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
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
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

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}.

\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}.


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