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## Solving Linear Systems with constant coefficients

Suppose $$x_1,x_2,\ldots,x_n$$ are $$n$$ functions of $$t$$. Consider the following system of $$n$$ linear ODEs: $$$\overrightarrow{x}'=A(t)\overrightarrow{x}+\overrightarrow{g}(t), \;\;\;\;(51)$$$ where $$\overrightarrow{x}=[x_1,\; x_2,\cdots,\; x_n]^T$$.

Theorem. The linear system (51) has a solution $$\overrightarrow{x}=\overrightarrow{\phi}(t)$$ on an interval where $$A(t)$$ and $$\overrightarrow{g}(t)$$ are continuous. Now consider the corresponding homogeneous system: $$$\overrightarrow{x}'=A(t)\overrightarrow{x}. \;\;\;\;(52)$$$
Theorem. If $$\overrightarrow{x_1},\ldots,\overrightarrow{x_n}$$ are solutions of (52), then so is $\overrightarrow{x}=c_1\overrightarrow{x_1}+\cdots+c_n\overrightarrow{x_n},$ for all scalars $$c_1,\ldots,c_n$$. (Principle of superposition)

Theorem. $$\overrightarrow{x_1},\ldots,\overrightarrow{x_n}$$ are linearly independent solutions of (52) iff $W(\overrightarrow{x_1},\ldots,\overrightarrow{x_n})=\det[\overrightarrow{x_1},\ldots,\overrightarrow{x_n}]\neq 0$ for some (actually all) value of $$t$$.

Theorem. If $$\overrightarrow{x_1},\ldots,\overrightarrow{x_n}$$ are linearly independent solutions of (52), then any solution $$\overrightarrow{x}$$ of (52) is a linear combination of $$\overrightarrow{x_1},\ldots,\overrightarrow{x_n}$$, i.e., $\overrightarrow{x}=c_1\overrightarrow{x_1}+\cdots+c_n\overrightarrow{x_n},$ for some scalars $$c_1,\ldots,c_n$$.

Any $$n$$ linearly independent solutions of (52) are called fundamental solutions of (52) and $\overrightarrow{x}=c_1\overrightarrow{x_1}+\cdots+c_n\overrightarrow{x_n}$ is called the general solution of (52).

How to solve (52) when entries of $$A$$ are constants?
Recall that $$x'=\lambda x$$ has the general solution $$x=ce^{\lambda t}$$. So a solution of (52) may have a similar form $$\overrightarrow{x}=e^{\lambda t}\overrightarrow{v}$$ for some constant vector $$\overrightarrow{v}$$. Let's verify that. \begin{align*} \overrightarrow{x}'=A\overrightarrow{x} &\implies \lambda e^{\lambda t}\overrightarrow{v}=Ae^{\lambda t}\overrightarrow{v}\\ &\implies \lambda \overrightarrow{v}=A\overrightarrow{v}\\ &\implies A\overrightarrow{v}=\lambda\overrightarrow{v}\\ &\implies \overrightarrow{v} \text{ is an eigenvector of $$A$$ corresponding to eigenvalue }\lambda.\\ \end{align*} Theorem. If $$\overrightarrow{v_1},\ldots,\overrightarrow{v_n}$$ are $$n$$ linearly independent eigenvectors of $$A$$ corresponding to eigenvalues $$\lambda_1,\ldots,\lambda_n$$ respectively, then $$e^{\lambda_1t}\overrightarrow{v_1},\ldots,e^{\lambda_nt}\overrightarrow{v_n}$$ are $$n$$ linearly independent solutions of $\overrightarrow{x}'=A\overrightarrow{x}$ and the general solution is $\overrightarrow{x}=c_1e^{\lambda_1t}\overrightarrow{v_1}+\cdots+c_ne^{\lambda_nt}\overrightarrow{v_n},$ for arbitrary scalars $$c_1,\ldots,c_n$$.

