The ODE enjoys a symmetry: under the substitution

$$v(t)=t^{\frac{n+2}{n}}\tilde{v}(1/t),$$

we have

$$(n+2)vv^{\prime\prime}-2v^{\prime 2}=t^{\frac{4}{n}-2}((n+2)\tilde{v}^{\prime\prime}\tilde{v}-2\tilde{v}^{\prime 2}),$$

$$\frac{n+2}{n}v+(1-t)v^{\prime}=t^{2/n}\{\frac{n+2}{n}\tilde{v}+(1-t^{-1})\tilde{v}^{\prime}\},$$

so the function $\tilde{v}$ is another solution of the same ODE. The geometric origin of this symmetry is that the NA MA equation is symmetric in $x_{1},x_{2}$, up to the minor issue of $d_{1},d_{2}$ which disappears after trivial changes of variables.