Solution Verification and Linearly Independence

Question

I have the following problem \begin{equation} \begin{cases}\frac{dx}{dt}= -6x+5y \\ \frac{dy}{dt} = -5x+4y \end{cases}\end{equation}. I solved and did all the work I came up with this solution: \begin{equation} c_1e^{-t}\begin{pmatrix}1\\1\end{pmatrix}+c_2e^{-t}(t\begin{pmatrix}1\\1\end{pmatrix}+\begin{pmatrix}0\\\frac15\end{pmatrix})\end{equation} Symbolab comes up with this answer: \begin{equation} c_1e^{-t}\begin{pmatrix}1\\1\end{pmatrix}+c_2e^{-t}(t\begin{pmatrix}1\\1\end{pmatrix}+\begin{pmatrix}-\frac15\\ 0\end{pmatrix})\end{equation} Is my solution wrong or am I right? this question stems from because after an extensive goolge search I am seeing that my last matrix vector is linearly independent to the Symbolab one. So Does this linear independence mean that my answer is a solution? Also could you explain what this matrix linear independence mean in Matrices?

Answer 1

Both answers are correct. Since $\lambda =-1 $ is a repeated eigenvalue: $$\pmatrix {-5 & 5 \\ -5 &5}\pmatrix {w_1 \\w_2}=\pmatrix {0 \\0}$$ We get our first eigenvecctor: $$w_1=w_2 \implies w=\pmatrix {1 \\1}$$ For the second eigenvector we have: $$\pmatrix {-5 & 5 \\ -5 &5}\pmatrix {v_1 \\v_2}=\pmatrix {1\\1}$$ $$-5v_1+5v_2=1$$ for the soft the eigenvector is $$v_2=0 \implies v= \left(-\frac 15,0 \right)$$ $$X(t)=\begin{equation} c_1e^{-t}\begin{pmatrix}1\\1\end{pmatrix}+c_2e^{-t}(t\begin{pmatrix}1\\1\end{pmatrix}+\begin{pmatrix}-\dfrac 15\\ 0\end{pmatrix})\end{equation}$$ And for you it's: $$-5v_1+5v_2=1$$ $$v_1=0 \implies v=\left(0,\dfrac 15 \right)$$ $$X(t)=\begin{equation} c_1e^{-t}\begin{pmatrix}1\\1\end{pmatrix}+c_2e^{-t}(t\begin{pmatrix}1\\1\end{pmatrix}+\begin{pmatrix}0\\ \dfrac 1 5\end{pmatrix})\end{equation}$$

Answer 2

Get into the habit of checking your work: substitute your solution into the original system and see if it is satisfied. It does, so your solution is correct.

Just as the choice of eigenvectors is not unique, neither is the choice of generalized eigenvectors. You chose $(0,1/5)^T$, while the software used $(-1/5,0)^T$ instead. If you multiple either of them by $(A+I)$, you get $(1,1)^T$, so both are indeed valid generalized eigenvectors. In fact, any vector that isn’t a multiple of $(1,1)^T$ will do for a generalized eigenvector, since $(A+I)$ will map any such vector to a multiple of $(1,1)^T$.

Now, to see the equivalence of the two solutions, rewrite both expressions to collect terms $$e^{-t}\left(c_1\pmatrix{1\\1}+c_2\pmatrix{0\\\frac15}\right) + c_2te^{-1}\pmatrix{1\\1} \\ e^{-t}\left(c_1'\pmatrix{1\\1}+c_2'\pmatrix{-\frac15\\0}\right) + c_2'te^{-1}\pmatrix{1\\1}$$ and compare coefficients. Obviously $c_2=c_2'$, and setting $c_1'=c_1+\frac15c_2$ produces identical expressions. In other words, we can always choose these arbitrary constants so as to make the two solutions identical.