Ordinary non-linear differential equation [closed]
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Given the following differential equation:
$$ddot{x}(t) + x(t) = frac{1}{x(t)}$$
can your solution be explained in a closed form?
real-analysis differential-equations
asked Nov 13 at 14:00
TeM
390215
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Jens, Kelvin Lois, A. Pongrácz, rtybase Nov 13 at 19:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Jens, Kelvin Lois, A. Pongrácz, rtybase
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up vote
-1
down vote
favorite
Given the following differential equation:
$$ddot{x}(t) + x(t) = frac{1}{x(t)}$$
can your solution be explained in a closed form?
real-analysis differential-equations
asked Nov 13 at 14:00
TeM
390215
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Jens, Kelvin Lois, A. Pongrácz, rtybase Nov 13 at 19:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Jens, Kelvin Lois, A. Pongrácz, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Given the following differential equation:
$$ddot{x}(t) + x(t) = frac{1}{x(t)}$$
can your solution be explained in a closed form?
real-analysis differential-equations
asked Nov 13 at 14:00
TeM
390215
Given the following differential equation:
$$ddot{x}(t) + x(t) = frac{1}{x(t)}$$
can your solution be explained in a closed form?
real-analysis differential-equations
real-analysis differential-equations
asked Nov 13 at 14:00
TeM
390215
asked Nov 13 at 14:00
TeM
390215
asked Nov 13 at 14:00
TeM
390215
asked Nov 13 at 14:00
TeM
390215
asked Nov 13 at 14:00
TeM
390215
390215
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Jens, Kelvin Lois, A. Pongrácz, rtybase Nov 13 at 19:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Jens, Kelvin Lois, A. Pongrácz, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Jens, Kelvin Lois, A. Pongrácz, rtybase Nov 13 at 19:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Jens, Kelvin Lois, A. Pongrácz, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
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1
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Use $ddot{x}=dot{x}, ddot{x}/dx$, so then $dot{x}, ddot{x}/dx = x^{-1}-x$, and then integrate. This is a general procedure for dealing with equations with no explicit time dependence.
answered Nov 13 at 14:04
Richard Martin
1,3938
So it is correct to get to $t = int frac{1}{ln x - frac{x^2}{2}},text{d}x$ ?
– TeM
Nov 13 at 14:12
1
No, $dot{x}^2/2=ln x - x^2/2$, prov $x>0$.Then take square root and integrate (which cannot be done in closed form)
– Richard Martin
Nov 13 at 14:17
I realized immediately after the mistake. Thank you very much and sorry for the silly question.
– TeM
Nov 13 at 14:18
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Use $ddot{x}=dot{x}, ddot{x}/dx$, so then $dot{x}, ddot{x}/dx = x^{-1}-x$, and then integrate. This is a general procedure for dealing with equations with no explicit time dependence.
answered Nov 13 at 14:04
Richard Martin
1,3938
So it is correct to get to $t = int frac{1}{ln x - frac{x^2}{2}},text{d}x$ ?
– TeM
Nov 13 at 14:12
1
No, $dot{x}^2/2=ln x - x^2/2$, prov $x>0$.Then take square root and integrate (which cannot be done in closed form)
– Richard Martin
Nov 13 at 14:17
I realized immediately after the mistake. Thank you very much and sorry for the silly question.
– TeM
Nov 13 at 14:18
add a comment |
up vote
1
down vote
accepted
Use $ddot{x}=dot{x}, ddot{x}/dx$, so then $dot{x}, ddot{x}/dx = x^{-1}-x$, and then integrate. This is a general procedure for dealing with equations with no explicit time dependence.
answered Nov 13 at 14:04
Richard Martin
1,3938
So it is correct to get to $t = int frac{1}{ln x - frac{x^2}{2}},text{d}x$ ?
– TeM
Nov 13 at 14:12
1
No, $dot{x}^2/2=ln x - x^2/2$, prov $x>0$.Then take square root and integrate (which cannot be done in closed form)
– Richard Martin
Nov 13 at 14:17
I realized immediately after the mistake. Thank you very much and sorry for the silly question.
– TeM
Nov 13 at 14:18
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Use $ddot{x}=dot{x}, ddot{x}/dx$, so then $dot{x}, ddot{x}/dx = x^{-1}-x$, and then integrate. This is a general procedure for dealing with equations with no explicit time dependence.
answered Nov 13 at 14:04
Richard Martin
1,3938
Use $ddot{x}=dot{x}, ddot{x}/dx$, so then $dot{x}, ddot{x}/dx = x^{-1}-x$, and then integrate. This is a general procedure for dealing with equations with no explicit time dependence.
answered Nov 13 at 14:04
Richard Martin
1,3938
answered Nov 13 at 14:04
Richard Martin
1,3938
answered Nov 13 at 14:04
Richard Martin
1,3938
answered Nov 13 at 14:04
Richard Martin
1,3938
1,3938
So it is correct to get to $t = int frac{1}{ln x - frac{x^2}{2}},text{d}x$ ?
– TeM
Nov 13 at 14:12
1
No, $dot{x}^2/2=ln x - x^2/2$, prov $x>0$.Then take square root and integrate (which cannot be done in closed form)
– Richard Martin
Nov 13 at 14:17
I realized immediately after the mistake. Thank you very much and sorry for the silly question.
– TeM
Nov 13 at 14:18
add a comment |
So it is correct to get to $t = int frac{1}{ln x - frac{x^2}{2}},text{d}x$ ?
– TeM
Nov 13 at 14:12
1
No, $dot{x}^2/2=ln x - x^2/2$, prov $x>0$.Then take square root and integrate (which cannot be done in closed form)
– Richard Martin
Nov 13 at 14:17
I realized immediately after the mistake. Thank you very much and sorry for the silly question.
– TeM
Nov 13 at 14:18
So it is correct to get to $t = int frac{1}{ln x - frac{x^2}{2}},text{d}x$ ?
– TeM
Nov 13 at 14:12
So it is correct to get to $t = int frac{1}{ln x - frac{x^2}{2}},text{d}x$ ?
– TeM
Nov 13 at 14:12
1
1
No, $dot{x}^2/2=ln x - x^2/2$, prov $x>0$.Then take square root and integrate (which cannot be done in closed form)
– Richard Martin
Nov 13 at 14:17
No, $dot{x}^2/2=ln x - x^2/2$, prov $x>0$.Then take square root and integrate (which cannot be done in closed form)
– Richard Martin
Nov 13 at 14:17
I realized immediately after the mistake. Thank you very much and sorry for the silly question.
– TeM
Nov 13 at 14:18
I realized immediately after the mistake. Thank you very much and sorry for the silly question.
– TeM
Nov 13 at 14:18
add a comment |
