Evaluating differential equations with degree higher than 1.












1














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?










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  • 1




    @W.mu. Did you try with the first case ?
    – Claude Leibovici
    Jan 4 at 8:59










  • 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.
    – LutzL
    Jan 4 at 10:44










  • @LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
    – JHT
    Jan 4 at 16:41












  • 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.
    – LutzL
    Jan 4 at 16:46
















1














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?










share|cite|improve this question


















  • 1




    @W.mu. Did you try with the first case ?
    – Claude Leibovici
    Jan 4 at 8:59










  • 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.
    – LutzL
    Jan 4 at 10:44










  • @LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
    – JHT
    Jan 4 at 16:41












  • 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.
    – LutzL
    Jan 4 at 16:46














1












1








1


1





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?










share|cite|improve this question













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?







differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 8:33









JHTJHT

83




83








  • 1




    @W.mu. Did you try with the first case ?
    – Claude Leibovici
    Jan 4 at 8:59










  • 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.
    – LutzL
    Jan 4 at 10:44










  • @LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
    – JHT
    Jan 4 at 16:41












  • 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.
    – LutzL
    Jan 4 at 16:46














  • 1




    @W.mu. Did you try with the first case ?
    – Claude Leibovici
    Jan 4 at 8:59










  • 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.
    – LutzL
    Jan 4 at 10:44










  • @LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
    – JHT
    Jan 4 at 16:41












  • 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.
    – LutzL
    Jan 4 at 16:46








1




1




@W.mu. Did you try with the first case ?
– Claude Leibovici
Jan 4 at 8:59




@W.mu. Did you try with the first case ?
– Claude Leibovici
Jan 4 at 8:59












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.
– LutzL
Jan 4 at 10:44




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.
– LutzL
Jan 4 at 10:44












@LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
– JHT
Jan 4 at 16:41






@LutzL. By rationalize I mean removing the fractional powers of derivatives. Honestly I am a bit confused about the concept of degree.
– JHT
Jan 4 at 16:41














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.
– LutzL
Jan 4 at 16:46




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.
– LutzL
Jan 4 at 16:46










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