Second order non- linear differential equation solution [closed]

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Please suggest a substitution for solving:

$$y'' cot( y ) = (y')^2 +c $$

Thanks in advance.

differential-equations

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edited Jul 27 at 12:47

Dylan

11.4k31026

asked Jul 26 at 7:45

Narasimham

20.2k51957

closed as off-topic by Servaes, Jyrki Lahtonen, Dylan, Xander Henderson, José Carlos Santos Jul 27 at 14:33

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." – Servaes, Jyrki Lahtonen, Dylan, Xander Henderson, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Still trying to guess.. in the direction of log sec/ cosec $phi,phi^'$ etc. so far no luck.
  – Narasimham
  Jul 26 at 8:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is that $cot(yy'')$ or $cot(y)y''$?
  – LutzL
  Jul 26 at 8:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The latter; $ y^''cot (y) $
  – Narasimham
  Jul 26 at 8:25
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

2

Please suggest a substitution for solving:

$$y'' cot( y ) = (y')^2 +c $$

Thanks in advance.

differential-equations

share|cite|improve this question

edited Jul 27 at 12:47

Dylan

11.4k31026

asked Jul 26 at 7:45

Narasimham

20.2k51957

closed as off-topic by Servaes, Jyrki Lahtonen, Dylan, Xander Henderson, José Carlos Santos Jul 27 at 14:33

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." – Servaes, Jyrki Lahtonen, Dylan, Xander Henderson, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Still trying to guess.. in the direction of log sec/ cosec $phi,phi^'$ etc. so far no luck.
  – Narasimham
  Jul 26 at 8:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is that $cot(yy'')$ or $cot(y)y''$?
  – LutzL
  Jul 26 at 8:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The latter; $ y^''cot (y) $
  – Narasimham
  Jul 26 at 8:25
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

2

up vote
4
down vote

favorite

2

2

Please suggest a substitution for solving:

$$y'' cot( y ) = (y')^2 +c $$

Thanks in advance.

differential-equations

share|cite|improve this question

edited Jul 27 at 12:47

Dylan

11.4k31026

asked Jul 26 at 7:45

Narasimham

20.2k51957

Please suggest a substitution for solving:

$$y'' cot( y ) = (y')^2 +c $$

Thanks in advance.

differential-equations

share|cite|improve this question

edited Jul 27 at 12:47

Dylan

11.4k31026

asked Jul 26 at 7:45

Narasimham

20.2k51957

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 12:47

Dylan

11.4k31026

edited Jul 27 at 12:47

Dylan

11.4k31026

edited Jul 27 at 12:47

Dylan

11.4k31026

11.4k31026

asked Jul 26 at 7:45

Narasimham

20.2k51957

asked Jul 26 at 7:45

Narasimham

20.2k51957

asked Jul 26 at 7:45

Narasimham

20.2k51957

20.2k51957

closed as off-topic by Servaes, Jyrki Lahtonen, Dylan, Xander Henderson, José Carlos Santos Jul 27 at 14:33

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." – Servaes, Jyrki Lahtonen, Dylan, Xander Henderson, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Servaes, Jyrki Lahtonen, Dylan, Xander Henderson, José Carlos Santos Jul 27 at 14:33

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." – Servaes, Jyrki Lahtonen, Dylan, Xander Henderson, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Still trying to guess.. in the direction of log sec/ cosec $phi,phi^'$ etc. so far no luck.
  – Narasimham
  Jul 26 at 8:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is that $cot(yy'')$ or $cot(y)y''$?
  – LutzL
  Jul 26 at 8:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The latter; $ y^''cot (y) $
  – Narasimham
  Jul 26 at 8:25
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Still trying to guess.. in the direction of log sec/ cosec $phi,phi^'$ etc. so far no luck.
  – Narasimham
  Jul 26 at 8:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is that $cot(yy'')$ or $cot(y)y''$?
  – LutzL
  Jul 26 at 8:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The latter; $ y^''cot (y) $
  – Narasimham
  Jul 26 at 8:25
  

  
  
  
  

Still trying to guess.. in the direction of log sec/ cosec $phi,phi^'$ etc. so far no luck.
– Narasimham
Jul 26 at 8:06

