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+--- Thread: A differential equation (/showthread.php?tid=867)
A differential equation - tommy1729 - 05/15/2014
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
RE: A differential equation - tommy1729 - 05/15/2014
Bernouilli differential equations are strongly related ...
I intend to use them for tetration. But first I need to think about stuff.
regards
tommy1729
RE: A differential equation - MphLee - 05/15/2014
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...)?
RE: A differential equation - tommy1729 - 05/15/2014
@ 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
RE: A differential equation - MphLee - 05/16/2014
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(-)"?
RE: A differential equation - tommy1729 - 05/16/2014
int just means integral.
further df = df/dx
regards
tommy1729