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Patrick1182
science forum addict
Joined: 01 Feb 2006
Posts: 55
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 Posted: Tue Jun 20, 2006 11:55 am Post subject:
Seperable Differential Equation?
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Hi,
In example 1.5 of differential equations demystified the
author asks:
----------------------------------
Consider the differential equation
y' + 2xy = x
Find a complete solution.
SOLUTION:
This equation is plainly not separable (try it and convince
yourself that this is so). [ actual solution ...]
----------------------------------
My question is, why can't we do this:
y' + 2xy = x
y' + 2xy - x = 0
y' + x(2y - 1) = 0
x(2y - 1) = -y'
x = -y' / (2y - 1)
Now this is separated, right?
The dependent and independent
variables have been isolated
on each side.
If I integrate both sides don't
I get the right answer?
Thanks. |
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William Elliot
science forum Guru
Joined: 24 Mar 2005
Posts: 1906
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| Posted: Tue Jun 20, 2006 12:33 pm Post subject:
Re: Seperable Differential Equation?
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On Tue, 20 Jun 2006, Patrick wrote:
| Quote: | Consider the differential equation
y' + 2xy = x
Find a complete solution.
This equation is plainly not separable (try it and convince
yourself that this is so). [ actual solution ...]
My question is, why can't we do this:
Because y'all be making a fool of the author? |
| Quote: | y' + 2xy = x
y' + 2xy - x = 0
y' + x(2y - 1) = 0
x(2y - 1) = -y'
x = -y' / (2y - 1)
Now this is separated, right? The dependent and independent variables
have been isolated on each side.
Yah. |
| Quote: | If I integrate both sides don't I get the right answer?
Don't know. You didn't show your work, your answer, nor |
did you check your answer. Maybe author wanted to pretend
equation wasn't separable to demostrate another method. |
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Patrick1182
science forum addict
Joined: 01 Feb 2006
Posts: 55
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| Posted: Tue Jun 20, 2006 2:06 pm Post subject:
Re: Seperable Differential Equation?
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William Elliot wrote:
| Quote: | On Tue, 20 Jun 2006, Patrick wrote:
Consider the differential equation
y' + 2xy = x
Find a complete solution.
This equation is plainly not separable (try it and convince
yourself that this is so). [ actual solution ...]
My question is, why can't we do this:
Because y'all be making a fool of the author?
y' + 2xy = x
y' + 2xy - x = 0
y' + x(2y - 1) = 0
x(2y - 1) = -y'
x = -y' / (2y - 1)
Now this is separated, right? The dependent and independent variables
have been isolated on each side.
Yah.
If I integrate both sides don't I get the right answer?
Don't know. You didn't show your work, your answer, nor
did you check your answer. Maybe author wanted to pretend
equation wasn't separable to demostrate another method.
|
When I integrate I get
(-x^2)/2 = 1/2 * ln |2y - 1| + C
-x^2 = ln |2y - 1| + 2C
e^(-x^2) = e^(2C) * (2y - 1)
e^(-x^2) / e^(2C) = 2y - 1
[ e^(-x^2) / 2e^(2C) ] + 1/2 = y
Let K = 1/2e^(2C) and the answer is
y = K * e^(-x^2) + 1/2
The answer given in the book is
y = 1/2 + C * e^(-x^2)
But maybe I did mine the wrong way and got the right answer.
The author did mean to discuss a new technique, the
reason I paid extra attention to the non-separability
of the equation was because the author asked us to
"convince ourselves that this is so." |
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Dave L. Renfro
science forum Guru
Joined: 29 Apr 2005
Posts: 570
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| Posted: Tue Jun 20, 2006 2:08 pm Post subject:
Re: Seperable Differential Equation?
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Patrick wrote:
| Quote: | Consider the differential equation
y' + 2xy = x
Find a complete solution.
SOLUTION:
This equation is plainly not separable (try it
and convince yourself that this is so).
[ actual solution ...]
----------------------------------
My question is, why can't we do this:
|
[snip]
| Quote: | Now this is separated, right?
|
As William Elliot already said, you're correct.
I thought I'd point out that anytime you can
rewrite the equation in the form y' = f(x)*g(y),
the equation is separable. This is clearly possible
for your equation, since y' = x - 2xy = x(1 - 2y).
Dave L. Renfro |
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Dave L. Renfro
science forum Guru
Joined: 29 Apr 2005
Posts: 570
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| Posted: Tue Jun 20, 2006 2:29 pm Post subject:
Re: Seperable Differential Equation?
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Patrick wrote:
| Quote: | When I integrate I get
(-x^2)/2 = 1/2 * ln |2y - 1| + C
-x^2 = ln |2y - 1| + 2C
e^(-x^2) = e^(2C) * (2y - 1)
|
The second right-hand side factor should be |2y - 1|.
This introduces a +/- sign later on.
| Quote: | e^(-x^2) / e^(2C) = 2y - 1
[ e^(-x^2) / 2e^(2C) ] + 1/2 = y
Let K = 1/2e^(2C) and the answer is
y = K * e^(-x^2) + 1/2
The answer given in the book is
y = 1/2 + C * e^(-x^2)
But maybe I did mine the wrong way and got
the right answer.
|
They're the same, except you should have
K = +/- (1/2)*e^(2C), which leads us to
K being any nonzero number. A quick check
shows that y = 1/2 (i.e. K=0) also satisfies
the differential equation, so in fact K can
be any real number, which is the same set
of functions the book gives. (I'm making the
assumption that when the text answer doesn't
give any restrictions on 'C', then 'C' can
be any real number.)
