How to find an exact solution to a certain linear second-order non-homogeneous differential equationadvance techniques for converting second-order linear ODEs with rational function coefficients to some known ODE typesNumerical integration of nonlinear second order equationnon-homogeneous constant co-efficient 2nd order linear differential equationNeed some help with a second-order non-linear differential equationsolving second order linear differential equationNon-homogeneous solution to the second order differential equationSecond Order Non-Linear Differential Equation - AirDragSolving non-homogeneous second order differential equationExact Solution of Second Order Non-Linear Differential EquationSecond order and non-linear differential equationHow to find the solution of this second order differential equation with the time-varying coefficients?
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How to find an exact solution to a certain linear second-order non-homogeneous differential equation
advance techniques for converting second-order linear ODEs with rational function coefficients to some known ODE typesNumerical integration of nonlinear second order equationnon-homogeneous constant co-efficient 2nd order linear differential equationNeed some help with a second-order non-linear differential equationsolving second order linear differential equationNon-homogeneous solution to the second order differential equationSecond Order Non-Linear Differential Equation - AirDragSolving non-homogeneous second order differential equationExact Solution of Second Order Non-Linear Differential EquationSecond order and non-linear differential equationHow to find the solution of this second order differential equation with the time-varying coefficients?
$begingroup$
The equation below came up in some work I'm doing and I'm at a loss as to how to get a non-numerical solution:
$ddot x + k_1 tan( k_1 t) sec(k_1 t)dot x+k_2cos(k_1t) x=k_3cos(k_1t)$
ordinary-differential-equations
edited Mar 28 at 19:45
Jean Marie
31.2k42255
asked Mar 28 at 19:07
Joe CataniaJoe Catania
61
New contributor
Joe Catania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 2 more comments
$begingroup$
The equation below came up in some work I'm doing and I'm at a loss as to how to get a non-numerical solution:
$ddot x + k_1 tan( k_1 t) sec(k_1 t)dot x+k_2cos(k_1t) x=k_3cos(k_1t)$
ordinary-differential-equations
edited Mar 28 at 19:45
Jean Marie
31.2k42255
asked Mar 28 at 19:07
Joe CataniaJoe Catania
61
New contributor
Joe Catania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
A more detailed explanation of how it "came up in some work" might be interesting in itself and motivate why you want "a non-numerical solution". In other words, please add more context to avoid giving Readers the impression you think of them as a crank to be turned.
$endgroup$
– hardmath
Mar 28 at 19:17
1
$begingroup$
I'd break up the tangent and secant terms in the expressions that involve $sin$ and $cos$ to simplify first.
$endgroup$
– Rebellos
Mar 28 at 19:18
1
$begingroup$
Mathematica doesn't give an answer - I suspect there's no way to get a non-numerical solution.
$endgroup$
– Adrian Keister
Mar 28 at 19:35
1
$begingroup$
Have you noticed that there is a particular solution which is $x(t)=k_3/k_2=const.$. It suffices now to find the general solution to the same diff. equation with zero RHS and add to this constant...
$endgroup$
– Jean Marie
Mar 28 at 19:49
$begingroup$
@hardmath- I have a graphical plot that was generated by numerical iteration but it might be nice if there was a closed form solution available to check the result. It also would be nice if there was lookup table for these things.
$endgroup$
– Joe Catania
Mar 28 at 21:45
|
show 2 more comments
$begingroup$
The equation below came up in some work I'm doing and I'm at a loss as to how to get a non-numerical solution:
$ddot x + k_1 tan( k_1 t) sec(k_1 t)dot x+k_2cos(k_1t) x=k_3cos(k_1t)$
ordinary-differential-equations
edited Mar 28 at 19:45
Jean Marie
31.2k42255
asked Mar 28 at 19:07
Joe CataniaJoe Catania
61
New contributor
Joe Catania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The equation below came up in some work I'm doing and I'm at a loss as to how to get a non-numerical solution:
$ddot x + k_1 tan( k_1 t) sec(k_1 t)dot x+k_2cos(k_1t) x=k_3cos(k_1t)$
ordinary-differential-equations
ordinary-differential-equations
edited Mar 28 at 19:45
Jean Marie
31.2k42255
asked Mar 28 at 19:07
Joe CataniaJoe Catania
61
New contributor
Joe Catania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 19:45
Jean Marie
31.2k42255
asked Mar 28 at 19:07
Joe CataniaJoe Catania
61
New contributor
Joe Catania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 19:45
Jean Marie
31.2k42255
edited Mar 28 at 19:45
Jean Marie
31.2k42255
edited Mar 28 at 19:45
Jean Marie
31.2k42255
31.2k42255
asked Mar 28 at 19:07
Joe CataniaJoe Catania
61
New contributor
Joe Catania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 19:07
Joe CataniaJoe Catania
61
asked Mar 28 at 19:07
Joe CataniaJoe Catania
61
61
New contributor
Joe Catania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Joe Catania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Joe Catania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
A more detailed explanation of how it "came up in some work" might be interesting in itself and motivate why you want "a non-numerical solution". In other words, please add more context to avoid giving Readers the impression you think of them as a crank to be turned.
