Calculating Eigenvalues of ODE's x1', x2

[cue928]

x1' = x1 - 5x2
x2' = x1 + 3x2
$$\begin{bmatrix} 1 & -5\\ 1 & 3\end{bmatrix} \begin{bmatrix} 1-\lambda & -5\\1 & 3-\lambda\end{bmatrix}$$

The eigenvalue I have is lambda = 2+/- 2i.
Using lambda = 2-2i, I get the following:
$$\begin{bmatrix} -1+2i & -5\\1 & 1+2i\end{bmatrix}$$
I get an eigenvector V=[5 -1+2i] which seems to check out for zero. So my first question is if I calculated this correctly.

 

[vela]

Yes, that's an eigenvector for that eigenvalue.

 

[cue928]

Yes, that's an eigenvector for that eigenvalue.

Okay, well at least I know how to do through that step. Where I'm running into problems is thereafter. I know the answer (from the book) but my signs are off. Can you explain how I get from having an eigenvector of V=[5 -1+2i] and $$\lambda=2-2i$$ to the following equations:
x1(t) = e^2t[-5 C1 cos(2t) - 5C2 sin(2t)]
x2(t) = e^2t[(c1+2c2) cos2t + (-2c1 +c2)sin2t]

My professor went through a bunch of steps during office hours today but I think he still glossed over a few things because the signs aren't matching up. For example, on equation 1, I had +5C1... and +5C2...so I don't know where I'm going wrong.

The equations I had were:
x1(t) = 5e^2t[c1 cos 2t + c2 sin 2t]
x2(t) = e^2t[-c1+c2) cos(2t) + (2c1 +c2) sin(2t)]
I was breaking down lambda as two values. lambda = 2-2i, so I set p = 2, q = -2. With the eigenvector of V=[5 -1+2i], I used a=5, b=-1, c=2 and then I substituted those in. I'm close, but the signs being off makes me think I'm not as far off as I would be if the numbers were completely wrong.

I understand cos being an even function and sin an odd function but I still don't see the sign problem. For example, on eq 1, the book says i should have:
e^2t(-5 c1 cos 2t - 5c2 sin 2t)
But with a q value of -2 and substituting that in, I got:
e^2t(5 C1 cos2t + 5 C2 sin 2t)

 

[vela]

The solution is

$$\vec{x} = d_1\begin{pmatrix} 5 \\ -1-2i \end{pmatrix}e^{(2+2i)t} + d_2\begin{pmatrix} 5 \\ -1+2i \end{pmatrix}e^{(2-2i)t}$$

where d₁ and d₂ are arbitrary constants. Each eigenvector is paired with the exponential containing the corresponding eigenvalue. Now it's just a bunch of algebra. Pull the common factor of e^2t out and use Euler's formula to expand the complex exponentials to get

$$\vec{x} = e^{2t}\left[d_1\begin{pmatrix} 5 \\ -1-2i \end{pmatrix}(\cos 2t + i \sin 2t) + d_2\begin{pmatrix} 5 \\ -1+2i \end{pmatrix}(\cos 2t - i \sin 2t)}\right]$$

Collect terms and simplify. You'll find d₁ and d₂ appear in two combinations that you can identify as c₁ and c₂.

 

[cue928]

The solution is

$$\vec{x} = d_1\begin{pmatrix} 5 \\ -1-2i \end{pmatrix}e^{(2+2i)t} + d_2\begin{pmatrix} 5 \\ -1+2i \end{pmatrix}e^{(2-2i)t}$$

where d₁ and d₂ are arbitrary constants. Each eigenvector is paired with the exponential containing the corresponding eigenvalue. Now it's just a bunch of algebra. Pull the common factor of e^2t out and use Euler's formula to expand the complex exponentials to get

$$\vec{x} = e^{2t}\left[d_1\begin{pmatrix} 5 \\ -1-2i \end{pmatrix}(\cos 2t + i \sin 2t) + d_2\begin{pmatrix} 5 \\ -1+2i \end{pmatrix}(\cos 2t - i \sin 2t)}\right]$$

Collect terms and simplify. You'll find d₁ and d₂ appear in two combinations that you can identify as c₁ and c₂.

How do you multiply that out, though?

 

[vela]

According to the usual rules of vector addition and scalar multiplication. Are you not familiar with those?

 

[cue928]

No, unfortunately not. Neither linear algebra or the section of calculus that introduces vectors were required for this class. Can you provide some insight?

 

[vela]

They're pretty simple. Scalar multiplication is

$$a\begin{pmatrix}x_1 \\ x_2\end{pmatrix} = \begin{pmatrix}ax_1 \\ ax_2\end{pmatrix}$$

When you multiply by a scalar, you just multiply all the components by the scalar.

Vector addition is

$$\begin{pmatrix}x_1 \\ x_2\end{pmatrix}+\begin{pmatrix}y_1 \\ y_2\end{pmatrix} = \begin{pmatrix}x_1+y_1 \\ x_2+y_2 \end{pmatrix}$$

The nth component of the sum is just the sum of the nth components.