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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?

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- Thread starter bobls86
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- #1

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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

HallsofIvy

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Your equation is equivalent to a first-order system, if we let [tex]F = u[/tex] and [tex]F' = v[/tex]:

[tex] \begin{cases}\dot{u} = v \\ \dot{v} = ku-v^2\end{cases}[/tex]

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

[tex] \begin{cases}\dot{u} = v \\ \dot{v}^2 = v^4 + kv^2-kE\end{cases}[/tex]

Where E is a constant determined by the initial conditions.

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

[tex] \begin{cases}\dot{u} = v \\ \dot{v} = ku-v^2\end{cases}[/tex]

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

[tex] \begin{cases}\dot{u} = v \\ \dot{v}^2 = v^4 + kv^2-kE\end{cases}[/tex]

Where E is a constant determined by the initial conditions.

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

Last edited:

- #4

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Your equation is equivalent to a first-order system, if we let [tex]F = u[/tex] and [tex]F' = v[/tex]:

[tex] \begin{cases}\dot{u} = v \\ \dot{v} = ku-v^2\end{cases}[/tex]

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

[tex] \begin{cases}\dot{u} = v \\ \dot{v}^2 = v^4 + kv^2-kE\end{cases}[/tex]

Where E is a constant determined by the initial conditions.

Thanks for your reply.

While I cannot see how you obtain the fourth equation: [tex]\dot{v} ^2 = v^4 + kv^2-kE\[/tex]

Would you please support more details?

- #5

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It follows from the definition that the gradient of E is everywhere normal to the field g:

[tex]\nabla E(y) \cdot g(y) = 0 \quad \forall y \in D[/tex].

In your equation, [tex]g = (v,ku-v^2)[/tex]. A field normal to that is [tex]f = (-ku+v^2,v)[/tex]. 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 ;)

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