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I am having some trouble drawing these phase portraits
[![enter image description here][1]][1]

For 2.a when solving $[dot{y_1},dot{y_2}]^T = C * [y_1,y_2]^T$

You 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

$y_1 = C_1sin(bx) + C_2cos(bx)$ and $y_2 = C_3sin(bx) + C_4cos(bx)$

The only way I am able to plot it on matlab correctly is if i take $C_2 = C_3=0$ and $C_1 = C_4$ (otherwise ill get an ellipse)

I understand why $sinx,cosx$ gives you a circle but that is only a specific combination of coefficients. The phase diagram for C when plotted using y_1 = C_1sin(bx) + C_2cos(bx)$ and $y_2 = C_3sin(bx) + C_4cos(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 dont 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

thanks

2a

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.

2b

This is trivial if you calculate the matrix as before. But if you are not comfortable doing that, try with solutions of the form $ysim e^{pm lambda t}$, where $lambda$ are the eigenvalues of $C$