Determine if this statement is true or false
$begingroup$
Is this statement true or false?
Statement: Each differential equation on the form $y '= f (y)$ with f having continuous derivative on the entire real axis $ℝ$ leads to that all solutions $y(x)$ defined on the entire $ℝ$.
So if the continuous derivative is defined for all real numbers, is all solutions $y(x)$ defined for all real numbers for $y'=f(y)$
I have tried to solve the equation an example like $y '= y ^ 2$ where the right-hand side has continuous derivatives and makes conclusions itself
but I am not sure what the answer is...
ordinary-differential-equations
edited Dec 6 '18 at 21:50
LutzL
59.4k42057
$endgroup$
add a comment |
$begingroup$
Is this statement true or false?
Statement: Each differential equation on the form $y '= f (y)$ with f having continuous derivative on the entire real axis $ℝ$ leads to that all solutions $y(x)$ defined on the entire $ℝ$.
So if the continuous derivative is defined for all real numbers, is all solutions $y(x)$ defined for all real numbers for $y'=f(y)$
I have tried to solve the equation an example like $y '= y ^ 2$ where the right-hand side has continuous derivatives and makes conclusions itself
but I am not sure what the answer is...
ordinary-differential-equations
edited Dec 6 '18 at 21:50
LutzL
59.4k42057
$endgroup$
$begingroup$
Identical question with discussion in comments: math.stackexchange.com/q/3029066/115115
$endgroup$
– LutzL
Dec 6 '18 at 21:48
$begingroup$
And yes, add the initial condition $y(0)=1$ to get a soluton $y(x)=1/(1-x)$ that is not defined on all of $Bbb R$.
$endgroup$
– LutzL
Dec 6 '18 at 21:49
$begingroup$
Re-posting a question is not the way to draw more attention to it (or to escape down- and close-votes). If you have context to add, add it to the previous question. This improves the question and can reverse the trend in negative votes.
$endgroup$
– Blue
Dec 6 '18 at 21:51
$begingroup$
Ah ... I see that this isn't merely a re-post of a previous question. You have created a new account for this purpose (distinct from "Jacob Andreasson") ... which is an even less appropriate way to escape down- and close-votes. This is a violation of community standards.
$endgroup$
– Blue
Dec 6 '18 at 21:58
add a comment |
$begingroup$
Is this statement true or false?
Statement: Each differential equation on the form $y '= f (y)$ with f having continuous derivative on the entire real axis $ℝ$ leads to that all solutions $y(x)$ defined on the entire $ℝ$.
So if the continuous derivative is defined for all real numbers, is all solutions $y(x)$ defined for all real numbers for $y'=f(y)$
I have tried to solve the equation an example like $y '= y ^ 2$ where the right-hand side has continuous derivatives and makes conclusions itself
but I am not sure what the answer is...
ordinary-differential-equations
edited Dec 6 '18 at 21:50
LutzL
59.4k42057
$endgroup$
Is this statement true or false?
Statement: Each differential equation on the form $y '= f (y)$ with f having continuous derivative on the entire real axis $ℝ$ leads to that all solutions $y(x)$ defined on the entire $ℝ$.
So if the continuous derivative is defined for all real numbers, is all solutions $y(x)$ defined for all real numbers for $y'=f(y)$
I have tried to solve the equation an example like $y '= y ^ 2$ where the right-hand side has continuous derivatives and makes conclusions itself
but I am not sure what the answer is...
ordinary-differential-equations
ordinary-differential-equations
edited Dec 6 '18 at 21:50
LutzL
59.4k42057
edited Dec 6 '18 at 21:50
LutzL
59.4k42057
edited Dec 6 '18 at 21:50
LutzL
59.4k42057
edited Dec 6 '18 at 21:50
LutzL
59.4k42057
edited Dec 6 '18 at 21:50
LutzL
59.4k42057
59.4k42057
asked Dec 6 '18 at 21:46
user623692
$begingroup$
Identical question with discussion in comments: math.stackexchange.com/q/3029066/115115
$endgroup$
– LutzL
Dec 6 '18 at 21:48
$begingroup$
And yes, add the initial condition $y(0)=1$ to get a soluton $y(x)=1/(1-x)$ that is not defined on all of $Bbb R$.
