# Thread: Reducing order of DFQ and solving

1. ## Reducing order of DFQ and solving

Initially I have a function like this:

f''' + 1 - (f')^2 = 0 ... and f' (is f prime)

To solve this I am told to multiply by an integrating factor f'' and then integrate. So I get:

1/2(f'')^2 + f' - 1/3(f')^3 = C

And then the problem says to make a substitution that G = f' so I get:
1/2(G')^2 + G - 1/3(G)^3 = C ...which is a a 1st order ODE but I don't know how to solve that. Can anyone help me?

Thanks

2. It looks already separable to me. Just treat $f'$ as a variable so that:

$f''+1-(f')^2=0\Rightarrow d(f')+1-(f')^2=0$

Separating I get:

$\frac{d(f')}{(f')^2-1}=dx$

Integrate, isolate $f'$ then integrate again.

3. Originally Posted by shawsend
It looks already separable to me. Just treat $f'$ as a variable so that:

$f''+1-(f')^2=0\Rightarrow d(f')+1-(f')^2=0$

Separating I get:

$\frac{d(f')}{(f')^2-1}=dx$

Integrate, isolate $f'$ then integrate again.
Sorry the initial equation is :

$f'''+1-(f')^2=0$ It was probably hard to read before

The problem says to first multiply by a integrating factor $f''$

So I get to:

$1/2 (f'')^2 + f' -1/3(f')^2 = C$

I am then told to make the substitution $G(n) = f'(n)$ solve this to show that

$f' = 3*tanh^2[n/root2 + 1.146] -2$

f in all of this is a function of n, and the boundary conditions are:

$f(0)= f'(0) =0$ , and $f'(n)$ approaches 1 as n approaches infinity

Please someone help