$endgroup$
– hardmath
Mar 28 at 19:17
1
$begingroup$
I'd break up the tangent and secant terms in the expressions that involve $sin$ and $cos$ to simplify first.
$endgroup$
– Rebellos
Mar 28 at 19:18
1
$begingroup$
Mathematica doesn't give an answer - I suspect there's no way to get a non-numerical solution.
$endgroup$
– Adrian Keister
Mar 28 at 19:35
1
$begingroup$
Have you noticed that there is a particular solution which is $x(t)=k_3/k_2=const.$. It suffices now to find the general solution to the same diff. equation with zero RHS and add to this constant...
$endgroup$
– Jean Marie
Mar 28 at 19:49
$begingroup$
@hardmath- I have a graphical plot that was generated by numerical iteration but it might be nice if there was a closed form solution available to check the result. It also would be nice if there was lookup table for these things.
$endgroup$
– Joe Catania
Mar 28 at 21:45
|
show 2 more comments
$begingroup$
A more detailed explanation of how it "came up in some work" might be interesting in itself and motivate why you want "a non-numerical solution". In other words, please add more context to avoid giving Readers the impression you think of them as a crank to be turned.
$endgroup$
– hardmath
Mar 28 at 19:17
1
$begingroup$
I'd break up the tangent and secant terms in the expressions that involve $sin$ and $cos$ to simplify first.
$endgroup$
– Rebellos
Mar 28 at 19:18
1
$begingroup$
Mathematica doesn't give an answer - I suspect there's no way to get a non-numerical solution.
$endgroup$
– Adrian Keister
Mar 28 at 19:35
1
$begingroup$
Have you noticed that there is a particular solution which is $x(t)=k_3/k_2=const.$. It suffices now to find the general solution to the same diff. equation with zero RHS and add to this constant...
$endgroup$
– Jean Marie
Mar 28 at 19:49
$begingroup$
@hardmath- I have a graphical plot that was generated by numerical iteration but it might be nice if there was a closed form solution available to check the result. It also would be nice if there was lookup table for these things.
$endgroup$
– Joe Catania
Mar 28 at 21:45
$begingroup$
A more detailed explanation of how it "came up in some work" might be interesting in itself and motivate why you want "a non-numerical solution". In other words, please add more context to avoid giving Readers the impression you think of them as a crank to be turned.
$endgroup$
– hardmath
Mar 28 at 19:17
$begingroup$
A more detailed explanation of how it "came up in some work" might be interesting in itself and motivate why you want "a non-numerical solution". In other words, please add more context to avoid giving Readers the impression you think of them as a crank to be turned.
$endgroup$
– hardmath
Mar 28 at 19:17
1
1
$begingroup$
I'd break up the tangent and secant terms in the expressions that involve $sin$ and $cos$ to simplify first.
$endgroup$
– Rebellos
Mar 28 at 19:18
$begingroup$
I'd break up the tangent and secant terms in the expressions that involve $sin$ and $cos$ to simplify first.
$endgroup$
– Rebellos
Mar 28 at 19:18
1
1
$begingroup$
Mathematica doesn't give an answer - I suspect there's no way to get a non-numerical solution.
$endgroup$
– Adrian Keister
Mar 28 at 19:35
$begingroup$
Mathematica doesn't give an answer - I suspect there's no way to get a non-numerical solution.
$endgroup$
– Adrian Keister
Mar 28 at 19:35
1
1
$begingroup$
Have you noticed that there is a particular solution which is $x(t)=k_3/k_2=const.$. It suffices now to find the general solution to the same diff. equation with zero RHS and add to this constant...