Verify: \begin{align*} \overrightarrow{x}'&=c_1e^{\lambda_1t}\lambda_1\overrightarrow{v_1}+\cdots+c_ne^{\lambda_nt}\lambda_n\overrightarrow{v_n}\\ A\overrightarrow{x}&=c_1e^{\lambda_1t}A\overrightarrow{v_1}+\cdots+c_ne^{\lambda_nt}A\overrightarrow{v_n}\\ &=c_1e^{\lambda_1t}\lambda_1\overrightarrow{v_1}+\cdots+c_ne^{\lambda_nt}\lambda_n\overrightarrow{v_n}. \end{align*} Thus $$\overrightarrow{x}'=A\overrightarrow{x}$$.

Example. Find the general solution of $$\overrightarrow{x}'=\left[\begin{array}{rr} 1&1\\4&-2\end{array} \right] \overrightarrow{x}$$.

Solution. The eigenvalues of the coefficient matrix are $$2$$ and $$-3$$ with corresponding eigenvectors $$\left[\begin{array}{r}1\\1\end{array} \right]$$ and $$\left[\begin{array}{r}1\\-4\end{array} \right]$$ respectively (show all the steps). So the general solution is $\overrightarrow{x}(t)= c_1e^{2t}\left[\begin{array}{r}1\\1\end{array} \right] +c_2e^{-3t}\left[\begin{array}{r}1\\-4\end{array} \right].\\$
Example. $\overrightarrow{x}'=\left[\begin{array}{rrr} 1&3&-3\\2&2&-3\\2&4&-5\end{array} \right] \overrightarrow{x}$ Find the general solution.

Solution. The eigenvalues of the coefficient matrix are $$-2,-1$$, and $$1$$ with corresponding eigenvectors $$\left[\begin{array}{r}1\\1\\2\end{array} \right]$$, $$\left[\begin{array}{r}0\\1\\1\end{array} \right]$$, and $$\left[\begin{array}{r}1\\1\\1\end{array} \right]$$ respectively (show all the steps). So the general solution is $\overrightarrow{x}(t)=c_1e^{-2t}\left[\begin{array}{r}1\\1\\2\end{array} \right] +c_2e^{-t}\left[\begin{array}{r}0\\1\\1\end{array} \right] +c_3e^{t}\left[\begin{array}{r}1\\1\\1\end{array} \right].$
Example. Solve the IVP $$\overrightarrow{x}'=\left[\begin{array}{rr} -1&1\\1&-1\end{array} \right] \overrightarrow{x},\; \overrightarrow{x}(0)=\left[\begin{array}{c} 3\\1\end{array} \right]$$.

Solution. The eigenvalues of the coefficient matrix are $$0$$ and $$-2$$ with corresponding eigenvectors $$\left[\begin{array}{r}1\\1\end{array} \right]$$ and $$\left[\begin{array}{r}-1\\1\end{array} \right]$$ respectively (show all the steps). So the general solution is $\overrightarrow{x}(t)=c_1\left[\begin{array}{r}1\\1\end{array} \right] +c_2e^{-2t}\left[\begin{array}{r}-1\\1\end{array} \right].$ \begin{align*} \overrightarrow{x}(0)=\left[\begin{array}{c}3\\1\end{array} \right] &\implies c_1\left[\begin{array}{r}1\\1\end{array} \right] +c_2\left[\begin{array}{r}-1\\1\end{array} \right]=\left[\begin{array}{c}3\\1\end{array} \right]\\ &\implies c_1-c_2=3,\; c_1+c_2=1\\ &\implies c_1=2,\; c_2=-1. \end{align*} So the solution is $\overrightarrow{x}(t)=2\left[\begin{array}{r}1\\1\end{array} \right] -e^{-2t}\left[\begin{array}{r}-1\\1\end{array} \right].$ Geometric view: $\overrightarrow{x}(t)=2\left[\begin{array}{r}1\\1\end{array} \right] -e^{-2t}\left[\begin{array}{r}-1\\1\end{array} \right] \implies \left[\begin{array}{r}x_1\\x_2\end{array} \right]=\left[\begin{array}{r}2+e^{-2t}\\2-e^{-2t}\end{array} \right] \implies x_1=2+e^{-2t},\; x_2=2-e^{-2t}.$ Eliminating $$t$$, we get $$x_1+x_2=4$$, the trajectory of the particle whose planar motion is described by the given IVP.

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