y1(t) = e^-t and y2(t) = te^-t are two linearly independant solutions.
homogeneous solution: y(t) = c1 y1(t) + c2 y2(t)
= c1 e^-t + c2 te^-t
particular solution: y(t) = v1(t) y1(t) + v2(t) y2(t)
= v1(t) e^-t + v2(t) te^-t
v1(t) = int[ (-te^-2t) / W(y1(t), y2(t))] dt
W(y1(t), y2(t)) = y1(t) y2'(t) - y2(t)y1'(t)
y1'(t) = -e^-t and y2'(t) = -t e^-t + e^-t
W = - e^(-2t)( 2t - 1)
v1(t) = int[ t / (2t -1) ] dt = 1/4( 2t -1 + ln |2t-1|)
v2(t) = int[ e^(-2t) / (-e^(-2t)(2t-1)] dt = - int[ dt / (2t-1) ] = -1/2 ln
|2t -1|
y(t) = particular solution + homogeneous solution:
y(t) = (e^-t / 4)( 2t -1 + ln |2t-1|) + (te^-t / 2) ln |2t - 1| +
c1 e^-t + c2 te^-t
Thanks for your time. This is a very challenging class for me and your help
means a lot. |
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Virgil
science forum Guru
Joined: 24 Mar 2005
Posts: 5536
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| Posted: Mon May 22, 2006 7:27 am Post subject:
Re: diff eq.
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In article <xabcg.5423$G95.5070@tornado.socal.rr.com>,
"standelds" <standelds@hawaii.rr.com> wrote:
| Quote: | Can someone please take the time to check my work:
Find a general solution to the DE using the method of variation of
parameters:
y" + 2y' + y = e^-t
homogeneous: y" + 2y' +y = 0
auxiliary equation: r^2 + 2r + 1 = 0
( r + 1)^2 = 0
r = -1 (repeated root)
y1(t) = e^-t and y2(t) = te^-t are two linearly independant solutions.
homogeneous solution: y(t) = c1 y1(t) + c2 y2(t)
= c1 e^-t + c2 te^-t
particular solution: y(t) = v1(t) y1(t) + v2(t) y2(t)
= v1(t) e^-t + v2(t) te^-t
v1(t) = int[ (-te^-2t) / W(y1(t), y2(t))] dt
W(y1(t), y2(t)) = y1(t) y2'(t) - y2(t)y1'(t)
y1'(t) = -e^-t and y2'(t) = -t e^-t + e^-t
W = - e^(-2t)( 2t - 1)
|
I get a different, simpler value for the Wronskian.
| Quote: |
v1(t) = int[ t / (2t -1) ] dt = 1/4( 2t -1 + ln |2t-1|)
v2(t) = int[ e^(-2t) / (-e^(-2t)(2t-1)] dt = - int[ dt / (2t-1) ] = -1/2 ln
|2t -1|
y(t) = particular solution + homogeneous solution:
y(t) = (e^-t / 4)( 2t -1 + ln |2t-1|) + (te^-t / 2) ln |2t - 1| +
c1 e^-t + c2 te^-t
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My particular solution is (x^2/2)*e^(-x), which works in the original
equation, but yours does not.
| Quote: |
Thanks for your time. This is a very challenging class for me and your help
means a lot. |
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standelds
science forum addict
Joined: 09 Sep 2005
Posts: 55
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| Posted: Mon May 22, 2006 2:08 pm Post subject:
Re: diff eq.