Still trying to guess.. in the direction of log sec/ cosec $phi,phi^'$ etc. so far no luck.
– Narasimham
Jul 26 at 8:06

is that $cot(yy'')$ or $cot(y)y''$?
– LutzL
Jul 26 at 8:21

is that $cot(yy'')$ or $cot(y)y''$?
– LutzL
Jul 26 at 8:21

The latter; $ y^''cot (y) $
– Narasimham
Jul 26 at 8:25

The latter; $ y^''cot (y) $
– Narasimham
Jul 26 at 8:25

add a comment | 

3 Answers
3

active

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votes

up vote
3
down vote



accepted

$$cot y , y^'' = y^'^2 +c$$
Substitute $y'=p implies y''=pfrac dpdy$
$$pp'cot y = p^2 +c$$
$$int frac 2p p^2 +cdp=2int tan(y) dy$$
$$ln|p^2 +c|=-2ln |cos(y)|+K_1$$
$$p^2 =frac K_1 cos^2(y)-c$$
$$y'=sqrtfrac K_1 cos^2(y)-c$$
$$int frac dysqrtfrac K_1 cos^2(y)-c=x+K_2$$
With $u=sin(y)$
$$int frac dusqrtK_1+cu^2=x+K_2$$

share|cite|improve this answer

edited Jul 26 at 15:01

answered Jul 26 at 11:03

Isham

10.5k3829

add a comment | 

up vote
3
down vote

Substituting $u=sin(y)$ one gets
$$
u''=(sin(y))''=(cos(y)y')'=cos(y)y''-sin(y)y'^2=csin(y)=cu
$$
which is, depending on the sign of $c$, an oscillation equation or an exponential function which is easily solvable for $u$ and thus for $y$.
$$
u=begincases
a_1cos(sqrtx)+a_2sin(sqrtx)&text for c<0,\
b_1+b_2x&text for c=0,\
c_1e^sqrtcx+c_2e^sqrtcx&text for c>0.
endcases
$$

share|cite|improve this answer

answered Jul 26 at 8:25

LutzL

49.8k31849

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Pretty nice way around it ! Fancier than my brutal substitution.
  – Rebellos
  Jul 26 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The form reminded me of the more common second derivative of $exp(y)$.
  – LutzL
  Jul 26 at 8:31
  

  
  
  
  

add a comment | 

up vote
2
down vote

HINTS :

If $y$ is a function of $x$, namingly $y(x)$ and $c$ is just a constant such that $c in mathbb R$, then a (very complicated) solution can be yielded after a string of operations using the substitution
$$v(y) = fracmathrmdy(x)mathrmdy$$
which gives
$$fracmathrmd^2y(x)mathrmdx=fracmathrmdmathrmdxbigg(fracmathrmdy(x)mathrmdxbigg)=fracmathrmdv(y)mathrmdx=fracmathrmdv(y)mathrmdyfracmathrmdymathrmdx=v(y)fracmathrmdv(y)mathrmdx$$
and thus integrating solving for $v(y)$ and substituting back to find an expression in terms only of $y(x)$ and $x$.

---

If $y$ is a function of $c$, then there isn't a solution in terms of standard functions. Specifically, the solution is

$$y(c) = arcsinbig[c_1big(c_2mathrmBi(c) + pi mathrmAi(c)big)big]$$

which gives some rather interesting plots-properties while sampling an initial $y(0)$, where $mathrmAi(x)$ is the Airy function and $mathrmBi(x)$ is the Airy Bi function.

$qquadqquadqquadqquadqquad$

share|cite|improve this answer

answered Jul 26 at 8:11

Rebellos

10k21039

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

$$cot y , y^'' = y^'^2 +c$$
Substitute $y'=p implies y''=pfrac dpdy$
$$pp'cot y = p^2 +c$$
$$int frac 2p p^2 +cdp=2int tan(y) dy$$
$$ln|p^2 +c|=-2ln |cos(y)|+K_1$$
$$p^2 =frac K_1 cos^2(y)-c$$
$$y'=sqrtfrac K_1 cos^2(y)-c$$
$$int frac dysqrtfrac K_1 cos^2(y)-c=x+K_2$$
With $u=sin(y)$
$$int frac dusqrtK_1+cu^2=x+K_2$$

