How can one solve $yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A$? [closed]
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How can one solve the second order nonlinear differential equation $yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A$? Here, $A,b,c$ are constants and $y(t)$ is a function.
ordinary-differential-equations
edited Dec 31 '18 at 2:04
YiFan
5,6702831
asked Dec 31 '18 at 1:54
alaloushalaloush
61
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closed as off-topic by Eevee Trainer, Abcd, Dylan, Lee David Chung Lin, RRL Dec 31 '18 at 7:08
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Dylan, Lee David Chung Lin, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How can one solve the second order nonlinear differential equation $yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A$? Here, $A,b,c$ are constants and $y(t)$ is a function.
ordinary-differential-equations
edited Dec 31 '18 at 2:04
YiFan
5,6702831
asked Dec 31 '18 at 1:54
alaloushalaloush
61
$endgroup$
closed as off-topic by Eevee Trainer, Abcd, Dylan, Lee David Chung Lin, RRL Dec 31 '18 at 7:08
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Dylan, Lee David Chung Lin, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I added mathjax to the post, and removed the second paragraph which I thought was unclear. If you think the second paragraph was important, please do put it back with an edit, making it clearer what you meant.
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– YiFan
Dec 31 '18 at 2:05
add a comment |
$begingroup$
How can one solve the second order nonlinear differential equation $yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A$? Here, $A,b,c$ are constants and $y(t)$ is a function.
ordinary-differential-equations
edited Dec 31 '18 at 2:04
YiFan
5,6702831
asked Dec 31 '18 at 1:54
alaloushalaloush
61
$endgroup$
How can one solve the second order nonlinear differential equation $yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A$? Here, $A,b,c$ are constants and $y(t)$ is a function.
ordinary-differential-equations
ordinary-differential-equations
edited Dec 31 '18 at 2:04
YiFan
5,6702831
asked Dec 31 '18 at 1:54
alaloushalaloush
61
edited Dec 31 '18 at 2:04
YiFan
5,6702831
asked Dec 31 '18 at 1:54
alaloushalaloush
61
edited Dec 31 '18 at 2:04
YiFan
5,6702831
edited Dec 31 '18 at 2:04
YiFan
5,6702831
edited Dec 31 '18 at 2:04
YiFan
5,6702831
5,6702831
asked Dec 31 '18 at 1:54
alaloushalaloush
61
asked Dec 31 '18 at 1:54
alaloushalaloush
61
asked Dec 31 '18 at 1:54
alaloushalaloush
61
61
closed as off-topic by Eevee Trainer, Abcd, Dylan, Lee David Chung Lin, RRL Dec 31 '18 at 7:08
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Dylan, Lee David Chung Lin, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, Abcd, Dylan, Lee David Chung Lin, RRL Dec 31 '18 at 7:08
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Dylan, Lee David Chung Lin, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I added mathjax to the post, and removed the second paragraph which I thought was unclear. If you think the second paragraph was important, please do put it back with an edit, making it clearer what you meant.
$endgroup$
– YiFan
Dec 31 '18 at 2:05
add a comment |
$begingroup$
I added mathjax to the post, and removed the second paragraph which I thought was unclear. If you think the second paragraph was important, please do put it back with an edit, making it clearer what you meant.
$endgroup$
– YiFan
Dec 31 '18 at 2:05
$begingroup$
I added mathjax to the post, and removed the second paragraph which I thought was unclear. If you think the second paragraph was important, please do put it back with an edit, making it clearer what you meant.
$endgroup$
– YiFan
Dec 31 '18 at 2:05
$begingroup$
I added mathjax to the post, and removed the second paragraph which I thought was unclear. If you think the second paragraph was important, please do put it back with an edit, making it clearer what you meant.
$endgroup$
– YiFan
Dec 31 '18 at 2:05
add a comment |
1 Answer
1
active
oldest
votes
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Since it's autonomous, you can write this as a first order equation for $y'$ in terms of $y$. You then get solutions
$$ int ^{y left( x right) }!{frac {sqrt {b-1}}{sqrt {2,c
left( b-1 right) {s}+ left( b-1 right) {s}^{2}+C
left( b-1 right) {s}^{1-b}+2,A}}}{ds}=Bpm x
$$
where $B$ and $C$ are arbitrary constants.