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"Virgil" <vmhjr2@comcast.net> wrote in message
news:vmhjr2-691728.01275922052006@news.usenetmonster.com...
| Quote: | In article <xabcg.5423$G95.5070@tornado.socal.rr.com>,
"standelds" <standelds@hawaii.rr.com> wrote:
Can someone please take the time to check my work:
Find a general solution to the DE using the method of variation of
parameters:
y" + 2y' + y = e^-t
homogeneous: y" + 2y' +y = 0
auxiliary equation: r^2 + 2r + 1 = 0
( r + 1)^2 = 0
r = -1 (repeated root)
y1(t) = e^-t and y2(t) = te^-t are two linearly independant solutions.
homogeneous solution: y(t) = c1 y1(t) + c2 y2(t)
= c1 e^-t + c2 te^-t
particular solution: y(t) = v1(t) y1(t) + v2(t) y2(t)
= v1(t) e^-t + v2(t) te^-t
v1(t) = int[ (-te^-2t) / W(y1(t), y2(t))] dt
W(y1(t), y2(t)) = y1(t) y2'(t) - y2(t)y1'(t)
y1'(t) = -e^-t and y2'(t) = -t e^-t + e^-t
W = - e^(-2t)( 2t - 1)
I get a different, simpler value for the Wronskian.
|
is this a correct Wronskian calculation then:
W = e^-t (-te^-t + e^-t) - te^-t * -e^-t
= -te^(-2t) + e^(-2t) - te^(-2t) = -2te^(-2t) + e^(-2t)
= - e^(-2t) (2t -1)
I am having a problem with another similar exercise as well, and if I am
doing these wrong that would explain a lot.
| Quote: |
v1(t) = int[ t / (2t -1) ] dt = 1/4( 2t -1 + ln |2t-1|)
v2(t) = int[ e^(-2t) / (-e^(-2t)(2t-1)] dt = - int[ dt / (2t-1) ] = -1/2
ln
|2t -1|
y(t) = particular solution + homogeneous solution:
y(t) = (e^-t / 4)( 2t -1 + ln |2t-1|) + (te^-t / 2) ln |2t - 1|
+
c1 e^-t + c2 te^-t
My particular solution is (x^2/2)*e^(-x), which works in the original
equation, but yours does not.
Thanks for your time. This is a very challenging class for me and your
help
means a lot. |
|
|
| Back to top |
|
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Lynn Kurtz
science forum Guru
Joined: 02 May 2005
Posts: 603
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| Posted: Mon May 22, 2006 6:03 pm Post subject:
Re: diff eq.
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On Mon, 22 May 2006 04:33:33 GMT, "standelds"
<standelds@hawaii.rr.com> wrote:
| Quote: | Can someone please take the time to check my work:
Find a general solution to the DE using the method of variation of
parameters:
y" + 2y' + y = e^-t
homogeneous: y" + 2y' +y = 0
auxiliary equation: r^2 + 2r + 1 = 0
( r + 1)^2 = 0
r = -1 (repeated root)
y1(t) = e^-t and y2(t) = te^-t are two linearly independant solutions.
homogeneous solution: y(t) = c1 y1(t) + c2 y2(t)
= c1 e^-t + c2 te^-t
particular solution: y(t) = v1(t) y1(t) + v2(t) y2(t)
= v1(t) e^-t + v2(t) te^-t
|
Just a suggestion. Instead of using variation of parameters, why don't
you use the method of undetermined coefficients, or better yet, the
method of annihilators if you have studied that? See if you can see
why you would expect a particular solution of the form C*t^2*e^(-t).
--Lynn |
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Virgil
science forum Guru
Joined: 24 Mar 2005
Posts: 5536
|
| Posted: Mon May 22, 2006 7:53 pm Post subject:
Re: diff eq.
|
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|
In article <OBjcg.5452$G95.4927@tornado.socal.rr.com>,
"standelds" <standelds@hawaii.rr.com> wrote:
| Quote: | "Virgil" <vmhjr2@comcast.net> wrote in message
news:vmhjr2-691728.01275922052006@news.usenetmonster.com...
