10.2: Solution by Diagonalization

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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}.