I would expect that for most values of $b$ this can not be done in "closed form".
answered Dec 31 '18 at 2:23
Robert IsraelRobert Israel
333k23223484
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thanks @Robert Israel
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– alaloush
Dec 31 '18 at 12:34
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Thank you Mr. @Robert Israel But I want to write the equation as follows (yy')^2 = F(y); F(y) = {y^4} + 2c{y^3} + A{y^2} + B{y^{3 - b}} How can I do this by once integrating this equation yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A
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– alaloush
Dec 31 '18 at 12:45
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@Robert Israel Can you provide me some more info about how you arrived at this integral?
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– cgiovanardi
Dec 31 '18 at 17:39
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@cgiovanardi you can arrived this integral by using Maple [> y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0; [> dsolve(y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0, {y(x)}); [>
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– alaloush
Dec 31 '18 at 19:43
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since it's autonomous, you can write this as a first order equation for $y'$ in terms of $y$. You then get solutions
$$ int ^{y left( x right) }!{frac {sqrt {b-1}}{sqrt {2,c
left( b-1 right) {s}+ left( b-1 right) {s}^{2}+C
left( b-1 right) {s}^{1-b}+2,A}}}{ds}=Bpm x
$$
where $B$ and $C$ are arbitrary constants.
I would expect that for most values of $b$ this can not be done in "closed form".
answered Dec 31 '18 at 2:23
Robert IsraelRobert Israel
333k23223484
$endgroup$
$begingroup$
thanks @Robert Israel
$endgroup$
– alaloush
Dec 31 '18 at 12:34
$begingroup$
Thank you Mr. @Robert Israel But I want to write the equation as follows (yy')^2 = F(y); F(y) = {y^4} + 2c{y^3} + A{y^2} + B{y^{3 - b}} How can I do this by once integrating this equation yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A
$endgroup$
– alaloush
Dec 31 '18 at 12:45
$begingroup$
@Robert Israel Can you provide me some more info about how you arrived at this integral?
$endgroup$
– cgiovanardi
Dec 31 '18 at 17:39
$begingroup$
@cgiovanardi you can arrived this integral by using Maple [> y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0; [> dsolve(y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0, {y(x)}); [>
$endgroup$
– alaloush
Dec 31 '18 at 19:43
add a comment |
$begingroup$
Since it's autonomous, you can write this as a first order equation for $y'$ in terms of $y$. You then get solutions
$$ int ^{y left( x right) }!{frac {sqrt {b-1}}{sqrt {2,c
left( b-1 right) {s}+ left( b-1 right) {s}^{2}+C
left( b-1 right) {s}^{1-b}+2,A}}}{ds}=Bpm x
$$
where $B$ and $C$ are arbitrary constants.
I would expect that for most values of $b$ this can not be done in "closed form".
answered Dec 31 '18 at 2:23
Robert IsraelRobert Israel
333k23223484
$endgroup$
$begingroup$
thanks @Robert Israel
$endgroup$
– alaloush
Dec 31 '18 at 12:34
$begingroup$
Thank you Mr. @Robert Israel But I want to write the equation as follows (yy')^2 = F(y); F(y) = {y^4} + 2c{y^3} + A{y^2} + B{y^{3 - b}} How can I do this by once integrating this equation yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A
$endgroup$
– alaloush
Dec 31 '18 at 12:45
$begingroup$
@Robert Israel Can you provide me some more info about how you arrived at this integral?
$endgroup$
– cgiovanardi
Dec 31 '18 at 17:39
$begingroup$
@cgiovanardi you can arrived this integral by using Maple [> y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0; [> dsolve(y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0, {y(x)}); [>
$endgroup$
– alaloush
Dec 31 '18 at 19:43
add a comment |
$begingroup$
Since it's autonomous, you can write this as a first order equation for $y'$ in terms of $y$. You then get solutions
$$ int ^{y left( x right) }!{frac {sqrt {b-1}}{sqrt {2,c
left( b-1 right) {s}+ left( b-1 right) {s}^{2}+C
left( b-1 right) {s}^{1-b}+2,A}}}{ds}=Bpm x
$$
where $B$ and $C$ are arbitrary constants.
I would expect that for most values of $b$ this can not be done in "closed form".
answered Dec 31 '18 at 2:23
Robert IsraelRobert Israel
333k23223484
$endgroup$
Since it's autonomous, you can write this as a first order equation for $y'$ in terms of $y$. You then get solutions
$$ int ^{y left( x right) }!{frac {sqrt {b-1}}{sqrt {2,c
left( b-1 right) {s}+ left( b-1 right) {s}^{2}+C
left( b-1 right) {s}^{1-b}+2,A}}}{ds}=Bpm x
$$
where $B$ and $C$ are arbitrary constants.