In article <xabcg.5423$G95.5070@tornado.socal.rr.com>,
"standelds" <standelds@hawaii.rr.com> wrote:
Can someone please take the time to check my work:
Find a general solution to the DE using the method of variation of
parameters:
y" + 2y' + y = e^-t
homogeneous: y" + 2y' +y = 0
auxiliary equation: r^2 + 2r + 1 = 0
( r + 1)^2 = 0
r = -1 (repeated root)
y1(t) = e^-t and y2(t) = te^-t are two linearly independant solutions.
homogeneous solution: y(t) = c1 y1(t) + c2 y2(t)
= c1 e^-t + c2 te^-t
particular solution: y(t) = v1(t) y1(t) + v2(t) y2(t)
= v1(t) e^-t + v2(t) te^-t
v1(t) = int[ (-te^-2t) / W(y1(t), y2(t))] dt
W(y1(t), y2(t)) = y1(t) y2'(t) - y2(t)y1'(t)
y1'(t) = -e^-t and y2'(t) = -t e^-t + e^-t
W = - e^(-2t)( 2t - 1)
I get a different, simpler value for the Wronskian.
is this a correct Wronskian calculation then:
W = e^-t (-te^-t + e^-t) - te^-t * -e^-t
= -te^(-2t) + e^(-2t) - te^(-2t) = -2te^(-2t) + e^(-2t)
^ |
Sign error---------------------- ^, neg times neg is pos.
| Quote: | = - e^(-2t) (2t -1)
I am having a problem with another similar exercise as well, and if I am
doing these wrong that would explain a lot.
v1(t) = int[ t / (2t -1) ] dt = 1/4( 2t -1 + ln |2t-1|)
v2(t) = int[ e^(-2t) / (-e^(-2t)(2t-1)] dt = - int[ dt / (2t-1) ] = -1/2
ln
|2t -1|
y(t) = particular solution + homogeneous solution:
y(t) = (e^-t / 4)( 2t -1 + ln |2t-1|) + (te^-t / 2) ln |2t - 1|
+
c1 e^-t + c2 te^-t
My particular solution is (x^2/2)*e^(-x), which works in the original
equation, but yours does not.
Thanks for your time. This is a very challenging class for me and your
help
means a lot. |
|
|
| Back to top |
|
|
standelds
science forum addict
Joined: 09 Sep 2005
Posts: 55
|
| Posted: Tue May 23, 2006 12:22 am Post subject:
Re: diff eq.
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|
"[Mr.] Lynn Kurtz" <kurtzDELETE-THIS@asu.edu> wrote in message
news:VfxxRO1xEIvwM4DSplOGcO4LgUOx@4ax.com...
| Quote: | On Mon, 22 May 2006 04:33:33 GMT, "standelds"
standelds@hawaii.rr.com> wrote:
Can someone please take the time to check my work:
Find a general solution to the DE using the method of variation of
parameters:
y" + 2y' + y = e^-t
homogeneous: y" + 2y' +y = 0
auxiliary equation: r^2 + 2r + 1 = 0
( r + 1)^2 = 0
r = -1 (repeated root)
y1(t) = e^-t and y2(t) = te^-t are two linearly independant solutions.
homogeneous solution: y(t) = c1 y1(t) + c2 y2(t)
= c1 e^-t + c2 te^-t
particular solution: y(t) = v1(t) y1(t) + v2(t) y2(t)
= v1(t) e^-t + v2(t) te^-t
Just a suggestion. Instead of using variation of parameters, why don't
you use the method of undetermined coefficients, or better yet, the
method of annihilators if you have studied that? See if you can see
why you would expect a particular solution of the form C*t^2*e^(-t).
--Lynn
|
Mr. Kurtz,
Thanks for the suggestion. I haven't studied annihilators, yet. I chose
the method I did primarily because the those were the directions for the
section in the book. Secondarily, my professor said the method of variation
of parameter's and more specifically Green's Function are the most important
things we study the whole semester of differential equations, and therefore
the people seeking mathematics degrees should work as many propblems as
possible using these tools.
I don't know whether he's correct in his assessment or not, but he did his
PhD thesis in operator theory so he uses it as a basis to motivate some of
the other methods. So it keeps popping up, like the Wronskian from linear
algebra does.
Respectfully,
Dustin |
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Lynn Kurtz
science forum Guru
Joined: 02 May 2005
Posts: 603
|
| Posted: Tue May 23, 2006 12:41 am Post subject:
Re: diff eq.
|
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|
On Tue, 23 May 2006 00:22:45 GMT, "standelds"
<standelds@hawaii.rr.com> wrote:
| Quote: |
"[Mr.] Lynn Kurtz" <kurtzDELETE-THIS@asu.edu> wrote in message
news:VfxxRO1xEIvwM4DSplOGcO4LgUOx@4ax.com...
