Show that $H(v, y) = \frac{v^2}{2} - F(y)$ is a first integral

Question

$H(v, y) = \frac{v^2}{2} - F(y)$ is a hamiltonian of $y' = v, v' = f(y)$

Linear first integrals are of the form $I(x) = b^Tx + c$ where $b \in ℝ^d $ and $ c \in ℝ$

Quadratic first integrals are of the form $I(x) = x^TMx+b^Tx+c$ where $M \in ℝ^{d,d}$, $b \in ℝ^d $ and $ c \in ℝ$

I am unsure how to get the function $H(v,y)$ in either form or somehow else show it is indeed a first integral

Could someone please help

Answer 1

Indeed your ODE is the Hamiltonian system $$ \dot y = H_v=v\\ \dot v = -H_y=F'(y) $$ For any Hamiltonian system you have along solutions $$ \frac{d}{dt}H(v(t),y(t))=H_v\dot v+H_y\dot y=0 $$ so that the Hamiltonian function is a first integral.