How to reduce the ODE to first order?

Question

How to reduce the following O.D.E. into first order?

$$(vv_{xxx}-v_{x}v_{xx})g^2 + v^3 v_x = 0, \tag{1}$$ where $v=v(x)$ and $v_{x}$ is the derivative w.r.t $x$, $g$ is a constant. After integration, equation (1) can be written as $$(vv_{xx}-v_{x}^2)g^2+\frac{v^4}{4} - c=0, \tag{2}$$ where $c$ is an integration constant. How to integrate further to get a first order O.D.E.?

Answer 1

Note that $$ v_{xx}=\frac{dv_x}{dx}=\frac{dv_x}{dv}\frac{dv}{dx}=v_x\frac{dv_x}{dv}=\frac{1}{2}\frac{dv_x^2}{dv}$$ Thus the equation (2) can be written as $$(\frac{1}{2}v\frac{dv_x^2}{dv}-v_x^2)g^2+\frac{v^2}{4}-c=0$$ which is a first order ODE about $v_x^2(v)$ and can be solved easily.

Answer 2

The equation becomes singular, formally or even actually, for $v=0$. Thus outside this plane we can set, following along the structure that you already found, $v_x=av$. Then $v_{xx}=a_xv+av_x=(a_x+a^2)v=bv$ and $$ vv_{xxx}-v_xv_{xx}=v^2(b_x+ab-ab) $$ and the equation in total now reads $$ g^2b_x+av^2=0 $$ This in total can be written as first-order system \begin{align} v_x&=av\\ a_x&=b-a^2\\ b_x&=-\frac{av^2}{g^2} \end{align}

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Of course you can also follow along the usual derivatives array and set $V=(v,v_x,v_{xx})$ and replace the third derivative in $V_x=(v_x,v_{xx},v_{xxx})$ via the given ODE.

Answer 3

$$(vv_{xxx}-v_{x}v_{xx})g^2 + v^3 v_x = 0, \tag{1}$$ $$\left(\dfrac {v_{xx}}v \right)'g^2 + \dfrac 12 (v^2)' = 0$$ $$\dfrac {v_{xx}}vg^2 + \dfrac 12 v^2 = C$$ Mutiply by $2v_xv$: $$ 2{v_xv_{xx}}g^2 + v_xv^3 = 2Cv_xv$$ Integrate. $$ (v_{x})^2g^2 +\dfrac 14 v^4 = C_1v^2+C_2$$