Procedure for 3 by 3 Non homogenous Linear systems (Differential Equations)

Question

Here is the problem I have. $$x^{'}(t)=Ax(t)+f(t)$$ where $A=\begin{pmatrix} 5&-3&-2\\8&-5&-4\\-4&3&3 \end{pmatrix} f(t)=\begin{pmatrix} -\sin (t)\\ 0 \\ 2 \end{pmatrix}$

I am trying to find the general solution to this problem. I found the eigenvalue to be 1 (it is repeated). Afterwards I solved the system $4x_1-3x_2-2x_1=0$ for $x_1$ and chose values so that I would not have decimals to be the following:

$$e_1= \begin{pmatrix} 2\\ 2\\ 1 \end{pmatrix}$$

I am kind of stuck at this point on what to do. Do I need to solve $$\begin{pmatrix}4&-3&-2\\8&-6&-4\\-4&3&2\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$$ but than where do I go from there?

Answer 1

Given:

$$A = \begin{bmatrix} 5 & -3 & -2 \\ 8 & -5 & -4 \\ -4 & 3 & 3 \end{bmatrix}, ~~ F[t] = \begin{bmatrix} - \sin t \\ 0 \\ 2 \end{bmatrix}$$

We can find the solution to this system using:

$$X(t) = e^{At}X_0 + \int_{t_0}^t e^{A(t-s)}F(s)~ds$$

If we find the Jordan Normal Form of the matrix, we have:

$$A = PJP^{-1} = \begin{bmatrix} 0 & -2 & 0 \\ 1 & -4 & 0 \\ -\dfrac{3}{2} & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} -2 & 1 & 0 \\ -\dfrac{1}{2} & 0 & 0 \\ -2 & \dfrac{3}{2} & 1 \end{bmatrix} $$

Update How did we find these eigenvectors?

$A$ has only two Jordan blocks for only one eigenvalue $\lambda = 1$, so:

$$A-I \ne 0 \\ (A-I)^2 = 0$$

Thus, we can take any vector $v_3$ such that $(A-I)v_3 \ne 0$, which becomes an eigenvector of rank $2$. So, take:

$$v_3 = (0,0,1) \implies (A-I)v_3 \ne 0$$

Now, we have:

$$v_2 = (A-I)v_3 = (-2,-4,2)$$

Lastly, we need a third linearly independent eigenvector such that:

$$(A-I)v_1 = 0 \implies v_1 = \left(0,1,-\dfrac{3}{2}\right)$$

Note: $P$ is made up of three generalized eigenvectors, $P = [v_1~|~v_2~|~v_3]$, from the single eigenvalue, since we need three linearly independent eigenvectors.

We can now find $e^{Jt}$ as:

$$E^{Jt} =e^t\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & t \\ 0 & 0 & 1 \end{bmatrix}$$

We can now write the fundamental matrix as:

$\phi(t) = Pe^{Jt}P^{-1} = \begin{bmatrix} 0 & -2 & 0 \\ 1 & -4 & 0 \\ -\dfrac{3}{2} & 2 & 1 \end{bmatrix} e^t\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & t \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -2 & 1 & 0 \\ -\dfrac{1}{2} & 0 & 0 \\ -2 & \dfrac{3}{2} & 1 \end{bmatrix}$

So,

$$ \phi(t) = e^t\begin{bmatrix} 4t+1 & -3t & -2t \\ 8t & 1-6t & -4t \\ -4t & 3t & 2t+1 \end{bmatrix}$$

We now have:

$$\phi^{-1}(t) = e^{-t}\begin{bmatrix} 1-4t & 3t & 2t \\ -8t & 6t+1 & 4t \\ 4t & -3t & 1-2t\end{bmatrix}$$

• Find $\phi^{-1}(t) \cdot F(t) = \begin{bmatrix} e^{-t} (4 t+(4 t-1) \sin (t)) \\ 8 e^{-t} t (\sin (t)+1) \\ e^{-t} (-4 \sin (t) t-4 t+2) \\ \end{bmatrix}$
• Now we integrate the previous result, that is $w = \int \phi^{-1}(t) F(t)$ and this gives us: $w$
• Next, we find:

$X(t) = X_h + X_p = \begin{bmatrix} x(t) \\ y(t) \\z(t) \end{bmatrix} = e^{At}X_0 + \phi(t) w = \begin{bmatrix} c_1 e^t (4 t+1)-3 c_2 e^t t-2 c_3 e^t t-4 t (2 t+t \sin (t)+(t+1) \cos (t)+1)+12 t (2 t+t \sin (t)+(t+1) \cos (t)+2)-\frac{1}{2} (4 t+1) (8 (t+1)+(4 t-1) \sin (t)+(4 t+3) \cos (t)) ~~~\\ ~~~8 c_1 e^t t-c_2 e^t (6 t-1)-4 c_3 e^t t-8 t (2 t+t \sin (t)+(t+1) \cos (t)+1)+4 (6 t-1) (2 t+t \sin (t)+(t+1) \cos (t)+2)-4 t (8 (t+1)+(4 t-1) \sin (t)+(4 t+3) \cos (t)) ~~~ \\ ~~~-4 c_1 e^t t+3 c_2 e^t t+c_3 e^t (2 t+1)+2 (2 t+1) (2 t+t \sin (t)+(t+1) \cos (t)+1)-12 t (2 t+t \sin (t)+(t+1) \cos (t)+2)+2 t (8 (t+1)+(4 t-1) \sin (t)+(4 t+3) \cos (t)) \end{bmatrix}$

Answer 2

Since you don't know about jordan normal form, since $A$ can't be diagonalized and since $A$ as only one eigenvalue, there's one little trick you can use.

For all $t\in \mathbb R$ it holds that $e^{At}=e^te^{(A-I)t}=e^{t}\sum \limits_{n=0}^\infty\left(\dfrac 1{n!}(A-I)^nt^n\right)$. Now note that $(A-I)^2$ is the null matrix.

Proceed with your favourite method. (This might help).