10.2: Solution by Diagonalization
Another way to view the problem of coupled first-order linear odes is from the perspective of matrix diagonalization. With
\[\dot{x}=A x \nonumber \]
we suppose A can be diagonalized using
\[\mathrm{S}^{-1} \mathrm{AS}=\Lambda, \nonumber \]
where \(\Lambda\) is the diagonal eigenvalue matrix, and \(S\) holds the eigenvectors. We can change variables in Equation \ref{10.7} using
\[x=S y \nonumber \]
and obtain
\[\text { Sy் }=\text { ASy. } \nonumber \]
Multiplication on the left by \(\mathrm{S}^{-1}\) and using Equation \ref{10.8} results in
\[\begin{aligned} \dot{\mathrm{y}} &=\mathrm{S}^{-1} \mathrm{ASy} \\ &=\Lambda \mathrm{y} . \end{aligned} \nonumber \]
The first-order differential equations in the \(y\) -variables are now uncoupled and can be immediately solved, and the x-variables can be recovered using Equation \ref{10.9}.
Example: Solve the previous example
\[\frac{d}{d t}\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{ll} 1 & 1 \\ 4 & 1 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) \nonumber \]
by the diagonalization method.
The eigenvalues and eigenvectors are known, and we have
\[\Lambda=\left(\begin{array}{rr} 3 & 0 \\ 0 & -1 \end{array}\right), \quad S=\left(\begin{array}{rr} 1 & 1 \\ 2 & -2 \end{array}\right) \nonumber \]
and the uncoupled \(y\) -equations are given by
\[\dot{y}_{1}=3 y_{1}, \quad \dot{y}_{2}=-y_{2}, \nonumber \]
with solution
\[y_{1}(t)=c_{1} e^{3 t}, \quad y_{2}=c_{2} e^{-t} \nonumber \]
Transforming back to the \(x\) -variables using Equation \ref{10.9}, we have
\[\begin{aligned} \left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) &=\left(\begin{array}{rr} 1 & 1 \\ 2 & -2 \end{array}\right)\left(\begin{array}{c} c_{1} e^{3 t} \\ c_{2} e^{-t} \end{array}\right) \\ &=\left(\begin{array}{c} c_{1} e^{3 t}+c_{2} e^{-t} \\ 2 c_{1} e^{3 t}-2 c_{2} e^{-t} \end{array}\right) \end{aligned} \nonumber \]
which agrees with solution Equation \ref{10.6}.