For a similar example, see my post at
The Math Forum, AP-Calculus, 13 April 2006
http://mathforum.org/kb/thread.jspa?messageID=4637781
Dave L. Renfro |
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William Elliot
science forum Guru
Joined: 24 Mar 2005
Posts: 1906
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| Posted: Wed Jun 21, 2006 6:00 am Post subject:
Re: Seperable Differential Equation?
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On Tue, 20 Jun 2006, Patrick wrote:
| Quote: | William Elliot wrote:
On Tue, 20 Jun 2006, Patrick wrote:
Consider the differential equation
y' + 2xy = x
This equation is plainly not separable (try it and convince
yourself that this is so). [ actual solution ...]
My question is, why can't we do this:
y' + 2xy = x
y' + 2xy - x = 0
y' + x(2y - 1) = 0
x(2y - 1) = -y'
x = -y' / (2y - 1)
|
Just a cotton picking nanosec.
You have two cases, y = 1/2 and y /= 1/2.
Notice y = 1/2 is a solution.
| Quote: | Now this is separated, right?
Yah.
If I integrate both sides don't I get the right answer?
Don't know. You didn't show your work, your answer, nor
did you check your answer. Maybe author wanted to pretend
equation wasn't separable to demostrate another method.
When I integrate I get
(-x^2)/2 = 1/2 * ln |2y - 1| + C
-x^2 = ln |2y - 1| + 2C
e^(-x^2) = e^(2C) * (2y - 1)
|
No, e^(-x^2) = e^(2C) * |2y - 1|
| Quote: | e^(-x^2) / e^(2C) = 2y - 1
|
No, e^(-x^2) / e^(2C) = |2y - 1|
| Quote: | [ e^(-x^2) / 2e^(2C) ] + 1/2 = y
|
Notice y = 1/2 is not a solution.
| Quote: | Let K = 1/2e^(2C) and the answer is
y = K * e^(-x^2) + 1/2
The answer given in the book is
y = 1/2 + C * e^(-x^2)
The difference I bet is that C in R while K > 0. |
So you didn't get all the solutions.
| Quote: | But maybe I did mine the wrong way and got the right answer.
The author did mean to discuss a new technique, the reason I paid extra
attention to the non-separability of the equation was because the author
asked us to "convince ourselves that this is so."
Are you convinced? By any chance, are you also dazzled or razzled? |
What you think? Separable or not?
y' + 2xy = x
y = e^(-x^2) is zero solution.
y = 1/2 is particular solution.
y = 1/2 + ce^(-x^2) for any c in R, is general solution. |
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Patrick1182
science forum addict
Joined: 01 Feb 2006
Posts: 55
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| Posted: Wed Jun 21, 2006 10:20 am Post subject:
Re: Seperable Differential Equation?
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William Elliot wrote:
| Quote: | On Tue, 20 Jun 2006, Patrick wrote:
William Elliot wrote:
On Tue, 20 Jun 2006, Patrick wrote:
Consider the differential equation
y' + 2xy = x
This equation is plainly not separable (try it and convince
yourself that this is so). [ actual solution ...]
My question is, why can't we do this:
y' + 2xy = x
y' + 2xy - x = 0
y' + x(2y - 1) = 0
x(2y - 1) = -y'
x = -y' / (2y - 1)
Just a cotton picking nanosec.
You have two cases, y = 1/2 and y /= 1/2.
Notice y = 1/2 is a solution.
Now this is separated, right?
Yah.
If I integrate both sides don't I get the right answer?
Don't know. You didn't show your work, your answer, nor
did you check your answer. Maybe author wanted to pretend
equation wasn't separable to demostrate another method.
When I integrate I get
(-x^2)/2 = 1/2 * ln |2y - 1| + C
-x^2 = ln |2y - 1| + 2C
e^(-x^2) = e^(2C) * (2y - 1)
No, e^(-x^2) = e^(2C) * |2y - 1|
|
I was being sloppy. thanks to you
and Dave for pointing this out.
| Quote: | e^(-x^2) / e^(2C) = 2y - 1
No, e^(-x^2) / e^(2C) = |2y - 1|
[ e^(-x^2) / 2e^(2C) ] + 1/2 = y
Notice y = 1/2 is not a solution.
|
Yes.
| Quote: | Let K = 1/2e^(2C) and the answer is
y = K * e^(-x^2) + 1/2
The answer given in the book is
y = 1/2 + C * e^(-x^2)
The difference I bet is that C in R while K > 0.
So you didn't get all the solutions.
But maybe I did mine the wrong way and got the right answer.
The author did mean to discuss a new technique, the reason I paid extra
attention to the non-separability of the equation was because the author
asked us to "convince ourselves that this is so."
Are you convinced? By any chance, are you also dazzled or razzled?
What you think? Separable or not?
|
I think it is separable.
| Quote: | y' + 2xy = x
y = e^(-x^2) is zero solution.
y = 1/2 is particular solution.
y = 1/2 + ce^(-x^2) for any c in R, is general solution. |
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