Evaluating differential equations with degree higher than 1.
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How would you rationalize the equation $y''+ sqrt{y'} + xy = 0$ ?
What about for a more abstract case like $y''^{1/m} + y'^{1/n} + C = 0$,
where m,n, and C are positive integers?
How do higher degrees effect the solution of differential equations?
ordinary-differential-equations
asked Jan 4 at 8:33
JHTJHT
83
$endgroup$
add a comment |
$begingroup$
How would you rationalize the equation $y''+ sqrt{y'} + xy = 0$ ?
What about for a more abstract case like $y''^{1/m} + y'^{1/n} + C = 0$,
where m,n, and C are positive integers?
How do higher degrees effect the solution of differential equations?
ordinary-differential-equations
asked Jan 4 at 8:33
JHTJHT
83
$endgroup$
1
$begingroup$
@W.mu. Did you try with the first case ?
$endgroup$
– Claude Leibovici
Jan 4 at 8:59
$begingroup$
For $y'>c>0$ you should be able to apply all standard theorems for non-linear ODE. What do you understand under "rationalize"? Where is the "higher degree", all powers are 1 or smaller than 1. It makes not much sense to apply the notion of "degree" to ODE that are not polynomials in the variables and their derivatives.
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– LutzL
Jan 4 at 10:44
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@LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
$endgroup$
– JHT
Jan 4 at 16:41
$begingroup$
Why would you do that? As it is, the second derivative is uniquely determined by value and first derivative, it is easy to form the first order system. Computing the square in the first equation to eliminate the square root introduces an additional value for $y''$, so it is not equivalent.
$endgroup$
– LutzL
Jan 4 at 16:46
add a comment |
$begingroup$
How would you rationalize the equation $y''+ sqrt{y'} + xy = 0$ ?
What about for a more abstract case like $y''^{1/m} + y'^{1/n} + C = 0$,
where m,n, and C are positive integers?
How do higher degrees effect the solution of differential equations?
ordinary-differential-equations
asked Jan 4 at 8:33
JHTJHT
83
$endgroup$
How would you rationalize the equation $y''+ sqrt{y'} + xy = 0$ ?
What about for a more abstract case like $y''^{1/m} + y'^{1/n} + C = 0$,
where m,n, and C are positive integers?
How do higher degrees effect the solution of differential equations?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 4 at 8:33
JHTJHT
83
asked Jan 4 at 8:33
JHTJHT
83
asked Jan 4 at 8:33
JHTJHT
83
asked Jan 4 at 8:33
JHTJHT
83
asked Jan 4 at 8:33
JHTJHT
83
83
1
$begingroup$
@W.mu. Did you try with the first case ?
$endgroup$
– Claude Leibovici
Jan 4 at 8:59
$begingroup$
For $y'>c>0$ you should be able to apply all standard theorems for non-linear ODE. What do you understand under "rationalize"? Where is the "higher degree", all powers are 1 or smaller than 1. It makes not much sense to apply the notion of "degree" to ODE that are not polynomials in the variables and their derivatives.
$endgroup$
– LutzL
Jan 4 at 10:44
$begingroup$
@LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
$endgroup$
– JHT
Jan 4 at 16:41
$begingroup$
Why would you do that? As it is, the second derivative is uniquely determined by value and first derivative, it is easy to form the first order system. Computing the square in the first equation to eliminate the square root introduces an additional value for $y''$, so it is not equivalent.
$endgroup$
– LutzL
Jan 4 at 16:46
add a comment |
1
$begingroup$
@W.mu. Did you try with the first case ?
$endgroup$
– Claude Leibovici
Jan 4 at 8:59
$begingroup$
For $y'>c>0$ you should be able to apply all standard theorems for non-linear ODE. What do you understand under "rationalize"? Where is the "higher degree", all powers are 1 or smaller than 1. It makes not much sense to apply the notion of "degree" to ODE that are not polynomials in the variables and their derivatives.
$endgroup$
– LutzL
Jan 4 at 10:44
$begingroup$
@LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
$endgroup$
– JHT
Jan 4 at 16:41
$begingroup$
Why would you do that? As it is, the second derivative is uniquely determined by value and first derivative, it is easy to form the first order system. Computing the square in the first equation to eliminate the square root introduces an additional value for $y''$, so it is not equivalent.
