Mixing
Finding multiple solutions to the ODE $x'(t) = frac{x(t)}{t^2}$
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I am trying to find a way to characterize infinitely many solutions to the ODE $x'(t)=frac{x(t)}{t^2}$ with $x(0) = 0$. I don't necessarily want an answer, as much as I would like guidance in my approach. I found that $$x(t) = begin{cases}0 & t=0 \
e^{-frac{1}{t}} & t > 0 end{cases}$$
solves this system. My idea to classify infinitely many cases is to find a family of functions $x_s$ that are $0$ until time $s$ and end up similar to $e^{-frac{1}{t}}$ after time $s$. I thought about doing this by defining $$x_s(t) = begin{cases}0 & tleq s \
e^{-frac{1}{t-s}} & t > s end{cases}$$
The problem is that we actually have $x_s'(t) = frac{x_s(t)}{(t-s)^2}$ which doesn't quite solve the system, but is really close. I really think we need the $t-s$ term in the denominator, otherwise $x$ wouldn't be continuous but I don't know how to deal with the fact that this causes it to not solve the system. Can anyone help point me in the correct direction?
ordinary-differential-equations
asked Jan 13 at 22:58
J. PistachioJ. Pistachio
488212
$endgroup$
add a comment |
$begingroup$
I am trying to find a way to characterize infinitely many solutions to the ODE $x'(t)=frac{x(t)}{t^2}$ with $x(0) = 0$. I don't necessarily want an answer, as much as I would like guidance in my approach. I found that $$x(t) = begin{cases}0 & t=0 \
e^{-frac{1}{t}} & t > 0 end{cases}$$
solves this system. My idea to classify infinitely many cases is to find a family of functions $x_s$ that are $0$ until time $s$ and end up similar to $e^{-frac{1}{t}}$ after time $s$. I thought about doing this by defining $$x_s(t) = begin{cases}0 & tleq s \
e^{-frac{1}{t-s}} & t > s end{cases}$$
The problem is that we actually have $x_s'(t) = frac{x_s(t)}{(t-s)^2}$ which doesn't quite solve the system, but is really close. I really think we need the $t-s$ term in the denominator, otherwise $x$ wouldn't be continuous but I don't know how to deal with the fact that this causes it to not solve the system. Can anyone help point me in the correct direction?
ordinary-differential-equations
asked Jan 13 at 22:58
J. PistachioJ. Pistachio
488212
$endgroup$
add a comment |
$begingroup$
I am trying to find a way to characterize infinitely many solutions to the ODE $x'(t)=frac{x(t)}{t^2}$ with $x(0) = 0$. I don't necessarily want an answer, as much as I would like guidance in my approach. I found that $$x(t) = begin{cases}0 & t=0 \
e^{-frac{1}{t}} & t > 0 end{cases}$$
solves this system. My idea to classify infinitely many cases is to find a family of functions $x_s$ that are $0$ until time $s$ and end up similar to $e^{-frac{1}{t}}$ after time $s$. I thought about doing this by defining $$x_s(t) = begin{cases}0 & tleq s \
e^{-frac{1}{t-s}} & t > s end{cases}$$
The problem is that we actually have $x_s'(t) = frac{x_s(t)}{(t-s)^2}$ which doesn't quite solve the system, but is really close. I really think we need the $t-s$ term in the denominator, otherwise $x$ wouldn't be continuous but I don't know how to deal with the fact that this causes it to not solve the system. Can anyone help point me in the correct direction?
ordinary-differential-equations
asked Jan 13 at 22:58
J. PistachioJ. Pistachio
488212
$endgroup$
I am trying to find a way to characterize infinitely many solutions to the ODE $x'(t)=frac{x(t)}{t^2}$ with $x(0) = 0$. I don't necessarily want an answer, as much as I would like guidance in my approach. I found that $$x(t) = begin{cases}0 & t=0 \
e^{-frac{1}{t}} & t > 0 end{cases}$$
solves this system. My idea to classify infinitely many cases is to find a family of functions $x_s$ that are $0$ until time $s$ and end up similar to $e^{-frac{1}{t}}$ after time $s$. I thought about doing this by defining $$x_s(t) = begin{cases}0 & tleq s \
e^{-frac{1}{t-s}} & t > s end{cases}$$
The problem is that we actually have $x_s'(t) = frac{x_s(t)}{(t-s)^2}$ which doesn't quite solve the system, but is really close. I really think we need the $t-s$ term in the denominator, otherwise $x$ wouldn't be continuous but I don't know how to deal with the fact that this causes it to not solve the system. Can anyone help point me in the correct direction?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 13 at 22:58
J. PistachioJ. Pistachio
488212
asked Jan 13 at 22:58
J. PistachioJ. Pistachio
488212
asked Jan 13 at 22:58
J. PistachioJ. Pistachio
488212
asked Jan 13 at 22:58
J. PistachioJ. Pistachio
488212
asked Jan 13 at 22:58
J. PistachioJ. Pistachio
488212
488212
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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How about the infinite set of solutions
$$x(t) = begin{cases} 0 & tleq 0 \
c, e^{-1/t} & t>0 end{cases}$$ with $c$ an arbitrary constant?
