My first question is, if I do the following, are all the voltages in
my differential equations the same voltage?

Yes, there is just one voltage in this circuit, the voltage across the parallel connected elements.

$$v_R = v_L = v_C = V $$

Second off, if this does all solve for the same voltage, then it is
problematic because I'm getting a complex value

The solutions to this homogeneous 2nd order linear differential equation are of the form:

$$Ae^{s_1t} + Be^{s_2t}$$

where \$s_1\$ and \$s_2\$ are the roots of the quadratic equation:

$$s^2 + \dfrac{1}{RC}s + \dfrac{1}{LC} = 0 $$

Now, there are three possibilities:

- There are two distinct real roots - this is the overdamped case
- There are two identical real roots - this is the critically damped
case
- There are two complex conjugate roots - this is the underdamped
case

Can you take it from here?

Hint: recall Euler's formula: \$e^{st} =e^{\alpha t}(\cos\omega t + j\sin\omega t\$) where \$s = \alpha + j \omega\$

Recall that, for node voltage analysis, a floating voltage source (a voltage source that does not connect to the GND node) poses a problem since you cannot write an equation relating the current through to the voltage across.

What you must do then is enclose the floating voltage source in a supernode, which reduces the number of KCL equations by one, and add the equation relating the voltage difference between the nodes the voltage source is connected to.

Now, the *dual* of node voltage analysis is mesh current analysis and here we have the dual problem when we have a current source common to two meshes - we can't write an equation relating the current through to the voltage across a current source.

What must be done then is to form a supermesh which reduces the number of KVL equations by one and add the equation relating the difference of the mesh currents to the common current source.

So, write KVL counter-clockwise around the supermesh consisting of the two voltage sources and the two resistors

$$V_1 = I_aR_1 + V_2 + (I_b - I_c)R_2$$

You have, by inspection (no KVL required for this mesh - this is dual to no KCL required for the node connected to a non-floating voltage source)

$$I_c = -1.25A $$

You need one more equation which is the equation relating to difference of the two mesh currents with the common current source.

$$3A = I_a - I_b $$

Now, you have 3 independent equations and 3 unknowns.

## Best Answer

\$u(t)\$ is the Unit Step Function. More than a mathematical function is a tool that allows us to model activations and deactivations in electrical circuits and abrupt changes of state in general. This function is defined as

$$ u(t) = \left\lbrace \begin{matrix} 0 \qquad t<0 \\ 1 \qquad t>0 \end{matrix} \right. $$

For your question, the variable \$t\$ has been transformed by a reflection with respect to \$t = 0\$. This means that

$$ u(-t) = \left\lbrace \begin{matrix} 0 \qquad t > 0 \\ 1 \qquad t < 0 \end{matrix} \right. $$

What does this mean? It means that the power supply has been active from \$t = -\infty\$ to \$t = 0\$, when it was deactivated. In other words, this source

should only be taken into account in determining the initial conditionsin the analysis of the circuit.