share|cite|improve this answer

edited Jul 26 at 15:01

answered Jul 26 at 11:03

Isham

10.5k3829

add a comment | 

up vote
3
down vote



accepted

$$cot y , y^'' = y^'^2 +c$$
Substitute $y'=p implies y''=pfrac dpdy$
$$pp'cot y = p^2 +c$$
$$int frac 2p p^2 +cdp=2int tan(y) dy$$
$$ln|p^2 +c|=-2ln |cos(y)|+K_1$$
$$p^2 =frac K_1 cos^2(y)-c$$
$$y'=sqrtfrac K_1 cos^2(y)-c$$
$$int frac dysqrtfrac K_1 cos^2(y)-c=x+K_2$$
With $u=sin(y)$
$$int frac dusqrtK_1+cu^2=x+K_2$$

share|cite|improve this answer

edited Jul 26 at 15:01

answered Jul 26 at 11:03

Isham

10.5k3829

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

$$cot y , y^'' = y^'^2 +c$$
Substitute $y'=p implies y''=pfrac dpdy$
$$pp'cot y = p^2 +c$$
$$int frac 2p p^2 +cdp=2int tan(y) dy$$
$$ln|p^2 +c|=-2ln |cos(y)|+K_1$$
$$p^2 =frac K_1 cos^2(y)-c$$
$$y'=sqrtfrac K_1 cos^2(y)-c$$
$$int frac dysqrtfrac K_1 cos^2(y)-c=x+K_2$$
With $u=sin(y)$
$$int frac dusqrtK_1+cu^2=x+K_2$$

share|cite|improve this answer

edited Jul 26 at 15:01

answered Jul 26 at 11:03

Isham

10.5k3829

$$cot y , y^'' = y^'^2 +c$$
Substitute $y'=p implies y''=pfrac dpdy$
$$pp'cot y = p^2 +c$$
$$int frac 2p p^2 +cdp=2int tan(y) dy$$
$$ln|p^2 +c|=-2ln |cos(y)|+K_1$$
$$p^2 =frac K_1 cos^2(y)-c$$
$$y'=sqrtfrac K_1 cos^2(y)-c$$
$$int frac dysqrtfrac K_1 cos^2(y)-c=x+K_2$$
With $u=sin(y)$
$$int frac dusqrtK_1+cu^2=x+K_2$$

share|cite|improve this answer

edited Jul 26 at 15:01

answered Jul 26 at 11:03

Isham

10.5k3829

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 15:01

edited Jul 26 at 15:01

edited Jul 26 at 15:01

answered Jul 26 at 11:03

Isham

10.5k3829

answered Jul 26 at 11:03

Isham

10.5k3829

answered Jul 26 at 11:03

Isham

10.5k3829

10.5k3829

add a comment | 

add a comment | 

up vote
3
down vote

Substituting $u=sin(y)$ one gets
$$
u''=(sin(y))''=(cos(y)y')'=cos(y)y''-sin(y)y'^2=csin(y)=cu
$$
which is, depending on the sign of $c$, an oscillation equation or an exponential function which is easily solvable for $u$ and thus for $y$.
$$
u=begincases
a_1cos(sqrtx)+a_2sin(sqrtx)&text for c<0,\
b_1+b_2x&text for c=0,\
c_1e^sqrtcx+c_2e^sqrtcx&text for c>0.
endcases
$$

share|cite|improve this answer

answered Jul 26 at 8:25

LutzL

49.8k31849

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Pretty nice way around it ! Fancier than my brutal substitution.
  – Rebellos
  Jul 26 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The form reminded me of the more common second derivative of $exp(y)$.
  – LutzL
  Jul 26 at 8:31
  

  
  
