$$x(t)=RC\frac{ dy(t)}{dt}+LC\frac{d^2y(t)}{dt}+y(t)$$ By applying Laplace Transform to both sides, and using the differentiation property, we get, $$x(s)=(RC) s Y(s)+ (LC) s^2 Y(s) + Y(s)$$ $$H(s)=\frac{Y(s)}{X(s)} =\frac{( 1/LC)}{s^{2} +( R/L) \ s+( 1/LC)}$$ Where $\omega$ and $\zeta$ have the relationship, $$\left(\omega =\sqrt{\frac{1}{\text{LC}}}\right) \left(\zeta =\frac{1}{2} R \sqrt{\frac{C}{L}}\right)$$ $$\text{c1}=\sqrt{\zeta ^2-1} \omega -\zeta \omega$$ $$\text{c2}=-\sqrt{\zeta ^2-1} \omega -\zeta \omega$$ $$H(s)=\frac{\omega ^2}{(s-\text{c1}) (s-\text{c2})}$$
By getting partial fractions, $$H(s)=\frac{\omega ^{2}}{\text{c1} -\text{c2}} \left(\frac{1}{s-\text{c1}} -\frac{1}{s-\text{c2}}\right)$$ from this simplified notation we can derive the impulse response in the time domain, $$ \begin{array}{l} h(t)=\frac{\omega ^{2}}{\text{c1} -\text{c2}} \left( e^{\text{c1} t} u( t) -e^{\text{c2t}} u( t)\right)\\ \\ s(t)=\frac{\omega ^{2}}{\text{c1} -\text{c2}} \left(\frac{e^{\text{c1} t} -1}{\text{c1}} -\frac{e^{\text{c2t}} -1}{\text{c2}}\right) \end{array}$$ Now we have s(t) which is the unit step respose! (derivative of s(t) is h(t))