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Finding the limit of a function given the differential equation
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Unsure how to go about this question. I have written it out as: $y(x)=f(x)+2$ since $y(0)=2$, but I am assuming that $f(0)=2$. This doesn't really help me though as it is said that the differential equation must not be solved.
differential-equations
edited Jul 31 at 4:31
zipirovich
9,94011630
asked Jul 31 at 3:42
J-Dorman
555
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up vote
1
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Unsure how to go about this question. I have written it out as: $y(x)=f(x)+2$ since $y(0)=2$, but I am assuming that $f(0)=2$. This doesn't really help me though as it is said that the differential equation must not be solved.
differential-equations
edited Jul 31 at 4:31
zipirovich
9,94011630
asked Jul 31 at 3:42
J-Dorman
555
3
It is an autonomous first order equation with a unique solution. At long times these can only blow up or go to an equilibrium point. To see which, check the sign of the derivative at the starting time and move in that direction from the starting position until you hit an equilibrium or have surpassed all equilibria.
– Ian
Jul 31 at 3:45
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Unsure how to go about this question. I have written it out as: $y(x)=f(x)+2$ since $y(0)=2$, but I am assuming that $f(0)=2$. This doesn't really help me though as it is said that the differential equation must not be solved.
differential-equations
edited Jul 31 at 4:31
zipirovich
9,94011630
asked Jul 31 at 3:42
J-Dorman
555
Unsure how to go about this question. I have written it out as: $y(x)=f(x)+2$ since $y(0)=2$, but I am assuming that $f(0)=2$. This doesn't really help me though as it is said that the differential equation must not be solved.
differential-equations
edited Jul 31 at 4:31
zipirovich
9,94011630
asked Jul 31 at 3:42
J-Dorman
555
edited Jul 31 at 4:31
zipirovich
9,94011630
edited Jul 31 at 4:31
zipirovich
9,94011630
edited Jul 31 at 4:31
zipirovich
9,94011630
9,94011630
asked Jul 31 at 3:42
J-Dorman
555
asked Jul 31 at 3:42
J-Dorman
555
asked Jul 31 at 3:42
J-Dorman
555
555
3
It is an autonomous first order equation with a unique solution. At long times these can only blow up or go to an equilibrium point. To see which, check the sign of the derivative at the starting time and move in that direction from the starting position until you hit an equilibrium or have surpassed all equilibria.
– Ian
Jul 31 at 3:45
add a comment |Â
3
It is an autonomous first order equation with a unique solution. At long times these can only blow up or go to an equilibrium point. To see which, check the sign of the derivative at the starting time and move in that direction from the starting position until you hit an equilibrium or have surpassed all equilibria.
– Ian
Jul 31 at 3:45
3
3
It is an autonomous first order equation with a unique solution. At long times these can only blow up or go to an equilibrium point. To see which, check the sign of the derivative at the starting time and move in that direction from the starting position until you hit an equilibrium or have surpassed all equilibria.
– Ian
Jul 31 at 3:45
It is an autonomous first order equation with a unique solution. At long times these can only blow up or go to an equilibrium point. To see which, check the sign of the derivative at the starting time and move in that direction from the starting position until you hit an equilibrium or have surpassed all equilibria.
– Ian
Jul 31 at 3:45
add a comment |Â
2 Answers
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up vote
3
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Let $f(y)$ be the right hand side of the ODE. Note that $f(y) <0$ for $y in (1,7)$. Note that $1$ is an equilibrium (stable, but that does not matter).
In particular, the solution must satisfy $y > 1$ for all $x ge 0$.
Hence if we start at $y(0) = 2$ then $y$ will decrease and is bounded below
by $1$, hence has a limit $y^*$. We must have $y^* = 1$, otherwise we have $y'(x) le - delta$ for some $delta>0$ for all $xge 0$ which is a contradiction.
answered Jul 31 at 4:03
copper.hat
122k557155
add a comment |Â
up vote
0
down vote
For small values of $x, y'(x) < 0$ Which is going to make y(x) a decreasing function. But as $y(x)$ approaches $1, y'(x)$ approaches $0.$ And what happens after that? $y(x)$ becomes stationary.
answered Jul 31 at 3:51
Doug M
39k31749
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let $f(y)$ be the right hand side of the ODE. Note that $f(y) <0$ for $y in (1,7)$. Note that $1$ is an equilibrium (stable, but that does not matter).
