<s>[INST] Solve the system of differential equations and plot the curves given the initial conditions. We are given the following system of ordinary differential equations:
$$\dot{x} = x - 4y \quad \dot{y} = x - 2y -4.$$
Thus, the slope of the trajectories is given by
$$\frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{x - 2y -4}{x-4y}.$$
We have the initial conditions $x(0) = 15$ and $y(0) = 5$.
How do I solve this system? The book does not give the solution but claims that we can solve for the trajectories. It does not suggest a method. It also asks what happens for other initial conditions.
I tried treating each separate equation as a linear equation and solved them via integrating factors, but that of course resulted in an integral equation on the right-hand side. I was not sure if that was the right way to go about it.
 [/INST] Here is an approach, from the second equation, we have
$$x = y' + 2 y + 4, ~~\mbox{so}~~ x' = y'' + 2 y'$$
Substituting these two into the first equation
$$ y'' + y'+2 y = 4$$
For initial conditions, we are given $y(0) = 5$ and use the second equation to find $y'(0) = x(0) - 2 y(0) - 4 = 1$.
This DEQ and ICs lets you find $y$, which you then use to find $x$.
We can also set this up and solve a non-homogeneous system using many methods
$$ Y' = A Y + g = \begin{pmatrix} 1 & -4 \\ 1 &- 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}+ \begin{pmatrix} 0 \\ - 4 \end{pmatrix}$$
You should get

 $$x(t)= -\frac{3 e^{-\frac{t}{2}} \sin \left(\frac{\sqrt{7} t}{2}\right)}{\sqrt{7}}+7 e^{-\frac{t}{2}} \cos \left(\frac{\sqrt{7} t}{2}\right)+8 \\y(t)= \frac{5 e^{-\frac{t}{2}} \sin \left(\frac{\sqrt{7} t}{2}\right)}{\sqrt{7}}+3 e^{-\frac{t}{2}} \cos \left(\frac{\sqrt{7} t}{2}\right)+2$$

You can also look at a phase portrait

Here is that same plot with a huge number of plot points

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