answered Jan 13 at 23:06
FabianFabian
19.8k3674
$endgroup$
$begingroup$
Of course! That is so much simpler than I was anticipating.
$endgroup$
– J. Pistachio
Jan 13 at 23:16
add a comment |
$begingroup$
Your "shifted" solutions are not right. Your equation is separable.
$$
frac{dx}{x}=frac{dt}{t^2}.
$$
After integration you get
$$
ln|x|=-frac 1{t}+c,
$$
or
$$
x=Ce^{-1/t},qquad C>0.
$$
You also have the special solution $x(t)=0$.
The formula of general solutions works both for $t<0$ and $t>0$. Solutions for $t<0$ approach infinity as $tto 0^-$ except for $x(t)=0$. The solutions for $t>0$ have limit zero as $tto 0^+$. You can formally paste the zero solution for $t<0$ with either solution of the form $x=Ce^{-1/t}$ for $t>0$ by defining $x(0)=0$ but this pasted solution is not a solution of the given equation, since the latter does not make sense at $t=0$.
Summing up, instead of a shift, you have a dilation parameter $C>0$.
answered Jan 13 at 23:16
GReyesGReyes
1,25915
$endgroup$
$begingroup$
The first derivative (of the pasted solution) has a limit of $0$ at $t to 0^+$ as well.
$endgroup$
– Dylan
Jan 14 at 8:46
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
How about the infinite set of solutions
$$x(t) = begin{cases} 0 & tleq 0 \
c, e^{-1/t} & t>0 end{cases}$$ with $c$ an arbitrary constant?
answered Jan 13 at 23:06
FabianFabian
19.8k3674
$endgroup$
$begingroup$
Of course! That is so much simpler than I was anticipating.
$endgroup$
– J. Pistachio
Jan 13 at 23:16
add a comment |
$begingroup$
How about the infinite set of solutions
$$x(t) = begin{cases} 0 & tleq 0 \
c, e^{-1/t} & t>0 end{cases}$$ with $c$ an arbitrary constant?
answered Jan 13 at 23:06
FabianFabian
19.8k3674
$endgroup$
$begingroup$
Of course! That is so much simpler than I was anticipating.
$endgroup$
– J. Pistachio
Jan 13 at 23:16
add a comment |
$begingroup$
How about the infinite set of solutions
$$x(t) = begin{cases} 0 & tleq 0 \
c, e^{-1/t} & t>0 end{cases}$$ with $c$ an arbitrary constant?
answered Jan 13 at 23:06
FabianFabian
19.8k3674
$endgroup$
How about the infinite set of solutions
$$x(t) = begin{cases} 0 & tleq 0 \
c, e^{-1/t} & t>0 end{cases}$$ with $c$ an arbitrary constant?
answered Jan 13 at 23:06
FabianFabian
19.8k3674
answered Jan 13 at 23:06
FabianFabian
19.8k3674
answered Jan 13 at 23:06
FabianFabian
19.8k3674
answered Jan 13 at 23:06
FabianFabian
19.8k3674
19.8k3674
$begingroup$
Of course! That is so much simpler than I was anticipating.
$endgroup$
– J. Pistachio
Jan 13 at 23:16
add a comment |
$begingroup$
Of course! That is so much simpler than I was anticipating.
$endgroup$
– J. Pistachio
Jan 13 at 23:16
$begingroup$
Of course! That is so much simpler than I was anticipating.
$endgroup$
– J. Pistachio
Jan 13 at 23:16
$begingroup$
Of course! That is so much simpler than I was anticipating.
$endgroup$
– J. Pistachio
Jan 13 at 23:16
add a comment |
$begingroup$
Your "shifted" solutions are not right. Your equation is separable.
$$
frac{dx}{x}=frac{dt}{t^2}.
$$
After integration you get
$$
ln|x|=-frac 1{t}+c,
$$
or
$$
x=Ce^{-1/t},qquad C>0.
$$
You also have the special solution $x(t)=0$.
