I am having some trouble drawing these phase portraits
For 2.a when solving $[\dot{y_1},\dot{y_2}]^T = C * [y_1,y_2]^T$
We get $\dot{y_1} = -b*y_2$ and $\dot{y_2} = b*y_1$ and thus can solve
$\ddot{y_1} = -b^2y_1$ and $\ddot{y_2} = -b^2y_2$ to find
$y_1 = C_1 \sin(bx) + C_2 \cos(bx)$ and $y_2 = C_3 \sin(bx) + C_4 \cos(bx)$
The only way I am able to plot it on MATLAB correctly is when I take $C_2 = C_3=0$ and $C_1 = C_4$ (otherwise I will get an ellipse)
I understand why $\sin x,\cos x$ gives a circle but that is only a specific combination of coefficients. The phase diagram for C when plotted using $y_1 = C_1 \sin(bx) + C_2\cos(bx)$ and $y_2 = C_3\sin(bx) + C_4\cos(bx)$ gives me ellipses. How can we say that the phase diagram of $C$ are circles if this is only true in specific cases and no boundary conditions have been given.
Also for 2.b I don't know how to solve that system so could someone please explain how it corresponds to a spiral so I can try it on MATLAB
I have attached my code and would be grateful is someone could show me the MATLAB code for the spiral.
I'm going to start where you left, but hopefully you will see that this can be done in fewer steps than you did
\begin{align} y_1(t) &= C_1 \sin bt + C_2 \cos b t \\ y_2(t) &= C_3 \sin bt + C_4 \cos b t \tag{1}\\ \end{align}
As you say $\dot{y}_1 = -b y_2$, which translates to
$$ bC_1 \cos bt - bC_2 \sin b t = -b(C_3 \sin bt + C_4 \cos b t) ~~~\Rightarrow~~~ C_1 = -C_4, C_2 = C_3 $$
Replace that back in Eq. (1) and you get
\begin{align} y_1(t) &= C_1 \sin bt + C_2 \cos b t \\ y_2(t) &= C_2 \sin bt - C_1 \cos b t \tag{2}\\ \end{align}
Now evaluate these equations at $t=0$,
\begin{align} y_1(0) &= C_2 \\ y_2(0) &= -C_1\\ \end{align}
Again, if you replace that in Eq. (2) you get
\begin{align} y_1(t) &= -y_2(0) \sin bt + y_1(0) \cos b t \\ y_2(t) &= y_1(0) \sin bt + y_2(0) \cos b t \tag{3}\\ \end{align}
or in matrix form
$$ \pmatrix{y_1(t) \\ y_2(t)} = \pmatrix{\cos bt & -\sin bt \\ \sin bt & \cos bt} \pmatrix{y_1(0) \\ y_2 (0)} \tag{4} $$
Or more compact
$$ {\bf y}(t) = A {\bf y}(0) $$
turns out the matrix $A$ is the exponential matrix of $C$: $A = e^{tC}$. Which is calculated very simply for this case as $A = U^{-1}{\rm diag}(\lambda_1 t, \lambda_2 t) U$, where $U$ is the eigenvector matrix of $C$ and $\lambda$ its eigenvalues.
In any case, if you want to plot Eq. (4) you just need to select an initial point and multiply times a matrix.
This is trivial if you calculate the matrix as before. But if you are not comfortable doing that, try with solutions of the form $y\sim e^{\pm \lambda t}$, where $\lambda$ are the eigenvalues of $C$