Asymptotic solution to a differential equation

[eljose]

if we have the equation:

y^{n}= F(y, \dot y, \ddot y, \dddot y,...,y^{n-1} )

where F can be a very difficult expression in the sense that can be non-linear and so on..my question is ¿how could we get an asimptotyc solution
y(x) with x--->oo of the differential equation...thanks.

 

[anton]

http://djvu.504.com1.ru:8019/WWW/888e7f7367ba8f0f0eb381b0cadf0de4.pdf

p. 230

change the dependent variable from y(x) to u(y)=y'(x). Then the Order of the eq. will lower by 1.
Note: NOT u(x)=y'(x), BUT u(y)=y'(x)

 

[eljose]

and there is no form to know how the differential equation diverges?..for example let,s suppose that for big x y(x) \sim x^{a} where a is a real and positive exponent then my question is if there would be any way to calculate a..thank you.

 

[Clausius2]

and there is no form to know how the differential equation diverges?..for example let,s suppose that for big x y(x) \sim x^{a} where a is a real and positive exponent then my question is if there would be any way to calculate a..thank you.

There's no general method for working out the asymptotic behavior of non linear differential equations. When it is non linear you're on your own. We usually assume a dominant balance and afterwards check it out, or we assume a behavior as you did. Your "a" can be calculated substituting your expression (if suitable) in the differential equation.