We declare known battery voltage \(E(t)\).

Here are the \(y\) values.

\begin{equation} \begin{cases} \dv{x_1}{t} = y_{4}\\ \dv{x_2}{t} = y_{3}\\ \dv{x_3}{t} = y_{1}\\ \dv{x_4}{t} = y_{2}\\ \end{cases} \end{equation}

And here are some of the \(x\) values.

\begin{equation} \begin{cases} \dv{x_4}{t}=-\frac{2}{RC}x_2-\frac{1}{RC}x_{3}-\frac{2E(t)}{R} \\ \dv{y_1}{t}=-\frac{1}{LC}x_2-\frac{E(t)}{C} \\ \dv{y_4}{t} = -\frac{R}{L}y_2-\frac{2E(t)}{L} \end{cases} \end{equation}

Right off the bat, we can see that we can make one substitution. That, given:

\begin{equation} \begin{cases} \dv{x_4}{t}=-\frac{2}{RC}x_2-\frac{1}{RC}x_{3}-\frac{2E(t)}{R} \\ \dv{x_4}{t} = y_{2} \end{cases} \end{equation}

We need to solve this system:

\begin{equation} \begin{cases} \dv{I}{t} = -0.73U + 0.0438 + 0.4 \dv{M}{t} \\ \dv{U}{t} = 0.4I - 0.012 \\ \dv{G}{t} = \dv{M}{t}- I \\ M(t)=0.02\sin (1.15t + \phi) \end{cases} \end{equation}

To be able to work on this, let us create some functions:

# variable
t, dm = var("t dm")

# functions
I = function("_I")(t) # _I because i is imaginary
U = function("U")(t)
G = function("G")(t)

# parameter
phi = var("phi", latex_name="\phi")

# our equations
eqns = [
    diff(I,t) == -0.73*U + 0.0438 + 0.4*dm,
    diff(U,t) == 0.4*I - 0.012,
    diff(G,t) == dm - I
]

eqns
desolve(eqns, U, ivar=t, algorithm="fricas").expand()

Great, now, we will run the laplace transform upon these equations:

We have:

\begin{equation} \frac{2y^{2}}{9-x^{2}} + y \dv{y}{x} + \frac{3y}{2-x} = 0 \end{equation}

We want to get rid of things; let’s begin by dividing the whole thing by \(y\).

\begin{equation} \frac{2y}{9-x^{2}} + \dv{y}{x} + \frac{3}{2-x} = 0 \end{equation}

Finally, then, moving the right expression to the right, we have:

\begin{equation} \frac{2y}{9-x^{2}} + \dv{y}{x} = \frac{-3}{2-x} \end{equation}

In this case, we have functions:

\begin{equation} \begin{cases} P(x) = \frac{2}{9-x^{2}}\\ Q(x) = \frac{-3}{2-x}\\ \end{cases} \end{equation}

Considering the system:

\begin{equation} \begin{cases} \dv{x}{t} = -2x+y+(1-\sigma)z \\ \dv{y}{t} = 3x-y \\ \dv{z}{t} = (3-\sigma y)x-z\\ \end{cases} \end{equation}

with the initial locations \((x_0, y_0, z_0)= (-1,1,2)\).

We notice first that the top and bottom expressions as a factor in \(x\) multiplied by \(y\), which means that our system is not homogenous. Let’s expand all the expressions first.

\begin{equation} \begin{cases} \dv{x}{t} = -2x+y+(1-\sigma)z \\ \dv{y}{t} = 3x-y \\ \dv{z}{t} = 3x-\sigma yx-z\\ \end{cases} \end{equation}

Intersects:

\begin{equation} f(x) = (x+c)^{2} \end{equation}

\begin{equation} h(x) = c x \end{equation}

Doesn’t Intersect:

\begin{equation} g(x) = c e^{\frac{x^{4}}{4}}} \end{equation}

\begin{align} &h_1(x)-h_2(x) = c_1x-c_2x \\ \Rightarrow\ & 0 = c_1x-c_2x \\ \Rightarrow\ & 0 = x(c_1-c_2) \end{align}

\begin{align} &g_1(x)-g_2(x) = c_1e^{\frac{x^{4}}{4}} - c_2e^{\frac{x^{4}}{4}} \\ \Rightarrow\ & 0 = \qty(c_1 - c_2)e^{\frac{x^{4}}{4}} \\ \Rightarrow\ & 0 = e^{\frac{x^{4}}{4}}(c_1-c_2) \end{align}

\begin{align} & f_1(x)-f_2(x)=(x+c_1)^{2}-(x+c_2)^{2} \\ \Rightarrow\ & 0 = (x+c_1)^{2}-(x+c_2)^{2} \\ \Rightarrow\ & 0 = 2x(c_1-c_2)+{c_1}^{2}+{c_2}^{2} \end{align}