Please check my work. I need to solve the following problem but my answer varies from that of the book by a factor of $C$ for capacitance. A print screen of the problem is given below.
Problem to solve:
My setup:
The differential equation for the charge on the capacitor is...
$$R\cdot q'+\frac{1}{C}\cdot q=0$$
with initial condition $q(0)=v_0$.
My solution
$$\mathcal{L}\{q'(t)\}=sQ-v_0$$
$$(sQ-v_0)+\frac{1}{RC}\cdot Q=0$$
$$Q(s+\frac{1}{RC})=v_0$$
$$Q=\frac{v_0}{s+\frac{1}{RC}}$$
My answer:
$$q(t)=v_0e^{-\frac{1}{RC}\cdot t}$$
Answer in text:
$$q(t)=Cv_0e^{-\frac{1}{RC}\cdot t}$$
Where did that $C$ come from? Was my setup wrong?
This is more physics but look the answer is that your answer is correct if you used the initial condition $q(0)=q_0$.
$Cv_0=q_0$: this is the definition of capaciatance.
The initial voltage is $v_0$ but you'd call the initial charge by $q_0$... or initial charge
$$q_0=CV_0.$$