Solving a systemof eqns by given initial conditions

Question

I have the system

$x'(t)=2x+8y$

$y'(t)=-x-2y$

Which has the I.C. $X(0)=(6,-2)$. So I take that it means this:

$6=2x+8y$

$-2=-x-2y$

and then it is solved as any other system? We get eigenvector $(1,1/2)$, but since the eigenvalues of the system are two, $\pm 2i$, we have to find the second generalized eigenvector.

$$ \left( \begin{array}{cc} 2-2i & 8 \\ -1 & -2-2i \end{array} \right) % \left( \begin{array}{cc} 1\\ \frac{1}{2} \end{array} \right)=% \left( \begin{array}{cc} 6-2i \\ -2-i \end{array} \right) $$

So the second vector would be $(6-2i, -2-i)$. Using the general solutions for the system, we get

\begin{equation} y(t)=e^{2it}(1, \frac{1}{2})+e^{-2it}(6-2i, -2-i) \end{equation}

Would this be correct, or did I misinterpret the initial conditions?

Thanks

Answer 1

If$$A=\begin{bmatrix}2&8\\-1&-2\end{bmatrix},$$then$$\exp(tA)=\begin{bmatrix}\sin (2 t)+\cos (2 t) & 4 \sin (2 t) \\ -\frac{1}{2} \sin (2 t) & \cos (2 t)-\sin (2 t)\end{bmatrix}.$$So, if $f(t)=\exp(tA).(6,-2)$, you have $f(0)=(6,-2)$ and $f'(t)=A.f(t)$. So, take\begin{align}\bigl(x(t),y(t)\bigr)&=f(t)\\&=\bigl(6 \cos (2 t)-2 \sin (2 t),-\sin (2 t)-2 \cos (2 t)\bigr).\end{align}

Answer 2

In your case, the solution writes $$ \mathbf{x}(t) = a_1 e^{\lambda_1 t} \mathbf{v}_1 + a_2 e^{\lambda_2 t} \mathbf{v}_2 $$ using the eigenvector/eigenvalue of the matrix $\mathbf{A}$ in the relation $ \dot{\mathbf{x}}(t) = \mathbf{A} \mathbf{x}(t) $.

Both coefficients $a_1,a_2$ are found using the provided initial conditions $$ \mathbf{x}(0) = a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 $$ which is again a 2-by-2 linear system..

UPDATE With Matlab commands,

A=[2,8;-1,-2];[U,D]=eig(A);U\[6;-2]

I found $a_k = 3.1820 \pm 1.0607i$ and $\mathbf{v}_1 = [0.9428; -0.2357 + 0.2357i]$ Thus $$ \mathbf{x}(t) = a_1 e^{2i t} \mathbf{v}_1 + a_1^* e^{-2i t} \mathbf{v}_1^* = 2 \mathcal{R} [ a_1 e^{2i t} \mathbf{v}_1 ] $$

UPDATE 2 Because eigenvalues/eigenvectors are complex conjugate, it holds $$ \mathbf{x}(0) = a_1 \mathbf{v}_1 + a_1^* \mathbf{v}_1^*= 2 \mathcal{R}(a_1 \mathbf{v}_1) $$ $$ \mathbf{A} \mathbf{x}(0) = \lambda_1 a_1 \mathbf{v}_1 + \lambda_2 a_2 \mathbf{v}_2 = 4i \mathcal{I} (a_1 \mathbf{v}_1) $$ Thus $$ 2 a_1 \mathbf{v}_1= \mathbf{x}(0)- i \frac{\mathbf{A}\mathbf{x}(0)}{2} $$

From here simplifications occur \begin{eqnarray*} \mathbf{x}(t) &=& \mathcal{R}(2 a_1 \mathbf{v}_1 e^{2i t}) \\ &=& \mathbf{x}(0) \cos(2t) + \frac{\mathbf{A}\mathbf{x}(0)}{2} \sin(2t) \end{eqnarray*}