How to solve the differential equation $frac{mathrm{d}^2x}{mathrm{d}t^2}=frac{Kx}{sqrt{L^2-x^2}}$? [closed]
-2
$begingroup$
How to solve the differential equation $frac{mathrm{d}^2x}{mathrm{d}t^2}=frac{Kx}{sqrt{L^2-x^2}}$?
differential-equations
asked Nov 30 '18 at 8:01
tan9ptan9p
455313
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closed as off-topic by Saad, Dylan, Paul Frost, Adrian Keister, user10354138 Nov 30 '18 at 16:38
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." – Saad, Dylan, Paul Frost, Adrian Keister, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
-2
$begingroup$
How to solve the differential equation $frac{mathrm{d}^2x}{mathrm{d}t^2}=frac{Kx}{sqrt{L^2-x^2}}$?
differential-equations
asked Nov 30 '18 at 8:01
tan9ptan9p
455313
$endgroup$
closed as off-topic by Saad, Dylan, Paul Frost, Adrian Keister, user10354138 Nov 30 '18 at 16:38
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." – Saad, Dylan, Paul Frost, Adrian Keister, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
integrate twice
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:08
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The rhs is in terms of x @Guacho Perez
$endgroup$
– Ahmed S. Attaalla
Nov 30 '18 at 8:15
$begingroup$
@AhmedS.Attaalla true, my mistake. Then I doubt this has a nice closed form solution.
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:17
$begingroup$
Any equation of the form $d^2 x/dt^2=f(x)$ can be solved (in principle) by a standard trick: multiply by $dx/dt$ and then integrate with respect to $t$.
$endgroup$
– Hans Lundmark
Nov 30 '18 at 8:39
add a comment |
-2
-2
-2
1
$begingroup$
How to solve the differential equation $frac{mathrm{d}^2x}{mathrm{d}t^2}=frac{Kx}{sqrt{L^2-x^2}}$?
differential-equations
asked Nov 30 '18 at 8:01
tan9ptan9p
455313
$endgroup$
How to solve the differential equation $frac{mathrm{d}^2x}{mathrm{d}t^2}=frac{Kx}{sqrt{L^2-x^2}}$?
differential-equations
differential-equations
asked Nov 30 '18 at 8:01
tan9ptan9p
455313
asked Nov 30 '18 at 8:01
tan9ptan9p
455313
edited Nov 30 '18 at 8:07
tan9p
asked Nov 30 '18 at 8:01
tan9ptan9p
455313
asked Nov 30 '18 at 8:01
tan9ptan9p
455313
asked Nov 30 '18 at 8:01
tan9ptan9p
455313
455313
closed as off-topic by Saad, Dylan, Paul Frost, Adrian Keister, user10354138 Nov 30 '18 at 16:38
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." – Saad, Dylan, Paul Frost, Adrian Keister, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Dylan, Paul Frost, Adrian Keister, user10354138 Nov 30 '18 at 16:38
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." – Saad, Dylan, Paul Frost, Adrian Keister, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
integrate twice
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:08
$begingroup$
The rhs is in terms of x @Guacho Perez
$endgroup$
– Ahmed S. Attaalla
Nov 30 '18 at 8:15
$begingroup$
@AhmedS.Attaalla true, my mistake. Then I doubt this has a nice closed form solution.
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:17
$begingroup$
Any equation of the form $d^2 x/dt^2=f(x)$ can be solved (in principle) by a standard trick: multiply by $dx/dt$ and then integrate with respect to $t$.
$endgroup$
– Hans Lundmark
Nov 30 '18 at 8:39
add a comment |
$begingroup$
integrate twice
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:08
$begingroup$
The rhs is in terms of x @Guacho Perez
$endgroup$
– Ahmed S. Attaalla
Nov 30 '18 at 8:15
$begingroup$
@AhmedS.Attaalla true, my mistake. Then I doubt this has a nice closed form solution.
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:17
$begingroup$
Any equation of the form $d^2 x/dt^2=f(x)$ can be solved (in principle) by a standard trick: multiply by $dx/dt$ and then integrate with respect to $t$.
