$2 \times 2$ Systems and Second-Order Equations

Since many of you don't have a background in linear algebra, we'll concentrate on $2 \times 2$ systems. We spent a lot of effort solving second-order linear equations, so it's useful to know that these systems can be solved by reducing them to second-order equations.

For example, consider the system

\begin{displaymath}
\begin{array}{rl}
x_1' &= 2 x_1 + x_2 + t \\
x_2' &= x_1 + 2 x_2 + t^2
\end{array}\end{displaymath}

First solve the first equation for $x_2$ in terms of $x_1$, $x_1'$ and $t$:

\begin{displaymath}
x_2 = x_1' - 2 x_1 - t
\end{displaymath}

Differentiate to get $x_2'$:

\begin{displaymath}
x_2' = x_1'' - 2 x_1' - 1
\end{displaymath}

Then substitute these expressions for $x_2$ and $x_2'$ in the second equation:

\begin{displaymath}
x_1'' - 2 x_1' - 1 = x_1 + 2 (x_1' - 2 x_1 - t) + t^2
\end{displaymath}

or, simplifiying,

\begin{displaymath}
x_1'' - 4 x_1' + 3 x_1 = t^2 - 2 t + 1
\end{displaymath}

We could solve this using Undetermined Coefficients:

\begin{displaymath}
x_1 = \frac{1}{3} t^2 + \frac{2}{9} t + \frac{11}{27}
+ c_1 {\rm e}^{ t} + c_2 {\rm e}^{3t}
\end{displaymath}

and plug this in to the equation for $x_2$:

\begin{displaymath}
x_2 = -\frac{2}{3} t^2 - \frac{7}{9} t - \frac{16}{27} - c_1 {\rm e}^t
+ c_2 {\rm e}^{3t}
\end{displaymath}

The solution can be expressed in the language of vectors:

\begin{displaymath}
\pmatrix{x_1\cr x_2\cr} = \pmatrix{t^2/3 + 2 t/9 + 11/27\cr
...
...^t \pmatrix{1\cr -1\cr}
+ c_2 {\rm e}^{3t} \pmatrix{1\cr 1\cr}
\end{displaymath}

In general, with a system

\begin{displaymath}
\begin{array}{rl}
x_1' = a x_1 + b x_2 + f_1(t) \\
x_2' = c x_1 + d x_2 + f_2(t)
\end{array}\end{displaymath}

we can use the same method as long as $b \ne 0$. If $b = 0$ but $c \ne 0$, we would reverse the roles of $x_1$ and $x_2$: first solve the second equation for $x_1$, differentiate to get $x_1'$, and substitute in to the first equation to get a second-order linear differential equation for $x_2$.

Here's an example, this time homogeneous:


\begin{displaymath}
\begin{array}{rl}
x_1' &= 2 x_1 \cr
x_2' &= 3 x_1 + 4 x_2\cr
\end{array}\end{displaymath}


\begin{displaymath}
\displaylines{
x_1 = \displaystyle \frac{1}{3} x_2' - \frac{...
...matrix{-2/3\cr 1\cr}
+ c_2 {\rm e}^{4 t} \pmatrix{0\cr 1\cr}
}
\end{displaymath}

The one remaining case is where $b = 0$ and $c=0$. But this is actually the easiest case, where the two equations are ``uncoupled'', one only involving $x_1$ and the other only involving $x_2$. We can solve these two first-order linear equations separately. For example:


\begin{displaymath}
\begin{array}{rl}
x_1' = 2 x_1 \cr
x_2' = 3 x_2 \end{array}\end{displaymath}

Note that we use different names for the arbitrary constants in the two equations:

\begin{displaymath}
\displaylines{x_1 = c_1 {\rm e}^{2 t} \cr
x_2 = c_2 {\rm e}...
...pmatrix{1\cr 0\cr}
+ c_2 {\rm e}^{3t} \pmatrix{0\cr 1\cr}\cr
}
\end{displaymath}





Robert Israel
2002-03-12