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
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