In particular, the solution must satisfy $y > 1$ for all $x ge 0$.
Hence if we start at $y(0) = 2$ then $y$ will decrease and is bounded below
by $1$, hence has a limit $y^*$. We must have $y^* = 1$, otherwise we have $y'(x) le - delta$ for some $delta>0$ for all $xge 0$ which is a contradiction.
answered Jul 31 at 4:03
copper.hat
122k557155
add a comment |Â
up vote
3
down vote
Let $f(y)$ be the right hand side of the ODE. Note that $f(y) <0$ for $y in (1,7)$. Note that $1$ is an equilibrium (stable, but that does not matter).
In particular, the solution must satisfy $y > 1$ for all $x ge 0$.
Hence if we start at $y(0) = 2$ then $y$ will decrease and is bounded below
by $1$, hence has a limit $y^*$. We must have $y^* = 1$, otherwise we have $y'(x) le - delta$ for some $delta>0$ for all $xge 0$ which is a contradiction.
answered Jul 31 at 4:03
copper.hat
122k557155
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $f(y)$ be the right hand side of the ODE. Note that $f(y) <0$ for $y in (1,7)$. Note that $1$ is an equilibrium (stable, but that does not matter).
In particular, the solution must satisfy $y > 1$ for all $x ge 0$.
Hence if we start at $y(0) = 2$ then $y$ will decrease and is bounded below
by $1$, hence has a limit $y^*$. We must have $y^* = 1$, otherwise we have $y'(x) le - delta$ for some $delta>0$ for all $xge 0$ which is a contradiction.
answered Jul 31 at 4:03
copper.hat
122k557155
Let $f(y)$ be the right hand side of the ODE. Note that $f(y) <0$ for $y in (1,7)$. Note that $1$ is an equilibrium (stable, but that does not matter).
In particular, the solution must satisfy $y > 1$ for all $x ge 0$.
Hence if we start at $y(0) = 2$ then $y$ will decrease and is bounded below
by $1$, hence has a limit $y^*$. We must have $y^* = 1$, otherwise we have $y'(x) le - delta$ for some $delta>0$ for all $xge 0$ which is a contradiction.
answered Jul 31 at 4:03
copper.hat
122k557155
answered Jul 31 at 4:03
copper.hat
122k557155
answered Jul 31 at 4:03
copper.hat
122k557155
answered Jul 31 at 4:03
copper.hat
122k557155
122k557155
add a comment |Â
add a comment |Â
up vote
0
down vote
For small values of $x, y'(x) < 0$ Which is going to make y(x) a decreasing function. But as $y(x)$ approaches $1, y'(x)$ approaches $0.$ And what happens after that? $y(x)$ becomes stationary.
answered Jul 31 at 3:51
Doug M
39k31749
add a comment |Â
up vote
0
down vote
For small values of $x, y'(x) < 0$ Which is going to make y(x) a decreasing function. But as $y(x)$ approaches $1, y'(x)$ approaches $0.$ And what happens after that? $y(x)$ becomes stationary.
answered Jul 31 at 3:51
Doug M
39k31749
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For small values of $x, y'(x) < 0$ Which is going to make y(x) a decreasing function. But as $y(x)$ approaches $1, y'(x)$ approaches $0.$ And what happens after that? $y(x)$ becomes stationary.
answered Jul 31 at 3:51
Doug M
39k31749
For small values of $x, y'(x) < 0$ Which is going to make y(x) a decreasing function. But as $y(x)$ approaches $1, y'(x)$ approaches $0.$ And what happens after that? $y(x)$ becomes stationary.
answered Jul 31 at 3:51
Doug M
39k31749
answered Jul 31 at 3:51
Doug M
39k31749
answered Jul 31 at 3:51
Doug M
39k31749
answered Jul 31 at 3:51
Doug M
39k31749
39k31749
add a comment |Â
add a comment |Â
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3
It is an autonomous first order equation with a unique solution. At long times these can only blow up or go to an equilibrium point. To see which, check the sign of the derivative at the starting time and move in that direction from the starting position until you hit an equilibrium or have surpassed all equilibria.
– Ian
Jul 31 at 3:45