I would expect that for most values of $b$ this can not be done in "closed form".
answered Dec 31 '18 at 2:23
Robert IsraelRobert Israel
333k23223484
edited Dec 31 '18 at 5:33
answered Dec 31 '18 at 2:23
Robert IsraelRobert Israel
333k23223484
answered Dec 31 '18 at 2:23
Robert IsraelRobert Israel
333k23223484
answered Dec 31 '18 at 2:23
Robert IsraelRobert Israel
333k23223484
333k23223484
$begingroup$
thanks @Robert Israel
$endgroup$
– alaloush
Dec 31 '18 at 12:34
$begingroup$
Thank you Mr. @Robert Israel But I want to write the equation as follows (yy')^2 = F(y); F(y) = {y^4} + 2c{y^3} + A{y^2} + B{y^{3 - b}} How can I do this by once integrating this equation yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A
$endgroup$
– alaloush
Dec 31 '18 at 12:45
$begingroup$
@Robert Israel Can you provide me some more info about how you arrived at this integral?
$endgroup$
– cgiovanardi
Dec 31 '18 at 17:39
$begingroup$
@cgiovanardi you can arrived this integral by using Maple [> y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0; [> dsolve(y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0, {y(x)}); [>
$endgroup$
– alaloush
Dec 31 '18 at 19:43
add a comment |
$begingroup$
thanks @Robert Israel
$endgroup$
– alaloush
Dec 31 '18 at 12:34
$begingroup$
Thank you Mr. @Robert Israel But I want to write the equation as follows (yy')^2 = F(y); F(y) = {y^4} + 2c{y^3} + A{y^2} + B{y^{3 - b}} How can I do this by once integrating this equation yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A
$endgroup$
– alaloush
Dec 31 '18 at 12:45
$begingroup$
@Robert Israel Can you provide me some more info about how you arrived at this integral?
$endgroup$
– cgiovanardi
Dec 31 '18 at 17:39
$begingroup$
@cgiovanardi you can arrived this integral by using Maple [> y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0; [> dsolve(y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0, {y(x)}); [>
$endgroup$
– alaloush
Dec 31 '18 at 19:43
$begingroup$
thanks @Robert Israel
$endgroup$
– alaloush
Dec 31 '18 at 12:34
$begingroup$
thanks @Robert Israel
$endgroup$
– alaloush
Dec 31 '18 at 12:34
$begingroup$
Thank you Mr. @Robert Israel But I want to write the equation as follows (yy')^2 = F(y); F(y) = {y^4} + 2c{y^3} + A{y^2} + B{y^{3 - b}} How can I do this by once integrating this equation yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A
$endgroup$
– alaloush
Dec 31 '18 at 12:45
$begingroup$
Thank you Mr. @Robert Israel But I want to write the equation as follows (yy')^2 = F(y); F(y) = {y^4} + 2c{y^3} + A{y^2} + B{y^{3 - b}} How can I do this by once integrating this equation yy''+frac12(b-1)(y')^2 -frac12(b+1)y^2-bcy =A
$endgroup$
– alaloush
Dec 31 '18 at 12:45
$begingroup$
@Robert Israel Can you provide me some more info about how you arrived at this integral?
$endgroup$
– cgiovanardi
Dec 31 '18 at 17:39
$begingroup$
@Robert Israel Can you provide me some more info about how you arrived at this integral?
$endgroup$
– cgiovanardi
Dec 31 '18 at 17:39
$begingroup$
@cgiovanardi you can arrived this integral by using Maple [> y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0; [> dsolve(y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0, {y(x)}); [>
$endgroup$
– alaloush
Dec 31 '18 at 19:43
$begingroup$
@cgiovanardi you can arrived this integral by using Maple [> y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0; [> dsolve(y(x)*(diff(y(x), x, x, x))+b*(diff(y(x), x))*(diff(y(x), x, x))-(b+1)*y(x)*(diff(y(x), x))-bc*(diff(y(x), x)) = 0, {y(x)}); [>
$endgroup$
– alaloush
Dec 31 '18 at 19:43
add a comment |
$begingroup$
I added mathjax to the post, and removed the second paragraph which I thought was unclear. If you think the second paragraph was important, please do put it back with an edit, making it clearer what you meant.
$endgroup$
– YiFan
Dec 31 '18 at 2:05