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A differential equation
#1
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
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#2
Bernouilli differential equations are strongly related ...

I intend to use them for tetration. But first I need to think about stuff.

regards

tommy1729
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#3
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) a function and you have to solve( in your case) for the



in your case

Im getting it in the right way? But is the differentiation operator (or how is called...)?

MathStackExchange account:MphLee
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#4
@ 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
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#5
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
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#6
int just means integral.

further df = df/dx

regards

tommy1729
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