$endgroup$
– Hans Lundmark
Nov 30 '18 at 8:39
$begingroup$
integrate twice
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:08
$begingroup$
integrate twice
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:08
$begingroup$
The rhs is in terms of x @Guacho Perez
$endgroup$
– Ahmed S. Attaalla
Nov 30 '18 at 8:15
$begingroup$
The rhs is in terms of x @Guacho Perez
$endgroup$
– Ahmed S. Attaalla
Nov 30 '18 at 8:15
$begingroup$
@AhmedS.Attaalla true, my mistake. Then I doubt this has a nice closed form solution.
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:17
$begingroup$
@AhmedS.Attaalla true, my mistake. Then I doubt this has a nice closed form solution.
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:17
$begingroup$
Any equation of the form $d^2 x/dt^2=f(x)$ can be solved (in principle) by a standard trick: multiply by $dx/dt$ and then integrate with respect to $t$.
$endgroup$
– Hans Lundmark
Nov 30 '18 at 8:39
$begingroup$
Any equation of the form $d^2 x/dt^2=f(x)$ can be solved (in principle) by a standard trick: multiply by $dx/dt$ and then integrate with respect to $t$.
$endgroup$
– Hans Lundmark
Nov 30 '18 at 8:39
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
Hint: $$frac{xx'}{sqrt{L^2-x^2}} = -left(sqrt{L^2-x^2}right)'$$and
$$x''x' = frac 12 left((x')^2right)'$$
answered Nov 30 '18 at 14:04
TZakrevskiyTZakrevskiy
20.1k12354
$endgroup$
$begingroup$
Here missed a ' in $frac{1}{2}((x')^2)'$.I calculated in this way, and got $x'=pmsqrt{C-2Ksqrt{L^2-x^2}}$.then I don't know how to calculate that.
$endgroup$
– tan9p
Nov 30 '18 at 15:47
$begingroup$
@tan9p of course, fixed derivative it in the answer, thank you
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:49
$begingroup$
For the solution itself, if you do not have any additional info on the equation (e.g. $x(0)$ and $x'(0)$), the viable approach I see is to make a change of variables $x(t) = L sin (y(t))$. After that you can arrive to an expression for $y(t)$ as an implicit function.
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:59
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Hint: $$frac{xx'}{sqrt{L^2-x^2}} = -left(sqrt{L^2-x^2}right)'$$and
$$x''x' = frac 12 left((x')^2right)'$$
answered Nov 30 '18 at 14:04
TZakrevskiyTZakrevskiy
20.1k12354
$endgroup$
$begingroup$
Here missed a ' in $frac{1}{2}((x')^2)'$.I calculated in this way, and got $x'=pmsqrt{C-2Ksqrt{L^2-x^2}}$.then I don't know how to calculate that.
$endgroup$
– tan9p
Nov 30 '18 at 15:47
$begingroup$
@tan9p of course, fixed derivative it in the answer, thank you
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:49
$begingroup$
For the solution itself, if you do not have any additional info on the equation (e.g. $x(0)$ and $x'(0)$), the viable approach I see is to make a change of variables $x(t) = L sin (y(t))$. After that you can arrive to an expression for $y(t)$ as an implicit function.
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:59
add a comment |
1
$begingroup$
Hint: $$frac{xx'}{sqrt{L^2-x^2}} = -left(sqrt{L^2-x^2}right)'$$and
$$x''x' = frac 12 left((x')^2right)'$$
answered Nov 30 '18 at 14:04
TZakrevskiyTZakrevskiy
20.1k12354
$endgroup$
$begingroup$
Here missed a ' in $frac{1}{2}((x')^2)'$.I calculated in this way, and got $x'=pmsqrt{C-2Ksqrt{L^2-x^2}}$.then I don't know how to calculate that.
$endgroup$
– tan9p
Nov 30 '18 at 15:47
$begingroup$
@tan9p of course, fixed derivative it in the answer, thank you
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:49
$begingroup$
For the solution itself, if you do not have any additional info on the equation (e.g. $x(0)$ and $x'(0)$), the viable approach I see is to make a change of variables $x(t) = L sin (y(t))$. After that you can arrive to an expression for $y(t)$ as an implicit function.
