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I have the following system. $\vec{x^{'}}(t)=\begin{pmatrix}4&-2\\8&-4 \end{pmatrix} \vec{x}+ \begin{pmatrix}t^{-3}\\-t^{-2}\end{pmatrix}$

I get $\lambda=0$ and the eigenvector of $$4x_1=2x_2 \implies \begin{pmatrix} 1\\2 \end{pmatrix}$$

Next I solved $$\begin{pmatrix}4&-2\\8&-4\end{pmatrix}\begin{pmatrix}\vec{o_1}\\ \vec{o_2} \end{pmatrix}=\begin{pmatrix}1\\2 \end{pmatrix} \implies \vec{o_1}=\dfrac{1+2\vec{o_2}}{4}$$

So can choose a eigenvector of $\begin{pmatrix} 0\\\frac{-1}{2} \end{pmatrix}$

Therefore, since $\lambda=0$ the fundamental matrix will be just the eigenvectors coupled together that is $$\phi(t)=\begin{pmatrix} 1&t\\2&2t-\frac{1}{2} \end{pmatrix}\implies \phi^{-1}(t)=\begin{pmatrix}1-4t&2t\\4&-2\end{pmatrix}$$

I multiply this matrix by $g(t)$ to get $$\begin{pmatrix}t^{-3}-4t^{-2}-2t^{-1}\\4t^{-3}+2t^{-2}\end{pmatrix}$$

Taking the integral gives me $$\begin{pmatrix} \frac{-1}{2}t^{-2}+4t^{-1}-2\ln t\\-2t^{-2}-2t^{-1} \end{pmatrix}$$

I multiply by $\phi(t)$ to get $\begin{pmatrix}1&t\\2&2t-\frac{1}{2}\end{pmatrix}\begin{pmatrix} \frac{-1}{2}t^{-2}+4t^{-1}-2\ln t\\-2t^{-2}-2t^{-1} \end{pmatrix}=\begin{pmatrix}-\frac{1}{2}t^{-2}-2\ln t+2t^{-1} -2\\-4\ln t+5t^{-1}-4 \end{pmatrix}$

Why cant I get these right...

Therefore all that is left is to multiply by $\phi$ and add the homogenous solution which is simply $\begin{pmatrix}1\\2 \end{pmatrix}c_1+c_2\bigg[t\begin{pmatrix}1\\2\end{pmatrix}+\begin{pmatrix}\frac{1}{4}\\0 \end{pmatrix}\bigg]+ x_p$ where $x_p$ is the particular solution. In the book they got something way off that is $$\begin{pmatrix}1\\2 \end{pmatrix}c_1+c_2\bigg[\begin{pmatrix}1\\2\end{pmatrix}t-\frac{1}{2}\begin{pmatrix}0\\1 \end{pmatrix}\bigg]-2\begin{pmatrix}1\\2 \end{pmatrix} \ln t+\begin{pmatrix}2\\5 \end{pmatrix}t^{-1}-\begin{pmatrix}\frac{1}{2}\\0 \end{pmatrix}t^{-2}$$

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  • $\begingroup$ @Amzoti I posted my work can you tell me where I'm going wrong in my procedure? $\endgroup$ – adam Apr 24 '14 at 2:30
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Given:

$$\vec{x^{'}}(t)=\begin{pmatrix}4&-2\\8&-4 \end{pmatrix} \vec{x}+ \begin{pmatrix}t^{-3}\\-t^{-2}\end{pmatrix}$$

We have as eigenvalues and eigenvectors:

$$\lambda_{1,2} = 0, v_1 = (1,2), v_2 = \left(\dfrac{1}{4},0\right)$$

Update 1: Note that eigenvectors are not unique and your book chose $v_2 = \left( 0, -\dfrac{1}{2}\right)$.

From this, we can write the general solution as:

$$x(t) = e^{0t}\left(c_1 \begin{bmatrix} 1\\2 \end{bmatrix} + c_2\left(\begin{bmatrix} 0\\-\frac{1}{2} \end{bmatrix} + t\begin{bmatrix} 1\\2 \end{bmatrix} \right) \right) = \begin{bmatrix} c_1 + c_2 t \\2c_1 + c_2\left(-\frac{1}{2} +2 t \right)\end{bmatrix}$$

This means we have:

$$\phi(t) = \begin{bmatrix} 1 & t \\2 & -\frac{1}{2} + 2 t \end{bmatrix}$$

Now, to find the exponential, just take:

$$e^{At} = \phi(t) \phi^{-1}(0) = \begin{bmatrix} 1 + 4 t & -2 t \\ 8 t & 1 - 4 t \end{bmatrix}$$

Do you see where your $\phi(t)$ went astray? Recall, we have a repeated eigenvalue with a generalized eigenvector!

Update 2

Recall, we have:

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

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