The formula of general solutions works both for $t<0$ and $t>0$. Solutions for $t<0$ approach infinity as $tto 0^-$ except for $x(t)=0$. The solutions for $t>0$ have limit zero as $tto 0^+$. You can formally paste the zero solution for $t<0$ with either solution of the form $x=Ce^{-1/t}$ for $t>0$ by defining $x(0)=0$ but this pasted solution is not a solution of the given equation, since the latter does not make sense at $t=0$.
Summing up, instead of a shift, you have a dilation parameter $C>0$.
answered Jan 13 at 23:16
GReyesGReyes
1,25915
$endgroup$
$begingroup$
The first derivative (of the pasted solution) has a limit of $0$ at $t to 0^+$ as well.
$endgroup$
– Dylan
Jan 14 at 8:46
add a comment |
$begingroup$
Your "shifted" solutions are not right. Your equation is separable.
$$
frac{dx}{x}=frac{dt}{t^2}.
$$
After integration you get
$$
ln|x|=-frac 1{t}+c,
$$
or
$$
x=Ce^{-1/t},qquad C>0.
$$
You also have the special solution $x(t)=0$.
The formula of general solutions works both for $t<0$ and $t>0$. Solutions for $t<0$ approach infinity as $tto 0^-$ except for $x(t)=0$. The solutions for $t>0$ have limit zero as $tto 0^+$. You can formally paste the zero solution for $t<0$ with either solution of the form $x=Ce^{-1/t}$ for $t>0$ by defining $x(0)=0$ but this pasted solution is not a solution of the given equation, since the latter does not make sense at $t=0$.
Summing up, instead of a shift, you have a dilation parameter $C>0$.
answered Jan 13 at 23:16
GReyesGReyes
1,25915
$endgroup$
$begingroup$
The first derivative (of the pasted solution) has a limit of $0$ at $t to 0^+$ as well.
$endgroup$
– Dylan
Jan 14 at 8:46
add a comment |
$begingroup$
Your "shifted" solutions are not right. Your equation is separable.
$$
frac{dx}{x}=frac{dt}{t^2}.
$$
After integration you get
$$
ln|x|=-frac 1{t}+c,
$$
or
$$
x=Ce^{-1/t},qquad C>0.
$$
You also have the special solution $x(t)=0$.
The formula of general solutions works both for $t<0$ and $t>0$. Solutions for $t<0$ approach infinity as $tto 0^-$ except for $x(t)=0$. The solutions for $t>0$ have limit zero as $tto 0^+$. You can formally paste the zero solution for $t<0$ with either solution of the form $x=Ce^{-1/t}$ for $t>0$ by defining $x(0)=0$ but this pasted solution is not a solution of the given equation, since the latter does not make sense at $t=0$.
Summing up, instead of a shift, you have a dilation parameter $C>0$.
answered Jan 13 at 23:16
GReyesGReyes
1,25915
$endgroup$
Your "shifted" solutions are not right. Your equation is separable.
$$
frac{dx}{x}=frac{dt}{t^2}.
$$
After integration you get
$$
ln|x|=-frac 1{t}+c,
$$
or
$$
x=Ce^{-1/t},qquad C>0.
$$
You also have the special solution $x(t)=0$.
The formula of general solutions works both for $t<0$ and $t>0$. Solutions for $t<0$ approach infinity as $tto 0^-$ except for $x(t)=0$. The solutions for $t>0$ have limit zero as $tto 0^+$. You can formally paste the zero solution for $t<0$ with either solution of the form $x=Ce^{-1/t}$ for $t>0$ by defining $x(0)=0$ but this pasted solution is not a solution of the given equation, since the latter does not make sense at $t=0$.
Summing up, instead of a shift, you have a dilation parameter $C>0$.
answered Jan 13 at 23:16
GReyesGReyes
1,25915
answered Jan 13 at 23:16
GReyesGReyes
1,25915
answered Jan 13 at 23:16
GReyesGReyes
1,25915
answered Jan 13 at 23:16
GReyesGReyes
1,25915
1,25915
$begingroup$
The first derivative (of the pasted solution) has a limit of $0$ at $t to 0^+$ as well.
$endgroup$
– Dylan
Jan 14 at 8:46
add a comment |
$begingroup$
The first derivative (of the pasted solution) has a limit of $0$ at $t to 0^+$ as well.
$endgroup$
– Dylan
Jan 14 at 8:46
$begingroup$
The first derivative (of the pasted solution) has a limit of $0$ at $t to 0^+$ as well.
$endgroup$
– Dylan
Jan 14 at 8:46
$begingroup$
The first derivative (of the pasted solution) has a limit of $0$ at $t to 0^+$ as well.
$endgroup$
– Dylan
Jan 14 at 8:46
add a comment |
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