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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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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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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)

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...)?

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MphLee

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@ 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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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(-)"?

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MphLee

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int just means integral.

further df = df/dx

regards

tommy1729