Derive A Second Order Differential Equation for i1 - Circuits

Question

The equations are:

$$ \begin{cases} I_2 R +\dfrac{Q}{C} = V \\ L \dfrac{dI_1}{dt} = I_2 R \\ \dfrac{dQ}{dt} = I_1 + I_2 \\ \end{cases} $$

Answer 1

Starting from here: $$ \begin{cases} I_2 R =-\dfrac{Q}{C} +V \\ L \dfrac{dI_1}{dt} = I_2 R \\ \dfrac{dQ}{dt} = I_1 + I_2 \\ \end{cases} $$ Substitute $I_2$ in the last two equations: $$ \begin{cases} I_2 R =-\dfrac{Q}{C} +V \\ L \dfrac{dI_1}{dt} = -\dfrac{Q}{C} +V \\ R\dfrac{dQ}{dt} = RI_1 -\dfrac{Q}{C} +V \\ \end{cases} $$

$$L \dfrac{dI_1}{dt} = -\dfrac{Q}{C} +V \\$$ Differentiate: $$L \dfrac{d^2I_1}{dt^2} = -\dfrac 1 C\dfrac{dQ}{dt} \\$$ $$RL \dfrac{d^2I_1}{dt^2} = -\dfrac 1 C(RI_1 -\dfrac{Q}{C} +V )\\$$ $$L \dfrac{dI_1}{dt} = -\dfrac{Q}{C} +V \\$$ $$\implies -\dfrac{Q}{C} =L \dfrac{dI_1}{dt}-V \\$$ Therefore: $$\boxed {RLC \dfrac{d^2I_1}{dt^2}+ L \dfrac{dI_1}{dt}+RI_1 =0}$$ I hope that's what you wanted a differential with $I_1$ as the function.