  
  

add a comment | 

up vote
3
down vote

Substituting $u=sin(y)$ one gets
$$
u''=(sin(y))''=(cos(y)y')'=cos(y)y''-sin(y)y'^2=csin(y)=cu
$$
which is, depending on the sign of $c$, an oscillation equation or an exponential function which is easily solvable for $u$ and thus for $y$.
$$
u=begincases
a_1cos(sqrtx)+a_2sin(sqrtx)&text for c<0,\
b_1+b_2x&text for c=0,\
c_1e^sqrtcx+c_2e^sqrtcx&text for c>0.
endcases
$$

share|cite|improve this answer

answered Jul 26 at 8:25

LutzL

49.8k31849

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Pretty nice way around it ! Fancier than my brutal substitution.
  – Rebellos
  Jul 26 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The form reminded me of the more common second derivative of $exp(y)$.
  – LutzL
  Jul 26 at 8:31
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Substituting $u=sin(y)$ one gets
$$
u''=(sin(y))''=(cos(y)y')'=cos(y)y''-sin(y)y'^2=csin(y)=cu
$$
which is, depending on the sign of $c$, an oscillation equation or an exponential function which is easily solvable for $u$ and thus for $y$.
$$
u=begincases
a_1cos(sqrtx)+a_2sin(sqrtx)&text for c<0,\
b_1+b_2x&text for c=0,\
c_1e^sqrtcx+c_2e^sqrtcx&text for c>0.
endcases
$$

share|cite|improve this answer

answered Jul 26 at 8:25

LutzL

49.8k31849

Substituting $u=sin(y)$ one gets
$$
u''=(sin(y))''=(cos(y)y')'=cos(y)y''-sin(y)y'^2=csin(y)=cu
$$
which is, depending on the sign of $c$, an oscillation equation or an exponential function which is easily solvable for $u$ and thus for $y$.
$$
u=begincases
a_1cos(sqrtx)+a_2sin(sqrtx)&text for c<0,\
b_1+b_2x&text for c=0,\
c_1e^sqrtcx+c_2e^sqrtcx&text for c>0.
endcases
$$

share|cite|improve this answer

answered Jul 26 at 8:25

LutzL

49.8k31849

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 8:25

LutzL

49.8k31849

answered Jul 26 at 8:25

LutzL

49.8k31849

answered Jul 26 at 8:25

LutzL

49.8k31849

49.8k31849

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Pretty nice way around it ! Fancier than my brutal substitution.
  – Rebellos
  Jul 26 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The form reminded me of the more common second derivative of $exp(y)$.
  – LutzL
  Jul 26 at 8:31
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Pretty nice way around it ! Fancier than my brutal substitution.
  – Rebellos
  Jul 26 at 8:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The form reminded me of the more common second derivative of $exp(y)$.
  – LutzL
  Jul 26 at 8:31
  

  
  
  
  

Pretty nice way around it ! Fancier than my brutal substitution.
– Rebellos
Jul 26 at 8:30

Pretty nice way around it ! Fancier than my brutal substitution.
– Rebellos
Jul 26 at 8:30

The form reminded me of the more common second derivative of $exp(y)$.
– LutzL
Jul 26 at 8:31

The form reminded me of the more common second derivative of $exp(y)$.
– LutzL
Jul 26 at 8:31

add a comment | 

up vote
2
down vote

HINTS :

If $y$ is a function of $x$, namingly $y(x)$ and $c$ is just a constant such that $c in mathbb R$, then a (very complicated) solution can be yielded after a string of operations using the substitution
$$v(y) = fracmathrmdy(x)mathrmdy$$
which gives
$$fracmathrmd^2y(x)mathrmdx=fracmathrmdmathrmdxbigg(fracmathrmdy(x)mathrmdxbigg)=fracmathrmdv(y)mathrmdx=fracmathrmdv(y)mathrmdyfracmathrmdymathrmdx=v(y)fracmathrmdv(y)mathrmdx$$
and thus integrating solving for $v(y)$ and substituting back to find an expression in terms only of $y(x)$ and $x$.

---

If $y$ is a function of $c$, then there isn't a solution in terms of standard functions. Specifically, the solution is

$$y(c) = arcsinbig[c_1big(c_2mathrmBi(c) + pi mathrmAi(c)big)big]$$

which gives some rather interesting plots-properties while sampling an initial $y(0)$, where $mathrmAi(x)$ is the Airy function and $mathrmBi(x)$ is the Airy Bi function.

