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Math Help - more of finding the general solution...

  1. #1
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    more of finding the general solution...

    ok so my problem is...





    ok so when i started the problem i found that lambda was 0.

    well i found that lambda squared was 0? so do i pick any value for k1 and k2? im not real sure of where to go from here?

    any help would be appreciated...

    thanks in advance.
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  2. #2
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    Dear slapmaxwell1,

    Your value for lambda is correct. But can you please tell me what you mean by k1 and k2? Are they values of the eigenvector?
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  3. #3
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    Quote Originally Posted by slapmaxwell1 View Post
    ok so my problem is...





    ok so when i started the problem i found that lambda was 0.

    well i found that lambda squared was 0? so do i pick any value for k1 and k2? im not real sure of where to go from here?

    any help would be appreciated...

    thanks in advance.
    So you have the system

    \mathbf{X}'=\begin{pmatrix} 3 & -1 \\ 9 & -3 \end{pmatrix}\mathbf{X}

    So the characteristic polynomial is

    (3-\lambda)(-3-\lambda)-(-1)(9)=-9+\lambda^2+9 =\lambda^2=0

    So you have a repeated eigenvalue

    \begin{pmatrix} 3 & -1 \\ 9 & -3 \end{pmatrix} \to \begin{pmatrix} 1 & -\frac{1}{3} \\ 0 & 0 \end{pmatrix}

    This gives the one eigenvector

    \begin{pmatrix} 1 \\ 3\end{pmatrix}

    Now we need to find a generalized eigenvector

    \mathbf{P}=\begin{pmatrix}p_1 \\ p_2 \end{pmatrix}

    Such that

    \begin{pmatrix} 3 & -1 \\ 9 & -3 \end{pmatrix}\begin{pmatrix}p_1 \\ p_2 \end{pmatrix}=\begin{pmatrix} 1 \\ 3\end{pmatrix}

    Reducing this gives

    \begin{pmatrix} 3 & -1 & 1  \\ 9 & -3 & 3 \end{pmatrix} \to \begin{pmatrix} 1 & -\frac{1}{3} & \frac{1}{3}  \\ 0 & 0 & 0 \end{pmatrix}

    So we can pick p_2=-1 \implies p_1 =0

    and we get the vector


    \mathbf{P}=\begin{pmatrix}0 \\ -1 \end{pmatrix}

    So then the general solution is

    \mathbf{X}=c_1\begin{pmatrix} 1 \\ 3\end{pmatrix}+c_2\left[t\begin{pmatrix} 1 \\ 3\end{pmatrix}+ \begin{pmatrix}0 \\ -1 \end{pmatrix}\right]
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    So you have the system

    \mathbf{X}'=\begin{pmatrix} 3 & -1 \\ 9 & -3 \end{pmatrix}\mathbf{X}

    So the characteristic polynomial is

    (3-\lambda)(-3-\lambda)-(-1)(9)=-9+\lambda^2+9 =\lambda^2=0

    So you have a repeated eigenvalue

    \begin{pmatrix} 3 & -1 \\ 9 & -3 \end{pmatrix} \to \begin{pmatrix} 1 & -\frac{1}{3} \\ 0 & 0 \end{pmatrix}

    This gives the one eigenvector

    \begin{pmatrix} 1 \\ 3\end{pmatrix}

    Now we need to find a generalized eigenvector

    \mathbf{P}=\begin{pmatrix}p_1 \\ p_2 \end{pmatrix}

    Such that

    \begin{pmatrix} 3 & -1 \\ 9 & -3 \end{pmatrix}\begin{pmatrix}p_1 \\ p_2 \end{pmatrix}=\begin{pmatrix} 1 \\ 3\end{pmatrix}

    Reducing this gives

    \begin{pmatrix} 3 & -1 & 1  \\ 9 & -3 & 3 \end{pmatrix} \to \begin{pmatrix} 1 & -\frac{1}{3} & \frac{1}{3}  \\ 0 & 0 & 0 \end{pmatrix}

    So we can pick p_2=-1 \implies p_1 =0

    and we get the vector


    \mathbf{P}=\begin{pmatrix}0 \\ -1 \end{pmatrix}

    So then the general solution is

    \mathbf{X}=c_1\begin{pmatrix} 1 \\ 3\end{pmatrix}+c_2\left[t\begin{pmatrix} 1 \\ 3\end{pmatrix}+ \begin{pmatrix}0 \\ -1 \end{pmatrix}\right]

    thanks for that! i had a major brain fart!
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