$endgroup$
– Jean Marie
Mar 28 at 19:49
$begingroup$
Have you noticed that there is a particular solution which is $x(t)=k_3/k_2=const.$. It suffices now to find the general solution to the same diff. equation with zero RHS and add to this constant...
$endgroup$
– Jean Marie
Mar 28 at 19:49
$begingroup$
@hardmath- I have a graphical plot that was generated by numerical iteration but it might be nice if there was a closed form solution available to check the result. It also would be nice if there was lookup table for these things.
$endgroup$
– Joe Catania
Mar 28 at 21:45
$begingroup$
@hardmath- I have a graphical plot that was generated by numerical iteration but it might be nice if there was a closed form solution available to check the result. It also would be nice if there was lookup table for these things.
$endgroup$
– Joe Catania
Mar 28 at 21:45
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Assume $k_1,k_2neq0$ for the key case:
$ddot x+k_1tan(k_1t)sec(k_1t)dot x+k_2cos(k_1t)x=k_3cos(k_1t)$
$dfracd^2xdt^2+dfrack_1sin(k_1t)cos^2(k_1t)dfracdxdt+cos(k_1t)(k_2x-k_3)=0$
$cos^2(k_1t)dfracd^2xdt^2+k_1sin(k_1t)dfracdxdt+cos^3(k_1t)(k_2x-k_3)=0$
Let $u=x-dfrack_3k_2$ ,
Then $cos^2(k_1t)dfracd^2udt^2+k_1sin(k_1t)dfracdudt+k_2cos^3(k_1t)u=0$
Let $r=cos(k_1t)$ ,
Then $dfracdudt=dfracdudrdfracdrdt=-k_1sin(k_1t)dfracdudr$
$dfracd^2udt^2=dfracddtleft(-k_1sin(k_1t)dfracdudrright)=-k_1sin(k_1t)dfracddtleft(dfracdudrright)-k_1^2cos(k_1t)dfracdudr=-k_1sin(k_1t)dfracddrleft(dfracdudrright)dfracdrdt-k_1^2cos(k_1t)dfracdudr=-k_1sin(k_1t)dfracd^2udr^2(-k_1sin(k_1t))-k_1^2cos(k_1t)dfracdudr=k_1^2sin^2(k_1t)dfracd^2udr^2-k_1^2cos(k_1t)dfracdudr$
$thereforecos^2(k_1t)left(k_1^2sin^2(k_1t)dfracd^2udr^2-k_1^2cos(k_1t)dfracdudrright)-k_1^2sin^2(k_1t)dfracdudr+k_2cos^3(k_1t)u=0$
$k_1^2sin^2(k_1t)cos^2(k_1t)dfracd^2udr^2-k_1^2(cos^3(k_1t)+sin^2(k_1t))dfracdudr+k_2cos^3(k_1t)u=0$
$k_1^2r^2(1-r^2)dfracd^2udr^2-k_1^2(r^3-r^2+1)dfracdudr+k_2r^3u=0$
$k_1^2r^2(r^2-1)dfracd^2udr^2+k_1^2(r^2(r-1)+1)dfracdudr-k_2r^3u=0$
$dfracd^2udr^2+left(dfrac1r+1+dfrac1r^2(r+1)(r-1)right)dfracdudr-dfrack_2rk_1^2(r+1)(r-1)u=0$
$dfracd^2udr^2+left(dfrac12(r+1)+dfrac12(r-1)-dfrac1r^2right)dfracdudr-dfrack_22k_1^2left(dfrac1r+1+dfrac1r-1right)u=0$
answered Mar 29 at 4:33
doraemonpauldoraemonpaul
12.9k31761
$endgroup$
$begingroup$
Thanks, I'll have to take a closer look at this.
