$\text{The relation between the input current}\left(I\right)\text{and the output voltage}\left(V\right)\text{of a circuit is governed by the equation:}$

$\mathrm{ C}\frac{dV}{dt}=I\left(t\right)-m\left(t\right).\text{The circuit is excited by}\mathrm{ I}\left(t\right)=q\delta \left(t\right),\text{where}\mathrm{ q }\text{is a real valued constant.}V\text{ at}\mathrm{ t}=0{}^{-}\text{is}{\mathrm{ V}}_0.$

$\text{Which of the following is an equivalent representation of the above case?}$

The correct option is:
$C\frac{dV}{dt}=-m\left(t\right),\text{with}V(t=0^{+})=V_0+\frac{q}{C}$

Step-by-Step Explanation:

We start with the governing differential equation of the circuit:
$C\frac{dV}{dt}=I\left(t\right)-m\left(t\right)$

We are given that the circuit is excited by an impulse current:
$I\left(t\right)=q\delta \left(t\right)$
where $\delta \left(t\right)$ is the Dirac delta function, and the initial condition just before the impulse occurs (at $t=0^{-}$) is:
$V\left(0^{-}\right)=V_0$

To find the equivalent representation for $t>0$, we can study the effect of the delta function impulse at $t=0$ on the voltage $V\left(t\right)$.
We integrate both sides of the differential equation over an infinitesimally small interval containing the origin, from $t=0^{-}$ to $t=0^{+}$:
${\int }_{0^{-}}^{0^{+}}C\frac{dV}{dt}dt={\int }_{0^{-}}^{0^{+}}q\delta \left(t\right)dt-{\int }_{0^{-}}^{0^{+}}m\left(t\right)dt$

Evaluating each term:
1. The left side yields:
$C\left[V\right(0^{+})-V(0^{-}\left)\right]$
2. The integral of the delta function is:
$q{\int }_{0^{-}}^{0^{+}}\delta \left(t\right)dt=q$
3. Since $m\left(t\right)$ is a well-behaved (bounded) function, its integral over an infinitesimal duration is zero:
${\int }_{0^{-}}^{0^{+}}m\left(t\right)dt=0$

Substituting these results back into our integrated equation, we get:
$C\left[V\right(0^{+})-V(0^{-}\left)\right]=q$

Rearranging to solve for the voltage immediately after the impulse, $V\left(0^{+}\right)$:
$V\left(0^{+}\right)-V\left(0^{-}\right)=\frac{q}{C}$
$V\left(0^{+}\right)=V\left(0^{-}\right)+\frac{q}{C}$

Using the initial condition $V\left(0^{-}\right)=V_0$, we establish the new initial condition at $t=0^{+}$:
$V(t=0^{+})=V_0+\frac{q}{C}$

For all times $t>0$, the impulse current is zero ($I\left(t\right)=0$). Therefore, the differential equation simplifies to:
$C\frac{dV}{dt}=-m\left(t\right)$

Thus, the system's behavior for $t>0$ is equivalently represented by the differential equation $C\frac{dV}{dt}=-m\left(t\right)$ starting with the updated initial state $V(t=0^{+})=V_0+\frac{q}{C}$.