$endgroup$
– LutzL
Jan 4 at 16:46
1
1
$begingroup$
@W.mu. Did you try with the first case ?
$endgroup$
– Claude Leibovici
Jan 4 at 8:59
$begingroup$
@W.mu. Did you try with the first case ?
$endgroup$
– Claude Leibovici
Jan 4 at 8:59
$begingroup$
For $y'>c>0$ you should be able to apply all standard theorems for non-linear ODE. What do you understand under "rationalize"? Where is the "higher degree", all powers are 1 or smaller than 1. It makes not much sense to apply the notion of "degree" to ODE that are not polynomials in the variables and their derivatives.
$endgroup$
– LutzL
Jan 4 at 10:44
$begingroup$
For $y'>c>0$ you should be able to apply all standard theorems for non-linear ODE. What do you understand under "rationalize"? Where is the "higher degree", all powers are 1 or smaller than 1. It makes not much sense to apply the notion of "degree" to ODE that are not polynomials in the variables and their derivatives.
$endgroup$
– LutzL
Jan 4 at 10:44
$begingroup$
@LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
$endgroup$
– JHT
Jan 4 at 16:41
$begingroup$
@LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
$endgroup$
– JHT
Jan 4 at 16:41
$begingroup$
Why would you do that? As it is, the second derivative is uniquely determined by value and first derivative, it is easy to form the first order system. Computing the square in the first equation to eliminate the square root introduces an additional value for $y''$, so it is not equivalent.
$endgroup$
– LutzL
Jan 4 at 16:46
$begingroup$
Why would you do that? As it is, the second derivative is uniquely determined by value and first derivative, it is easy to form the first order system. Computing the square in the first equation to eliminate the square root introduces an additional value for $y''$, so it is not equivalent.
$endgroup$
– LutzL
Jan 4 at 16:46
add a comment |
1 Answer
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$begingroup$
For $y''+sqrt{y'}+xy=0$ , you can consider as the ODE of the type http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=429
answered Jan 9 at 12:27
doraemonpauldoraemonpaul
12.7k31660
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1 Answer
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$begingroup$
For $y''+sqrt{y'}+xy=0$ , you can consider as the ODE of the type http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=429
answered Jan 9 at 12:27
doraemonpauldoraemonpaul
12.7k31660
$endgroup$
add a comment |
$begingroup$
For $y''+sqrt{y'}+xy=0$ , you can consider as the ODE of the type http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=429
answered Jan 9 at 12:27
doraemonpauldoraemonpaul
12.7k31660
$endgroup$
add a comment |
$begingroup$
For $y''+sqrt{y'}+xy=0$ , you can consider as the ODE of the type http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=429
answered Jan 9 at 12:27
doraemonpauldoraemonpaul
12.7k31660
$endgroup$
For $y''+sqrt{y'}+xy=0$ , you can consider as the ODE of the type http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=429
answered Jan 9 at 12:27
doraemonpauldoraemonpaul
12.7k31660
answered Jan 9 at 12:27
doraemonpauldoraemonpaul
12.7k31660
answered Jan 9 at 12:27
doraemonpauldoraemonpaul
12.7k31660
answered Jan 9 at 12:27
doraemonpauldoraemonpaul
12.7k31660
12.7k31660
add a comment |
add a comment |
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$begingroup$
@W.mu. Did you try with the first case ?
$endgroup$
– Claude Leibovici
Jan 4 at 8:59
$begingroup$
For $y'>c>0$ you should be able to apply all standard theorems for non-linear ODE. What do you understand under "rationalize"? Where is the "higher degree", all powers are 1 or smaller than 1. It makes not much sense to apply the notion of "degree" to ODE that are not polynomials in the variables and their derivatives.
$endgroup$
– LutzL
Jan 4 at 10:44
$begingroup$
@LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
$endgroup$
– JHT
Jan 4 at 16:41
$begingroup$
Why would you do that? As it is, the second derivative is uniquely determined by value and first derivative, it is easy to form the first order system. Computing the square in the first equation to eliminate the square root introduces an additional value for $y''$, so it is not equivalent.
$endgroup$
– LutzL
Jan 4 at 16:46