Multiply \( u^{\prime\prime}+\omega^2u=0 \) by \( u^{\prime} \) and integrate: $$ \int_0^T u^{\prime\prime}u^{\prime} dt + \int_0^T\omega^2 u u^{\prime} dt = 0\tp$$ Observing that $$ u^{\prime\prime}u^{\prime} = \frac{d}{dt}\half(u^{\prime})^2,\quad uu^{\prime} = \frac{d}{dt} {\half}u^2,$$ we get $$ \int_0^T (\frac{d}{dt}\half(u^{\prime})^2 + \frac{d}{dt} \half\omega^2u^2)dt = E(T) - E(0), $$ where $$ E(t) = \half(u^{\prime})^2 + \half\omega^2u^2 $$