On Mon, 22 May 2006 04:33:33 GMT, "standelds"
standelds@hawaii.rr.com> wrote:
Can someone please take the time to check my work:
Find a general solution to the DE using the method of variation of
parameters:
y" + 2y' + y = e^-t
homogeneous: y" + 2y' +y = 0
auxiliary equation: r^2 + 2r + 1 = 0
( r + 1)^2 = 0
r = -1 (repeated root)
y1(t) = e^-t and y2(t) = te^-t are two linearly independant solutions.
homogeneous solution: y(t) = c1 y1(t) + c2 y2(t)
= c1 e^-t + c2 te^-t
particular solution: y(t) = v1(t) y1(t) + v2(t) y2(t)
= v1(t) e^-t + v2(t) te^-t
Just a suggestion. Instead of using variation of parameters, why don't
you use the method of undetermined coefficients, or better yet, the
method of annihilators if you have studied that? See if you can see
why you would expect a particular solution of the form C*t^2*e^(-t).
--Lynn
Mr. Kurtz,
Thanks for the suggestion. I haven't studied annihilators, yet. I chose
the method I did primarily because the those were the directions for the
section in the book. Secondarily, my professor said the method of variation
of parameter's and more specifically Green's Function are the most important
things we study the whole semester of differential equations, and therefore
the people seeking mathematics degrees should work as many propblems as
possible using these tools.
I don't know whether he's correct in his assessment or not, but he did his
PhD thesis in operator theory so he uses it as a basis to motivate some of
the other methods. So it keeps popping up, like the Wronskian from linear
algebra does.
Respectfully,
Dustin
|
I don't disagree with his assessment, and it is true that variation of
parameters is a more general method than undetermined coefficients.
And practice is good.
However, for constant coefficient equations like this one, it is
overkill.
--Lynn |
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Patrick O'Sullivan
science forum beginner
Joined: 24 May 2006
Posts: 1
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| Posted: Wed May 24, 2006 3:52 am Post subject:
Re: diff eq.
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|
Look up something about Laplace transforms and you'll be very happy. ;-)
standelds wrote:
| Quote: | Can someone please take the time to check my work:
Find a general solution to the DE using the method of variation of
parameters:
y" + 2y' + y = e^-t
homogeneous: y" + 2y' +y = 0
auxiliary equation: r^2 + 2r + 1 = 0
( r + 1)^2 = 0
r = -1 (repeated root)
y1(t) = e^-t and y2(t) = te^-t are two linearly independant solutions.
homogeneous solution: y(t) = c1 y1(t) + c2 y2(t)
= c1 e^-t + c2 te^-t
particular solution: y(t) = v1(t) y1(t) + v2(t) y2(t)
= v1(t) e^-t + v2(t) te^-t
v1(t) = int[ (-te^-2t) / W(y1(t), y2(t))] dt
W(y1(t), y2(t)) = y1(t) y2'(t) - y2(t)y1'(t)
y1'(t) = -e^-t and y2'(t) = -t e^-t + e^-t
W = - e^(-2t)( 2t - 1)
v1(t) = int[ t / (2t -1) ] dt = 1/4( 2t -1 + ln |2t-1|)
v2(t) = int[ e^(-2t) / (-e^(-2t)(2t-1)] dt = - int[ dt / (2t-1) ] = -1/2 ln
|2t -1|
y(t) = particular solution + homogeneous solution:
y(t) = (e^-t / 4)( 2t -1 + ln |2t-1|) + (te^-t / 2) ln |2t - 1| +
c1 e^-t + c2 te^-t
Thanks for your time. This is a very challenging class for me and your help
means a lot.
|
--
Patrick O'Sullivan
Rutgers University Network Operations
patrick.osullivan@rutgers.edu |
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