M555: Ordinary Differential Equations

3.4 Repeated Roots

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3.4.1 The Characteristic Equation

We showed in the sections 3.1 , 3.2 and 3.3 that if one wants to solve a 2nd-order linear homogeneous differential equation then the general form of the differential equation is given by

$$P(x)y'' + Q(x)y' + R(x)y = 0$$
and if the equation has constant coefficients the differential equation is given by

$$ay'' + by' + cy = 0.$$
We also showed that if one writes the equation in operator format, we can use $D$ or $d$ to represent the derivative operator and the form of the differential equation becomes

$$ad^2[y] + bd[y] + c[y] = 0.$$
This linear combination of the linear derivative operator gives rise to a linear operator form

$$\left(ad^2 + bd + c\right)[y] = 0.$$
Certainly $y=0$ is a solution to any linear homogeneous differential equation. In our search to find the non-trivial solutions we recognize that it is the operator that is equal to the zero operator when $y(t)$ is non-trivial and we get the characteristic operator

$$ad^2 + bd + c$$
and the characteristic equation

$$ad^2 + bd + c = 0.$$
I realize that the author uses $r$ but the use of a different letter than $d$ for the derivative just seems wrong. In our first two sections all of our characteristic equations had two real roots $r_1$ and $r_2$ (that's where the $r$ comes from); and these roots gave us a pair of linearly independent solutions

$$y_1(t) = e^{r_1t},\qquad\text{and}\qquad y_2(t) = e^{r_2t}.$$
We learned in Section 3.2 that the solution space of all possible solutions to our 2nd-order linear homogeneous differential equation is a two-dimensional subspace of the vector space $C^2[a,b]$ and that these two linearly independent solutions provide us with a basis for our two-dimensional subspace. This pair of basis vectors are called the fundamental solutions . A set of fundamental solutions is far from unique but using the two solutions we obtain from the roots of the characteristic polynomial is certainly the most economical set of fundamental solutions to find.

Quadratic equations will always have two roots, counting multiplicity. We may determine the nature of these roots by checking the discriminant $b^2 - 4ac$. The "standard" possibilities are

• $b^2-4ac \gt 0$: Real distinct roots
• $b^2-4ac \lt 0$: Complex conjugate roots
• $b^2-4ac = 0$: Real repeated roots

There are however more possibilities than this and we discuss the associated vocabulary below.

3.4.2 Undamped

If $\ b=0\ $ then the form of the differential operator is

$$ay'' + cy = \left(ad^2 + c\right)[y] $$

and the sign of the discriminant will be determined by the signs of $a$ and $c$. In this case we call the 2nd-order linear differential equation undamped.

1. If $\ 4ac \lt 0$, that is the discriminant $\ -4ac \gt 0$, then we have two different real roots and two linearly independent solutions of the form
   
   $$y_1(t) = e^{r_1t},\qquad\text{and}\qquad y_2(t) = e^{r_2t}.$$
   
   where $\ r_1 = -\sqrt{\frac{c}{a}}\ $ and $\ r_2 = \sqrt{\frac{c}{a}}$.

2. If $\ 4ac \gt 0$. that is the discriminant $\ -4ac \lt 0$, then we have complex conjugate roots and two linearly independent solutions of the form

$$y_1(t) = e^{irt},\qquad\text{and}\qquad y_2(t) = e^{-irt}.$$

where $\ r = \sqrt{-\frac{c}{a}}\ $. We use Euler's formula and write the solutions

$$y_1(t) = \cos(rt)+i\sin(rt)\qquad\text{and}\qquad y_2(t) = \cos(rt)-i\sin(rt).$$

Our solutions take the form

$$y(t) = c_1\left(\cos(rt)+i\sin(rt)\right) + c_2\left(\cos(rt)-i\sin(rt)\right)$$

$$y(t) = c_1\cos(rt) + ic_1\sin(rt) + c_2\cos(rt) - ic_2\sin(rt)$$

$$y(t) = (c_1 + ic_2)\cos(rt) + (c_1-ic_2)\sin(rt)$$

$$y(t) = A\cos(rt) + B\sin(rt),$$

where $\ A = c_1 + ic_2\ $ and $\ B = c_1 - ic_2$. In our class, the initial conditions of our initial value problems will be real so our arbitrary constants will be real, so our solutions will be real-valued functions. It is important that in this class we never answer a question with the complex-valued fundamental solutions

$$y_1(t) = e^{irt}\qquad\text{and}\qquad y_2(t) = e^{-irt}.$$

Instead we always use the real-valued fundamental solutions

$$y_1(t) = \cos(rt)\qquad\text{and}\qquad y_2(t) = \sin(rt).$$

Now this illustrates that we could have considered complex-valued solutions to our differential equations from the beginning of our class; however, this is beyond the scope of our course. If you are interested in complex-valued solutions, particularly if you are a natural science or electrical engineering student, then I recommend either the course "Complex Analysis" or "Complex Analysis for Engineers".

Example 1

$$\begin{align*} y'' - 4y &= 0 \\ \\ \left(d^2 - 4\right)[y] &= 0 \\ \\ d^2 - 4 &= 0 \qquad\qquad\text{$a=1$, $b=0$ and $c=-4$} \\ \\ d^2 &= 4 \qquad\qquad -\frac{c}{a} \gt 0\\ \\ d &= \pm\,2 \\ \\ y_1(t) = e^{-2t}\qquad &\text{and} \qquad y_2(t) = e^{2t} \end{align*}$$

Example 2

$$\begin{align*} y'' + 4y &= 0 \\ \\ \left(d^2 + 4\right)[y] &= 0 \\ \\ d^2 + 4 &= 0 \qquad\qquad\text{$a=1$, $b=0$ and $c=4$} \\ \\ d^2 &= -4 \qquad\qquad -\dfrac{c}{a} \lt 0 \\ \\ d &= \pm\,2i \\ \\ y_1(t) = \cos(2t)\qquad &\text{and} \qquad y_2(t) = \sin{2t} \end{align*}$$

3.4.3 Damped

If $\ b\neq 0\ $ then the form of the differential operator is

$$ay''+ by' + cy = \left(ad^2 + bd + c\right)[y] $$

In this case we call the 2nd-order linear homogeneous differential equation damped and the sign of the discriminant $\ b^2-4ac\ $ determines the type of damped motion that will result.

