# First-order LSEs with constant coefficients: Repeated roots (continued)

In an earlier blog post, I showed 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. In particular, I considered 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, for the case where $(a-d)^2+4bc=0$ and $a\ne d$. Now I look at the case where $a=d$.

I assume the reader is familiar with the technique called the method of undetermined coefficients. (See, for example, the 8th edition of Elementary Differential Equations by Rainville, Bedient, and Bedient published in 1996 by Prentice Hall).

Given that $(a-d)^2+4bc=0$ and $a=d$, there are three possibilities: (1) $b=0$ and $c=0$, (2) $b\ne 0$ and $c=0$, and (3) $b=0$ and $c\ne 0$. For the first possibility, the solution is simple: $x=c_1e^{at}$ and $y=c_2e^{at}$ (since $a=d$), where $c_1$ and $c_2$ are arbitrary constants.

We now consider the case where $b\ne 0$ and $c=0$. From $\frac{\mathrm{d}y}{\mathrm{d}t}=cx+dy$ we have $y=c_2e^{at}$ (since $a=d$). From $\frac{\mathrm{d}x}{\mathrm{d}t}=ax+by$ we have $\frac{\mathrm{d}x}{\mathrm{d}t}-ax=bc_2e^{at}$.

We now use the method of undetermined coefficients to solve for $x$. The solution to the homogeneous equation $\frac{\mathrm{d}x_c}{\mathrm{d}t}-ax_c=0$ is $x_c=c_1e^{at}$, where $c_1$ is an arbitrary constant. (We will use this $x_c$ later.)

Now note that $R=bc_2e^{at}$ is a solution to the differential equation $\frac{\mathrm{d}R}{\mathrm{d}t}-aR=0$.

The equation $\frac{\mathrm{d}x}{\mathrm{d}t}-ax=bc_2e^{at}$ expressed using differential operators is $(D-a)x=R$. Applying $(D-a)$ on both sides of this yields $(D-a)^2x=(D-a)R=0$, where the last equality is explained by the previous paragraph. The solution to $(D-a)^2x=0$ is $x=c_1e^{at}+Ate^{at}$. Note that this is of the form $x=x_c+x_p$. From $x_c=c_1e^{at}$ which we got earlier, we get $x_p=Ate^{at}$, leading us to conclude that $A$ is a non-arbitrary constant.

We find the value of $A$ by plugging our solution into the original equation. That is, from $\frac{\mathrm{d}x}{\mathrm{d}t}-ax=bc_2e^{at}$ we get $\frac{\mathrm{d}}{\mathrm{d}t}\left(c_1e^{at}+Ate^{at}\right)-a\left(c_1e^{at}+Ate^{at}\right)=bc_2e^{at}$. This yields $A=bc_2$.

Thus, the solution to the system when $b\ne 0$ and $c=0$ is $x=c_1e^{at}+bc_2te^{at}$ and $y=c_2e^{at}$.

The case where $b=0$ and $c\ne 0$ is similar and is left to the reader as an exercise.