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:59
add a comment |
1
1
1
$begingroup$
Hint: $$frac{xx'}{sqrt{L^2-x^2}} = -left(sqrt{L^2-x^2}right)'$$and
$$x''x' = frac 12 left((x')^2right)'$$
answered Nov 30 '18 at 14:04
TZakrevskiyTZakrevskiy
20.1k12354
$endgroup$
Hint: $$frac{xx'}{sqrt{L^2-x^2}} = -left(sqrt{L^2-x^2}right)'$$and
$$x''x' = frac 12 left((x')^2right)'$$
answered Nov 30 '18 at 14:04
TZakrevskiyTZakrevskiy
20.1k12354
edited Nov 30 '18 at 15:49
answered Nov 30 '18 at 14:04
TZakrevskiyTZakrevskiy
20.1k12354
answered Nov 30 '18 at 14:04
TZakrevskiyTZakrevskiy
20.1k12354
answered Nov 30 '18 at 14:04
TZakrevskiyTZakrevskiy
20.1k12354
20.1k12354
$begingroup$
Here missed a ' in $frac{1}{2}((x')^2)'$.I calculated in this way, and got $x'=pmsqrt{C-2Ksqrt{L^2-x^2}}$.then I don't know how to calculate that.
$endgroup$
– tan9p
Nov 30 '18 at 15:47
$begingroup$
@tan9p of course, fixed derivative it in the answer, thank you
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:49
$begingroup$
For the solution itself, if you do not have any additional info on the equation (e.g. $x(0)$ and $x'(0)$), the viable approach I see is to make a change of variables $x(t) = L sin (y(t))$. After that you can arrive to an expression for $y(t)$ as an implicit function.
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:59
add a comment |
$begingroup$
Here missed a ' in $frac{1}{2}((x')^2)'$.I calculated in this way, and got $x'=pmsqrt{C-2Ksqrt{L^2-x^2}}$.then I don't know how to calculate that.
$endgroup$
– tan9p
Nov 30 '18 at 15:47
$begingroup$
@tan9p of course, fixed derivative it in the answer, thank you
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:49
$begingroup$
For the solution itself, if you do not have any additional info on the equation (e.g. $x(0)$ and $x'(0)$), the viable approach I see is to make a change of variables $x(t) = L sin (y(t))$. After that you can arrive to an expression for $y(t)$ as an implicit function.
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:59
$begingroup$
Here missed a ' in $frac{1}{2}((x')^2)'$.I calculated in this way, and got $x'=pmsqrt{C-2Ksqrt{L^2-x^2}}$.then I don't know how to calculate that.
$endgroup$
– tan9p
Nov 30 '18 at 15:47
$begingroup$
Here missed a ' in $frac{1}{2}((x')^2)'$.I calculated in this way, and got $x'=pmsqrt{C-2Ksqrt{L^2-x^2}}$.then I don't know how to calculate that.
$endgroup$
– tan9p
Nov 30 '18 at 15:47
$begingroup$
@tan9p of course, fixed derivative it in the answer, thank you
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:49
$begingroup$
@tan9p of course, fixed derivative it in the answer, thank you
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:49
$begingroup$
For the solution itself, if you do not have any additional info on the equation (e.g. $x(0)$ and $x'(0)$), the viable approach I see is to make a change of variables $x(t) = L sin (y(t))$. After that you can arrive to an expression for $y(t)$ as an implicit function.
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:59
$begingroup$
For the solution itself, if you do not have any additional info on the equation (e.g. $x(0)$ and $x'(0)$), the viable approach I see is to make a change of variables $x(t) = L sin (y(t))$. After that you can arrive to an expression for $y(t)$ as an implicit function.
$endgroup$
– TZakrevskiy
Nov 30 '18 at 15:59
add a comment |
$begingroup$
integrate twice
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:08
$begingroup$
The rhs is in terms of x @Guacho Perez
$endgroup$
– Ahmed S. Attaalla
Nov 30 '18 at 8:15
$begingroup$
@AhmedS.Attaalla true, my mistake. Then I doubt this has a nice closed form solution.
$endgroup$
– Guacho Perez
Nov 30 '18 at 8:17
$begingroup$
Any equation of the form $d^2 x/dt^2=f(x)$ can be solved (in principle) by a standard trick: multiply by $dx/dt$ and then integrate with respect to $t$.
$endgroup$
– Hans Lundmark
Nov 30 '18 at 8:39