$qquadqquadqquadqquadqquad$

share|cite|improve this answer

answered Jul 26 at 8:11

Rebellos

10k21039

add a comment | 

up vote
2
down vote

HINTS :

If $y$ is a function of $x$, namingly $y(x)$ and $c$ is just a constant such that $c in mathbb R$, then a (very complicated) solution can be yielded after a string of operations using the substitution
$$v(y) = fracmathrmdy(x)mathrmdy$$
which gives
$$fracmathrmd^2y(x)mathrmdx=fracmathrmdmathrmdxbigg(fracmathrmdy(x)mathrmdxbigg)=fracmathrmdv(y)mathrmdx=fracmathrmdv(y)mathrmdyfracmathrmdymathrmdx=v(y)fracmathrmdv(y)mathrmdx$$
and thus integrating solving for $v(y)$ and substituting back to find an expression in terms only of $y(x)$ and $x$.

---

If $y$ is a function of $c$, then there isn't a solution in terms of standard functions. Specifically, the solution is

$$y(c) = arcsinbig[c_1big(c_2mathrmBi(c) + pi mathrmAi(c)big)big]$$

which gives some rather interesting plots-properties while sampling an initial $y(0)$, where $mathrmAi(x)$ is the Airy function and $mathrmBi(x)$ is the Airy Bi function.

$qquadqquadqquadqquadqquad$

share|cite|improve this answer

answered Jul 26 at 8:11

Rebellos

10k21039

add a comment | 

up vote
2
down vote

up vote
2
down vote

HINTS :

If $y$ is a function of $x$, namingly $y(x)$ and $c$ is just a constant such that $c in mathbb R$, then a (very complicated) solution can be yielded after a string of operations using the substitution
$$v(y) = fracmathrmdy(x)mathrmdy$$
which gives
$$fracmathrmd^2y(x)mathrmdx=fracmathrmdmathrmdxbigg(fracmathrmdy(x)mathrmdxbigg)=fracmathrmdv(y)mathrmdx=fracmathrmdv(y)mathrmdyfracmathrmdymathrmdx=v(y)fracmathrmdv(y)mathrmdx$$
and thus integrating solving for $v(y)$ and substituting back to find an expression in terms only of $y(x)$ and $x$.

---

If $y$ is a function of $c$, then there isn't a solution in terms of standard functions. Specifically, the solution is

$$y(c) = arcsinbig[c_1big(c_2mathrmBi(c) + pi mathrmAi(c)big)big]$$

which gives some rather interesting plots-properties while sampling an initial $y(0)$, where $mathrmAi(x)$ is the Airy function and $mathrmBi(x)$ is the Airy Bi function.

$qquadqquadqquadqquadqquad$

share|cite|improve this answer

answered Jul 26 at 8:11

Rebellos

10k21039

HINTS :

If $y$ is a function of $x$, namingly $y(x)$ and $c$ is just a constant such that $c in mathbb R$, then a (very complicated) solution can be yielded after a string of operations using the substitution
$$v(y) = fracmathrmdy(x)mathrmdy$$
which gives
$$fracmathrmd^2y(x)mathrmdx=fracmathrmdmathrmdxbigg(fracmathrmdy(x)mathrmdxbigg)=fracmathrmdv(y)mathrmdx=fracmathrmdv(y)mathrmdyfracmathrmdymathrmdx=v(y)fracmathrmdv(y)mathrmdx$$
and thus integrating solving for $v(y)$ and substituting back to find an expression in terms only of $y(x)$ and $x$.

---

If $y$ is a function of $c$, then there isn't a solution in terms of standard functions. Specifically, the solution is

$$y(c) = arcsinbig[c_1big(c_2mathrmBi(c) + pi mathrmAi(c)big)big]$$

which gives some rather interesting plots-properties while sampling an initial $y(0)$, where $mathrmAi(x)$ is the Airy function and $mathrmBi(x)$ is the Airy Bi function.

$qquadqquadqquadqquadqquad$

share|cite|improve this answer

answered Jul 26 at 8:11

Rebellos

10k21039

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 8:11

Rebellos

10k21039

answered Jul 26 at 8:11

Rebellos

10k21039

answered Jul 26 at 8:11

Rebellos

10k21039

10k21039

add a comment | 

add a comment |