# How to solve this nonlinear ODE?

1. Dec 8, 2008

### bobls86

Hello,

How to solve the equation as follows:
F''+F'*F'-k*F=0, where k is a constant and k>0

Is there any analytical solution?

2. Dec 8, 2008

### HallsofIvy

Staff Emeritus
Certainly, there exist an analytic function that satisfies the equation. If by "analytical solution" you mean, rather, that there is a calculus method for getting an exact solution, I would suspect "no" simply because "almost all" non-linear differential equations cannot be solved that way.

3. Dec 8, 2008

### Matthaeus

Your equation is equivalent to a first-order system, if we let $$F = u$$ and $$F' = v$$:
$$\begin{cases}\dot{u} = v \\ \dot{v} = ku-v^2\end{cases}$$

From there you can quite easily find a first integral and then separate variables in the second equation:

$$\begin{cases}\dot{u} = v \\ \dot{v}^2 = v^4 + kv^2-kE\end{cases}$$

Where E is a constant determined by the initial conditions.

edit: the function I found is not a first integral :/

Last edited: Dec 9, 2008
4. Dec 8, 2008

### bobls86

While I cannot see how you obtain the fourth equation: $$\dot{v} ^2 = v^4 + kv^2-kE\$$
Would you please support more details?

5. Dec 9, 2008

### Matthaeus

I assumed you were familiar with the definition of first integral. A first integral for the system $$y' = g(y), g : D \subseteq \mathbb{R}^n \rightarrow \mathbb{R}^n$$ is a $$C^1$$ scalar function $$E : D \rightarrow \mathbb{R}$$ constant on every solution of the system. In other words, if $$\phi : I \rightarrow D$$ is a solution of the system, $$E(\phi(t)) = \mathrm{const.} \quad \forall t \in I$$.

It follows from the definition that the gradient of E is everywhere normal to the field g:
$$\nabla E(y) \cdot g(y) = 0 \quad \forall y \in D$$.

In your equation, $$g = (v,ku-v^2)$$. A field normal to that is $$f = (-ku+v^2,v)$$. The problem is, the equation I wrote yesterday is wrong because it is not so obvious to find a primitive of a conservative field parallel to this field f. I was too much in a hurry to check, sorry ;)