# RLC initial conditions

I am working on the following series RLC circuit:

Let the circuit have non-zero initial conditions $$\I(0^+)= 6\text{A}\$$ and a voltage across the capacitor of $$\V(0^+)=-12\text{V}\$$.

What are $$\\frac{dV(0^+)}{dt}\$$ and $$\\frac{dI(0^+)}{dt}\$$?

My solution is different from the text example, and I want to know why.

$$I=\frac{dv}{dt} ~~\rightarrow ~~ \frac{dV(0^+)}{dt}=\frac{1}{C}I(0^+) = \frac{6}{C}$$

and

$$v = L\frac{dI}{dt} ~~ \rightarrow ~~ \frac{dI(0^+)}{dt} = \frac{1}{L}V(0^+) =- \frac{12}{L}$$

The text, on the other hand, shows: $$\frac{dI(0^+)}{dt} = -\frac{1}{L}\Big[RI(0^+) + V(0^+)\Big]$$

What is going on here? What am I doing wrong?

$$v = \frac{dI}{dt} ~~ \rightarrow ~~ \frac{dI(0^+)}{dt} = \frac{1}{L}V(0^+) =- \frac{12}{L}$$