This is the first of three applications of eigenvectors and eigenvalues. Let $x(t),y(t)$
be functions of $t$
, and consider the simultaneous linear ordinary differential equations $$\dot{x}=ax+by\mathit{\hspace{1em}\hspace{1em}}\dot{y}=cx+dy,$$
where $\dot{x}$
denotes $dx/dt$
and $a,b,c,d$
are constants.

These equations are coupled: each equation involves $x$
and $y$
. Our goal is to decouple them, turning them into a pair of equations, each involving only one unknown function. We first rewrite the equations as $$\dot{v}=Av$$
using: $$v(t)=\left(\begin{array}{c}\hfill x(t)\hfill \\ \hfill y(t)\hfill \end{array}\right),\dot{v}=\left(\begin{array}{c}\hfill \dot{x}\hfill \\ \hfill \dot{y}\hfill \end{array}\right),A=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right).$$
Let ${\lambda}_{1},{\lambda}_{2}$
be the eigenvalues of $A$
and assume they are distinct; let ${u}_{1}$
and ${u}_{2}$
be eigenvectors for these eigenvalues. Write $v$
as $\alpha (t){u}_{1}+\beta (t){u}_{2}$
. In other words, $\alpha $
and $\beta $
are the components of $v$
when in the ${u}_{1}$
- and ${u}_{2}$
-directions (compare with $v=x\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)+y\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)$
).

Substituting $v=\alpha {u}_{1}+\beta {u}_{2}$
into the left-hand side of $\dot{v}=Av$
gives: $$\dot{v}=\dot{\alpha}{u}_{1}+\dot{\beta}{u}_{2}$$
(because ${u}_{1},{u}_{2}$
are constant).

Substituting it into the right-hand side gives $$Av=\alpha A{u}_{1}+\beta A{u}_{2}={\lambda}_{1}\alpha {u}_{1}+{\lambda}_{2}\beta {u}_{2}$$
(because ${u}_{k}$
is a ${\lambda}_{k}$
-eigenvector for $k=1,2$
).

Comparing the components in the ${u}_{1}$
-direction, we get $$\dot{\alpha}={\lambda}_{1}\alpha .$$
The ${u}_{2}$
-components tells us that $$\dot{\beta}={\lambda}_{2}\beta .$$
These equations now involve only $\alpha $
and $\beta $
separately. We have decoupled the equations.

An example in detail

Consider the equations: $$\dot{x}=y,\dot{y}=-x.$$
This is equivalent to the simple harmonic oscillator equation$\ddot{x}=-x$
. To see this, differentiate $\dot{x}=y$
to get $\ddot{x}=\dot{y}=-x$
.

In general, if you have a second-order equation like this, you can define $y=\dot{x}$
(which gives you one equation) and then use the second-order equation to express $\dot{y}$
in terms of $x$
and $y$
: this is a good trick for converting second-order equations into pairs of first order equations.

Remark:

The equation $\ddot{x}=-x$
describes physical situations like a particle on a spring:

one end of the spring is fixed at the origin;

the particle sits at a distance $x$
from the origin;

Newton's law tells us that if $m$
is the mass of the particle then $m\ddot{x}$
equals the force experience by the particle, which is $-kx$
by Hooke's law (for some constant $k$
), so the equation of motion is $m\ddot{x}=-kx$
;

if we work in units where $m=k=1$
then this gives us back our equation $\ddot{x}=-x$
.

Let's solve this equation. We have $A=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right)$
. This has characteristic polynomial $$det\left(\begin{array}{cc}\hfill -t\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill -t\hfill \end{array}\right)={t}^{2}+1,$$
which has roots ${\lambda}_{1}=i$
and ${\lambda}_{2}=-i$
.

The eigenvectors are:

for ${\lambda}_{1}=i$
, $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill i\hfill \end{array}\right)$
,

for ${\lambda}_{2}=-i$
, $\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -i\hfill \end{array}\right)$
.

(We're just picking particular eigenvectors, not writing down the general eigenvector).

We write $\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)$
in the form $$\alpha {u}_{1}+\beta {u}_{2}=\alpha \left(\begin{array}{c}\hfill 1\hfill \\ \hfill i\hfill \end{array}\right)+\beta \left(\begin{array}{c}\hfill 1\hfill \\ \hfill -i\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \alpha +\beta \hfill \\ \hfill i(\alpha -\beta )\hfill \end{array}\right).$$
Thus $\alpha =(x-iy)/2$
and $\beta =(x+iy)/2$
.

Our equations are now: $$\dot{\alpha}={\lambda}_{1}\alpha =i\alpha ,\dot{\beta}={\lambda}_{2}\beta =-i\beta .$$
Dividing the first equation by $\alpha $
gives $$\frac{\dot{\alpha}}{\alpha}=\frac{d}{dt}(\mathrm{log}\alpha )=i,$$
so $\mathrm{log}\alpha =it+\text{const}$
, so $\alpha (t)={C}_{1}{e}^{it}$
. Similarly, $\beta (t)={C}_{2}{e}^{-it}$
.

Therefore $$\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)={C}_{1}{e}^{it}\left(\begin{array}{c}\hfill 1\hfill \\ \hfill i\hfill \end{array}\right)+{C}_{2}{e}^{-it}\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -i\hfill \end{array}\right),$$
so $$x={C}_{1}{e}^{it}+{C}_{2}{e}^{-it}\mathit{\hspace{1em}}y=i({C}_{1}{e}^{it}-{C}_{2}{e}^{-it}).$$
This is the general solution to $\dot{x}=y$
, $\dot{y}=-x$
. You may be worried about the fact that there are $i$
s here: this is supposed to describe the motion of a particle on a spring, so $x$
and $y$
should be real numbers. The $i$
s will all cancel out if we pick appropriate initial conditions.

Suppose $x(0)=1$
and $y(0)=0$
(particle at rest at distance 1 from the origin). Substituting $t=0$
into our general solution, we get $$x(0)=1={C}_{1}+{C}_{2},y(0)=0=i({C}_{1}-{C}_{2}).$$
This implies ${C}_{1}={C}_{2}=1/2$
. Therefore $$x(t)=\frac{1}{2}({e}^{it}+{e}^{-it}),y(t)=\frac{i}{2}({e}^{it}-{e}^{-it}).$$
This means that $$x(t)=\mathrm{cos}(t),y(t)=-\mathrm{sin}(t)$$
using the formulae for trigonometric functions in terms of complex exponentials. The minus sign in $y$
is because our particle starts moving towards the origin as time increases.

The moral of this story is that you can use eigenvectors to decouple systems of linear differential equations.