First-order LSEs with constant coefficients: Repeated roots

I’m currently teaching an undergraduate course on ordinary differential equations using the 8th edition of Elementary Differential Equations by Rainville, Bedient, and Bedient (published in 1996 by Prentice Hall). I’ve always wanted to have a blog post containing some LaTeX (using the plug-in WP LaTeX), so in this blog post I’ll be showing my solution to one of the exercises in the book.

I will derive the solution to a first-order linear system of two (ordinary) differential equations with constant coefficients for the case where the characteristic equation of the (constant coefficient) square matrix has repeated roots.

Let the system $\begin{array}{rcl}\frac{\mathrm{d}x}{\mathrm{d}t}&=&ax+by\\ \frac{\mathrm{d}y}{\mathrm{d}t}&=&cx+dy\\\end{array}$, where $a,b,c,d$ are real constants and $x,y,t$ are real variables, be represented in matrix form as $\frac{\mathrm{d}}{\mathrm{d}t}\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}a & b\\c & d\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]$ or simply as $\mathbf{X}'=\mathbf{AX}$.

I will show that this system has the solution
$\mathbf{X}=c_1\left[\begin{array}{c}2b\\d-a\end{array}\right]e^{(a+d)t/2}+c_2\left(\left[\begin{array}{c}2b\\d-a\end{array}\right]t+\left[\begin{array}{c}0\\2\end{array}\right]\right)e^{(a+d)t/2}$
if $(a-d)^2+4bc=0$ and $a\ne d$.

I assume the reader is familiar with the solution to the case where the characteristic equation has distinct real roots.

From the characteristic equation of $\mathbf{A}$, we get
$\left|\mathbf{A}-m\mathbf{I}\right|=\left|\begin{array}{cc}a-m&b\\c&d-m\end{array}\right|=(a-m)(d-m)-bc=0$
and $m^2+(-a-d)m+(ad-bc)=0$ which has the solutions $m_1=\frac{1}{2}\left(-(-a-d)+\sqrt{(-a-d)^2-4(ad-bc)}\right)$ and $m_2=\frac{1}{2}\left(-(-a-d)-\sqrt{(-a-d)^2-4(ad-bc)}\right)$
Now, $m_1=m_2=(a+d)/2$ if $(-a-d)^2-4(ad-bc)=0$. That is, the characteristic equation has repeated roots if $(a-d)^2+4bc=0$ and $a\ne d$.

The two roots yield two solutions $\mathbf{X}_1$ and $\mathbf{X}_2$. (The complete solution would thus be $\mathbf{X}=c_1\mathbf{X_1}+c_2\mathbf{X_2}$.)

To find $\mathbf{X}_1$, recall that $(\mathbf{A}-m\mathbf{I})\mathbf{C}=\mathbf{0}$ then use $m=(a+d)/2$ to get $\left[\begin{array}{cc}a-(a+d)/2&b\\c&d-(a+d)/2\end{array}\right] \left[\begin{array}{c}k_1\\k_2\end{array}\right]=\left[\begin{array}{c} 0 \\ 0 \end{array}\right]$.

This yields two equivalent equations in $k_1$ and $k_2$. (This can be seen by using $c=-(a-d)^2/(4b)$ in the second equation.) We get $k_2=\frac{d-a}{2b}k_1$ and letting $k_1=2b$ and $k_2=d-a$ yields $\mathbf{X}_1=\left[\begin{array}{c}2b\\d-a\end{array}\right]e^{(a+d)t/2}$.

In finding $\mathbf{X}_2$, note that $\mathbf{X}_2\ne\left[\begin{array}{c}2b\\d-a\end{array}\right]te^{(a+d)t/2}$. Instead, assume that $\mathbf{X}_2=\left[\begin{array}{c}k_1(t)\\k_2(t)\end{array}\right]e^{(a+d)t/2}=\mathbf{C}(t)e^{(a+d)t/2}$.

Substituting this into $\mathbf{X}'=\mathbf{AX}$ yields $\mathbf{C}(t)me^{mt}+e^{mt}\mathbf{C}'(t)=\mathbf{AC}(t)e^{mt}$. Thus, $\left(\mathbf{AC}(t)-m\mathbf{C}(t)-\mathbf{C}'(t)\right)e^{mt}=\mathbf{0}$ and $\left(\mathbf{A}-m\mathbf{I}\right)\mathbf{C}(t)=\mathbf{C}'(t)$ since $e^{mt}\ne 0$ for real $m$.
$\left[\begin{array}{c}k_1'(t)\\k_2'(t)\end{array}\right]=\left[\begin{array}{cc}a-(a+d)/2&b\\c&d-(a+d)/2\end{array}\right] \left[\begin{array}{c}k_1(t)\\k_2(t)\end{array}\right]$
Using $c=-(a-d)^2/(4b)$ yields the system
$\begin{array}{rcl}k_1'(t)&=&\frac{a-d}{2}k_1(t)+bk_2(t)\\k_2'(t)&=&\frac{-(a-d)^2}{4b}k_1(t)-\frac{a-d}{2}k_2(t)\end{array}$.

After a little handwaving, we get $k_1(t)=2bt$ and $k_2(t)=(d-a)t+2$. Thus, $\mathbf{X}_2=\left[\begin{array}{c}2bt\\(d-a)t+2\end{array}\right]e^{(a+d)t/2}$.

Finally,
$\mathbf{X}=c_1\left[\begin{array}{c}2b\\d-a\end{array}\right]e^{(a+d)t/2}+c_2\left(\left[\begin{array}{c}2b\\d-a\end{array}\right]t+\left[\begin{array}{c} 0 \\ 2\end{array}\right]\right)e^{(a+d)t/2}$.