I am reading Rodrigues, Henrion, and Cantwell - Symmetries and analytical solutions of the Hamilton–Jacobi–Bellman equation for a class of optimal control problems, p.753.

Consider the following PDE: $$q_1x_2^2+q_2x_2^2+V_{x_1}x_2-\frac{V^2_{x_2}b^2}{4r}=0.$$

This PDE has the following dilation symmetry: $$\tilde{x}_1=e^sx_1,\, \tilde{x}_2=e^sx_2, \, \tilde{V}=e^{2s}V.$$ Note that $\tilde{V}_{\tilde{x}_1}=e^sV_{x_1}.$ So the above PDE in the tilde variables becomes $$e^{2s}\bigg(q_1x_2^2+q_2x_2^2+V_{x_1}x_2-\frac{V^2_{x_2}b^2}{4r}\bigg)=0.$$

So we can form the following characteristic equations $$ \frac{d x_1}{x_1}=\frac{d x_2}{x_2}=\frac{dV}{2V}.$$

my question is from the following statement,

Integrating and rearranging terms, the PDE is invariant under the change of variables: $\alpha= \frac{x_2}{x_1}$, $V=x_1^2 G(\alpha).$

How to understand the above statement? Does the dilation symmetry implies the above? How to see "being invariant"?