$endgroup$
– LutzL
Dec 6 '18 at 21:49
$begingroup$
Re-posting a question is not the way to draw more attention to it (or to escape down- and close-votes). If you have context to add, add it to the previous question. This improves the question and can reverse the trend in negative votes.
$endgroup$
– Blue
Dec 6 '18 at 21:51
$begingroup$
Ah ... I see that this isn't merely a re-post of a previous question. You have created a new account for this purpose (distinct from "Jacob Andreasson") ... which is an even less appropriate way to escape down- and close-votes. This is a violation of community standards.
$endgroup$
– Blue
Dec 6 '18 at 21:58
add a comment |
$begingroup$
Identical question with discussion in comments: math.stackexchange.com/q/3029066/115115
$endgroup$
– LutzL
Dec 6 '18 at 21:48
$begingroup$
And yes, add the initial condition $y(0)=1$ to get a soluton $y(x)=1/(1-x)$ that is not defined on all of $Bbb R$.
$endgroup$
– LutzL
Dec 6 '18 at 21:49
$begingroup$
Re-posting a question is not the way to draw more attention to it (or to escape down- and close-votes). If you have context to add, add it to the previous question. This improves the question and can reverse the trend in negative votes.
$endgroup$
– Blue
Dec 6 '18 at 21:51
$begingroup$
Ah ... I see that this isn't merely a re-post of a previous question. You have created a new account for this purpose (distinct from "Jacob Andreasson") ... which is an even less appropriate way to escape down- and close-votes. This is a violation of community standards.
$endgroup$
– Blue
Dec 6 '18 at 21:58
$begingroup$
Identical question with discussion in comments: math.stackexchange.com/q/3029066/115115
$endgroup$
– LutzL
Dec 6 '18 at 21:48
$begingroup$
Identical question with discussion in comments: math.stackexchange.com/q/3029066/115115
$endgroup$
– LutzL
Dec 6 '18 at 21:48
$begingroup$
And yes, add the initial condition $y(0)=1$ to get a soluton $y(x)=1/(1-x)$ that is not defined on all of $Bbb R$.
$endgroup$
– LutzL
Dec 6 '18 at 21:49
$begingroup$
And yes, add the initial condition $y(0)=1$ to get a soluton $y(x)=1/(1-x)$ that is not defined on all of $Bbb R$.
$endgroup$
– LutzL
Dec 6 '18 at 21:49
$begingroup$
Re-posting a question is not the way to draw more attention to it (or to escape down- and close-votes). If you have context to add, add it to the previous question. This improves the question and can reverse the trend in negative votes.
$endgroup$
– Blue
Dec 6 '18 at 21:51
$begingroup$
Re-posting a question is not the way to draw more attention to it (or to escape down- and close-votes). If you have context to add, add it to the previous question. This improves the question and can reverse the trend in negative votes.
$endgroup$
– Blue
Dec 6 '18 at 21:51
$begingroup$
Ah ... I see that this isn't merely a re-post of a previous question. You have created a new account for this purpose (distinct from "Jacob Andreasson") ... which is an even less appropriate way to escape down- and close-votes. This is a violation of community standards.
$endgroup$
– Blue
Dec 6 '18 at 21:58
$begingroup$
Ah ... I see that this isn't merely a re-post of a previous question. You have created a new account for this purpose (distinct from "Jacob Andreasson") ... which is an even less appropriate way to escape down- and close-votes. This is a violation of community standards.
$endgroup$
– Blue
Dec 6 '18 at 21:58
add a comment |
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$begingroup$
Identical question with discussion in comments: math.stackexchange.com/q/3029066/115115
$endgroup$
– LutzL
Dec 6 '18 at 21:48
$begingroup$
And yes, add the initial condition $y(0)=1$ to get a soluton $y(x)=1/(1-x)$ that is not defined on all of $Bbb R$.
$endgroup$
– LutzL
Dec 6 '18 at 21:49
$begingroup$
Re-posting a question is not the way to draw more attention to it (or to escape down- and close-votes). If you have context to add, add it to the previous question. This improves the question and can reverse the trend in negative votes.
$endgroup$
– Blue
Dec 6 '18 at 21:51
$begingroup$
Ah ... I see that this isn't merely a re-post of a previous question. You have created a new account for this purpose (distinct from "Jacob Andreasson") ... which is an even less appropriate way to escape down- and close-votes. This is a violation of community standards.
$endgroup$
– Blue
Dec 6 '18 at 21:58