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 conditions in the analysis of the circuit.