$endgroup$
– Joe Catania
Mar 31 at 13:54
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assume $k_1,k_2neq0$ for the key case:
$ddot x+k_1tan(k_1t)sec(k_1t)dot x+k_2cos(k_1t)x=k_3cos(k_1t)$
$dfracd^2xdt^2+dfrack_1sin(k_1t)cos^2(k_1t)dfracdxdt+cos(k_1t)(k_2x-k_3)=0$
$cos^2(k_1t)dfracd^2xdt^2+k_1sin(k_1t)dfracdxdt+cos^3(k_1t)(k_2x-k_3)=0$
Let $u=x-dfrack_3k_2$ ,
Then $cos^2(k_1t)dfracd^2udt^2+k_1sin(k_1t)dfracdudt+k_2cos^3(k_1t)u=0$
Let $r=cos(k_1t)$ ,
Then $dfracdudt=dfracdudrdfracdrdt=-k_1sin(k_1t)dfracdudr$
$dfracd^2udt^2=dfracddtleft(-k_1sin(k_1t)dfracdudrright)=-k_1sin(k_1t)dfracddtleft(dfracdudrright)-k_1^2cos(k_1t)dfracdudr=-k_1sin(k_1t)dfracddrleft(dfracdudrright)dfracdrdt-k_1^2cos(k_1t)dfracdudr=-k_1sin(k_1t)dfracd^2udr^2(-k_1sin(k_1t))-k_1^2cos(k_1t)dfracdudr=k_1^2sin^2(k_1t)dfracd^2udr^2-k_1^2cos(k_1t)dfracdudr$
$thereforecos^2(k_1t)left(k_1^2sin^2(k_1t)dfracd^2udr^2-k_1^2cos(k_1t)dfracdudrright)-k_1^2sin^2(k_1t)dfracdudr+k_2cos^3(k_1t)u=0$
$k_1^2sin^2(k_1t)cos^2(k_1t)dfracd^2udr^2-k_1^2(cos^3(k_1t)+sin^2(k_1t))dfracdudr+k_2cos^3(k_1t)u=0$
$k_1^2r^2(1-r^2)dfracd^2udr^2-k_1^2(r^3-r^2+1)dfracdudr+k_2r^3u=0$
$k_1^2r^2(r^2-1)dfracd^2udr^2+k_1^2(r^2(r-1)+1)dfracdudr-k_2r^3u=0$
$dfracd^2udr^2+left(dfrac1r+1+dfrac1r^2(r+1)(r-1)right)dfracdudr-dfrack_2rk_1^2(r+1)(r-1)u=0$
$dfracd^2udr^2+left(dfrac12(r+1)+dfrac12(r-1)-dfrac1r^2right)dfracdudr-dfrack_22k_1^2left(dfrac1r+1+dfrac1r-1right)u=0$
answered Mar 29 at 4:33
doraemonpauldoraemonpaul
12.9k31761
$endgroup$
$begingroup$
Thanks, I'll have to take a closer look at this.
$endgroup$
– Joe Catania
Mar 31 at 13:54
add a comment |
$begingroup$
Assume $k_1,k_2neq0$ for the key case:
$ddot x+k_1tan(k_1t)sec(k_1t)dot x+k_2cos(k_1t)x=k_3cos(k_1t)$
$dfracd^2xdt^2+dfrack_1sin(k_1t)cos^2(k_1t)dfracdxdt+cos(k_1t)(k_2x-k_3)=0$
$cos^2(k_1t)dfracd^2xdt^2+k_1sin(k_1t)dfracdxdt+cos^3(k_1t)(k_2x-k_3)=0$
Let $u=x-dfrack_3k_2$ ,
Then $cos^2(k_1t)dfracd^2udt^2+k_1sin(k_1t)dfracdudt+k_2cos^3(k_1t)u=0$
Let $r=cos(k_1t)$ ,
Then $dfracdudt=dfracdudrdfracdrdt=-k_1sin(k_1t)dfracdudr$
$dfracd^2udt^2=dfracddtleft(-k_1sin(k_1t)dfracdudrright)=-k_1sin(k_1t)dfracddtleft(dfracdudrright)-k_1^2cos(k_1t)dfracdudr=-k_1sin(k_1t)dfracddrleft(dfracdudrright)dfracdrdt-k_1^2cos(k_1t)dfracdudr=-k_1sin(k_1t)dfracd^2udr^2(-k_1sin(k_1t))-k_1^2cos(k_1t)dfracdudr=k_1^2sin^2(k_1t)dfracd^2udr^2-k_1^2cos(k_1t)dfracdudr$
$thereforecos^2(k_1t)left(k_1^2sin^2(k_1t)dfracd^2udr^2-k_1^2cos(k_1t)dfracdudrright)-k_1^2sin^2(k_1t)dfracdudr+k_2cos^3(k_1t)u=0$
$k_1^2sin^2(k_1t)cos^2(k_1t)dfracd^2udr^2-k_1^2(cos^3(k_1t)+sin^2(k_1t))dfracdudr+k_2cos^3(k_1t)u=0$
$k_1^2r^2(1-r^2)dfracd^2udr^2-k_1^2(r^3-r^2+1)dfracdudr+k_2r^3u=0$
$k_1^2r^2(r^2-1)dfracd^2udr^2+k_1^2(r^2(r-1)+1)dfracdudr-k_2r^3u=0$
$dfracd^2udr^2+left(dfrac1r+1+dfrac1r^2(r+1)(r-1)right)dfracdudr-dfrack_2rk_1^2(r+1)(r-1)u=0$
$dfracd^2udr^2+left(dfrac12(r+1)+dfrac12(r-1)-dfrac1r^2right)dfracdudr-dfrack_22k_1^2left(dfrac1r+1+dfrac1r-1right)u=0$
answered Mar 29 at 4:33
doraemonpauldoraemonpaul
12.9k31761
$endgroup$
$begingroup$
Thanks, I'll have to take a closer look at this.
