Let $0<a<b$. Consider the functional $S[y]=\int_{a}^{b}x^5(y'^2-\frac{2}{3}y^3)dx$. Show that $S[y]$ is invariant under the scale transformation $\bar{x}=\alpha x, \bar{y}=\beta y$, where $\alpha$ and $\beta$ are constants satisfying $\alpha^2\beta=1$.
Here's my work:
Consider the functional $S[y]=\int_{a}^{b}x^5(y'^2-\frac{2}{3}y^3)dx$ with $0<a<b$.
By substituting the scale transformation $\bar{x}=\alpha x$ and $\bar{y}=\beta y$, we obtain:
$\bar{S}[\bar{y}]=\int \bar{x}^5(\bar{y}^2(\frac{d\bar{y}}{d\bar{x}})^2-\frac{2}{3}\bar{y}^3)d\bar{x}$.
Note that $\bar{x}^5=\alpha^5 x^5, \bar{y}^2=\beta^2 y^2, \bar{y}^3=\beta^3 y^3$.
From here I'm stuck. How should I find $(\frac{d\bar{y}}{d\bar{x}})^2$? I know I should first find $\frac{d\bar{y}}{d\bar{x}}$ but how should I do so? How about $d\bar{x}$?
You can solve this with a simple chain rule. First, notice that
$$ d\bar{y} = \frac{d\bar{y}}{d\bar{x}}d\bar{x},$$ but also $d\bar{x} = \alpha\, dx$ and $d\bar{y} = \beta\,dy$. So, it follows that
$$ dy = \frac{\alpha}{\beta}\frac{d\bar{y}}{d\bar{x}}dx = \frac{dy}{dx}dx, $$ and therefore $d\bar{y}/d\bar{x} = (\beta/\alpha)dy/dx$.
Can you take it from here?