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 A differential equation tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/15/2014, 12:03 AM Im a bit rusty on differential equations. I believe this had a solution. Find f such that f ' (x) * x^A = g ( f(x) ) For a nonzero real A and a given entire g(x). In particular g(x) a polynomial. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/15/2014, 12:23 PM Bernouilli differential equations are strongly related ... I intend to use them for tetration. But first I need to think about stuff. regards tommy1729 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 05/15/2014, 02:32 PM (This post was last modified: 05/15/2014, 05:26 PM by MphLee.) I was never able to understand differential equations on wiki (probably because I'm not good with differentials ) but is it a kind of functional equation? Like you have a field with a third operation (i think is called composition algerba)$(F,+,\cdot,\circ)$ a function $':F \rightarrow F$ and you have to solve( in your case) for the $\chi$ $\chi' \cdot t=g\circ \chi$ in your case $t(x)=x^A$ Im getting it in the right way? But $'=D$ is the differentiation operator (or how is called...)? MathStackExchange account:MphLee tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/15/2014, 11:16 PM @ MphLee ... mainly ... Here is an example of how I solve the equation with g(x) = x^s where s is some real number. We can rewrite : Let df = f ' (x) and f(x) = f then the equation is equivalent to solving : df f^a = x^b (df f^a)^1/b = x w = f^c dw = c f^(c-1) df (dw)^q = c^q f^(cq - q) (df)^q ... (dw/c)^q = f^(cq -q) (df)^q (dw/c)^1/b = f^((c-1)/b) (df)^1/b ... c-1 = a dw = c x^b dw = (a+1) x^b w = int (a+1) x^b dx + C f = w^(1/c) = w^(1/(a+1)) = ( int (a+1) x^b dx + C )^(1/(a+1)) I think that is correct. I included " ... " to show a different way of thinking has started. I hope that helps. Informally : Integrals and derivatives are used to show how a function behaves for a given function. Differential equations are used to show how behaviour belongs to a function for a given behaviour. regards tommy1729 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 05/16/2014, 07:51 AM I don't want to waste your time explaining me (I should go study the basis) but I don't get some substitutions. You put the value of f(x)=f as and f'(x)=df but what is d? And what is "int(-)"? MathStackExchange account:MphLee tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/16/2014, 08:50 PM int just means integral. further df = df/dx regards tommy1729 « Next Oldest | Next Newest »

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