$endgroup$
– Joe Catania
Mar 31 at 13:54
add a comment |
$begingroup$
Assume $k_1,k_2neq0$ for the key case:
$ddot x+k_1tan(k_1t)sec(k_1t)dot x+k_2cos(k_1t)x=k_3cos(k_1t)$
$dfracd^2xdt^2+dfrack_1sin(k_1t)cos^2(k_1t)dfracdxdt+cos(k_1t)(k_2x-k_3)=0$
$cos^2(k_1t)dfracd^2xdt^2+k_1sin(k_1t)dfracdxdt+cos^3(k_1t)(k_2x-k_3)=0$
Let $u=x-dfrack_3k_2$ ,
Then $cos^2(k_1t)dfracd^2udt^2+k_1sin(k_1t)dfracdudt+k_2cos^3(k_1t)u=0$
Let $r=cos(k_1t)$ ,
Then $dfracdudt=dfracdudrdfracdrdt=-k_1sin(k_1t)dfracdudr$
$dfracd^2udt^2=dfracddtleft(-k_1sin(k_1t)dfracdudrright)=-k_1sin(k_1t)dfracddtleft(dfracdudrright)-k_1^2cos(k_1t)dfracdudr=-k_1sin(k_1t)dfracddrleft(dfracdudrright)dfracdrdt-k_1^2cos(k_1t)dfracdudr=-k_1sin(k_1t)dfracd^2udr^2(-k_1sin(k_1t))-k_1^2cos(k_1t)dfracdudr=k_1^2sin^2(k_1t)dfracd^2udr^2-k_1^2cos(k_1t)dfracdudr$
$thereforecos^2(k_1t)left(k_1^2sin^2(k_1t)dfracd^2udr^2-k_1^2cos(k_1t)dfracdudrright)-k_1^2sin^2(k_1t)dfracdudr+k_2cos^3(k_1t)u=0$
$k_1^2sin^2(k_1t)cos^2(k_1t)dfracd^2udr^2-k_1^2(cos^3(k_1t)+sin^2(k_1t))dfracdudr+k_2cos^3(k_1t)u=0$
$k_1^2r^2(1-r^2)dfracd^2udr^2-k_1^2(r^3-r^2+1)dfracdudr+k_2r^3u=0$
$k_1^2r^2(r^2-1)dfracd^2udr^2+k_1^2(r^2(r-1)+1)dfracdudr-k_2r^3u=0$
$dfracd^2udr^2+left(dfrac1r+1+dfrac1r^2(r+1)(r-1)right)dfracdudr-dfrack_2rk_1^2(r+1)(r-1)u=0$
$dfracd^2udr^2+left(dfrac12(r+1)+dfrac12(r-1)-dfrac1r^2right)dfracdudr-dfrack_22k_1^2left(dfrac1r+1+dfrac1r-1right)u=0$
answered Mar 29 at 4:33
doraemonpauldoraemonpaul
12.9k31761
$endgroup$
Assume $k_1,k_2neq0$ for the key case:
$ddot x+k_1tan(k_1t)sec(k_1t)dot x+k_2cos(k_1t)x=k_3cos(k_1t)$
$dfracd^2xdt^2+dfrack_1sin(k_1t)cos^2(k_1t)dfracdxdt+cos(k_1t)(k_2x-k_3)=0$
$cos^2(k_1t)dfracd^2xdt^2+k_1sin(k_1t)dfracdxdt+cos^3(k_1t)(k_2x-k_3)=0$
Let $u=x-dfrack_3k_2$ ,
Then $cos^2(k_1t)dfracd^2udt^2+k_1sin(k_1t)dfracdudt+k_2cos^3(k_1t)u=0$
Let $r=cos(k_1t)$ ,
Then $dfracdudt=dfracdudrdfracdrdt=-k_1sin(k_1t)dfracdudr$
$dfracd^2udt^2=dfracddtleft(-k_1sin(k_1t)dfracdudrright)=-k_1sin(k_1t)dfracddtleft(dfracdudrright)-k_1^2cos(k_1t)dfracdudr=-k_1sin(k_1t)dfracddrleft(dfracdudrright)dfracdrdt-k_1^2cos(k_1t)dfracdudr=-k_1sin(k_1t)dfracd^2udr^2(-k_1sin(k_1t))-k_1^2cos(k_1t)dfracdudr=k_1^2sin^2(k_1t)dfracd^2udr^2-k_1^2cos(k_1t)dfracdudr$
$thereforecos^2(k_1t)left(k_1^2sin^2(k_1t)dfracd^2udr^2-k_1^2cos(k_1t)dfracdudrright)-k_1^2sin^2(k_1t)dfracdudr+k_2cos^3(k_1t)u=0$
$k_1^2sin^2(k_1t)cos^2(k_1t)dfracd^2udr^2-k_1^2(cos^3(k_1t)+sin^2(k_1t))dfracdudr+k_2cos^3(k_1t)u=0$
$k_1^2r^2(1-r^2)dfracd^2udr^2-k_1^2(r^3-r^2+1)dfracdudr+k_2r^3u=0$
$k_1^2r^2(r^2-1)dfracd^2udr^2+k_1^2(r^2(r-1)+1)dfracdudr-k_2r^3u=0$
$dfracd^2udr^2+left(dfrac1r+1+dfrac1r^2(r+1)(r-1)right)dfracdudr-dfrack_2rk_1^2(r+1)(r-1)u=0$
$dfracd^2udr^2+left(dfrac12(r+1)+dfrac12(r-1)-dfrac1r^2right)dfracdudr-dfrack_22k_1^2left(dfrac1r+1+dfrac1r-1right)u=0$
answered Mar 29 at 4:33
doraemonpauldoraemonpaul
12.9k31761
edited Mar 29 at 4:55
answered Mar 29 at 4:33
doraemonpauldoraemonpaul
12.9k31761
answered Mar 29 at 4:33
doraemonpauldoraemonpaul
12.9k31761
answered Mar 29 at 4:33
doraemonpauldoraemonpaul
12.9k31761
12.9k31761
$begingroup$
Thanks, I'll have to take a closer look at this.
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– Joe Catania
Mar 31 at 13:54
add a comment |
$begingroup$
Thanks, I'll have to take a closer look at this.
$endgroup$
– Joe Catania
Mar 31 at 13:54
$begingroup$
Thanks, I'll have to take a closer look at this.
$endgroup$
– Joe Catania
Mar 31 at 13:54
$begingroup$
Thanks, I'll have to take a closer look at this.
$endgroup$
– Joe Catania
Mar 31 at 13:54
add a comment |
Joe Catania is a new contributor. Be nice, and check out our Code of Conduct.
Joe Catania is a new contributor. Be nice, and check out our Code of Conduct.
Joe Catania is a new contributor. Be nice, and check out our Code of Conduct.
Joe Catania is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
A more detailed explanation of how it "came up in some work" might be interesting in itself and motivate why you want "a non-numerical solution". In other words, please add more context to avoid giving Readers the impression you think of them as a crank to be turned.
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– hardmath
Mar 28 at 19:17
1
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I'd break up the tangent and secant terms in the expressions that involve $sin$ and $cos$ to simplify first.
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– Rebellos
Mar 28 at 19:18
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Mathematica doesn't give an answer - I suspect there's no way to get a non-numerical solution.
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– Adrian Keister
Mar 28 at 19:35
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Have you noticed that there is a particular solution which is $x(t)=k_3/k_2=const.$. It suffices now to find the general solution to the same diff. equation with zero RHS and add to this constant...
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– Jean Marie
Mar 28 at 19:49
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@hardmath- I have a graphical plot that was generated by numerical iteration but it might be nice if there was a closed form solution available to check the result. It also would be nice if there was lookup table for these things.
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– Joe Catania
Mar 28 at 21:45