1. Overdamped:$\ $ If $\ b^2-4ac \gt 0\ $ then we have two different real roots and two linearly independent solutions of the form

$$y_1(t) = e^{r_1t},\qquad\text{and}\qquad y_2(t) = e^{r_2t},$$

where $\ r_1 = \dfrac{-b -\sqrt{b^2-4ac}}{2a}\ $ and $\ r_2 = \dfrac{-b +\sqrt{b^2-4ac}}{2a}\ $.

1. Underdamped:$\ $ If $\ b^2-4ac \lt 0\ $ then we have complex conjugate roots and two linearly independent solutions of the form

$$y_1(t) = e^{(\mu+i\delta)t},\qquad\text{and}\qquad y_2(t) = e^{(\mu-i\delta)t},$$

where $\dfrac{-b\,\pm\,\sqrt{b^2-4ac}}{2a}=\mu\,\pm\,i\delta$. That is $\ \mu=-\dfrac{b}{2a}\ $ and $\ \delta = \dfrac{\sqrt{4ac-b^2}}{2a}$. This is unwieldy so we distribute the factor $e^{\mu t}$ and write our solutions.

$$y(t) = e^{\mu t}\left(c_1e^{i\delta t} + c_2e^{-i\delta t}\right).$$

We use Euler's equation as in the undamped case and obtain

$$y(t) = e^{\mu t}\left(A\cos(\delta t) + B\sin(\delta t)\right).$$

1. Critically Damped:$\ $ If $\ b^2-4ac = 0\ $ then we have a repeated root $d = -\dfrac{b}{2a}$. In this case the characteristic equation only gives us one of the two fundamental solutions we need.

Example 3

$$\begin{align*} y'' + 4y' + 3y &= 0 \\ \\ d^2 + 4d + 3 &= 0 \\ \\ (d + 3)(d + 1) &= 0 \\ \\ d = -3\qquad&\text{and}\qquad d=-1 \\ \\ y_1(t) = e^{-3t}\qquad&\text{and}\qquad y_2(t) = e^{-t} \\ \\ y(t) &= c_1e^{-3t} + c_2e^{-t} \\ \\ \end{align*}$$


$$\huge{\textbf{Overdamped}}$$

Example 4

$$\begin{align*} 8y'' + 4y' + 25y &= 0 \\ \\ 8d^2 + 4d + 25 &= 0 \\ \\ d &= \dfrac{-4\,\pm\sqrt{16 - 800}}{16} \\ \\ d &= \dfrac{-4\,\pm\sqrt{-784}}{16} \\ \\ d &= -\dfrac{1}{4}\,\pm\dfrac{7}{4}i \\ \\ y_1(t) = e^{-t/4}\cos\left(\frac{7t}{4}\right)\qquad&\text{and}\qquad y_2(t) = e^{-2t}\sin\left(\frac{7t}{4}\right) \\ \\ y(t) &= e^{-t/4}\left(A\cos\left(\frac{7t}{4}\right) + B\sin\left(\frac{7t}{4}\right)\right) \end{align*}$$


$$\huge{\textbf{Underdamped}}$$

Example 5

$$\begin{align*} y'' + 4y' + 4 &= 0 \\ \\ d^2 + 4d + 4 &= 0 \\ \\ d &= \dfrac{-4\,\pm\sqrt{16-16}}{2} = -2 \\ \\ y_1(t) &= e^{-2t} \\ \end{align*}$$

There are several methods for finding another linearly independent solution.



$$\huge{\textbf{Critically Damped}}$$

3.4.4 d'Alembert's Method

In d'Alembert's method we have an infinite family of solutions $cy_1(t)$. We replace our arbitrary constant by an arbitrary function $v(t)$ and solve for the unknown function. This is not the way we will usually solve a homogeneous problem, but it is an important technique we will need later for nonhomogeneous problems. So we define our solutions to be

$$y(t) = v(t)y_1(t)$$
so that

$$y'(t) = v'(t)y_1(t) + v(t)y_1'(t)$$
using the chain rule. Likewise

$$y''(t) = v''(t)y_1(t) + 2v'(t)y'(t) + v(t)y_1''(t).$$
In our example that gives us

$$\begin{align*} y(t) &= v(t)e^{-2t} \\ \\ y'(t) &= v'(t)e^{-2t} - 2v(t)e^{-2t} \\ \\ y''(t) &= v''(t)e^{-2t} - 4v'(t)e^{-2t} + 4ve^{-2t}. \\ \end{align*}$$
Substituting these expressions into our differential equation give us

$$\begin{align*} L\left[v(t)e^{-2t}\right] &= \left(v(t)e^{-2t}\right)'' + 4\left(v(t)e^{-2t}\right)' + 4\left(v(t)e^{-2t}\right) \\ \\ &= v''(t)e^{-2t} - 4v'(t)e^{-2t} + 4ve^{-2t} + 4\left(v'(t)e^{-2t} - 2v(t)e^{-2t}\right) + 4\left(v(t)e^{-2t}\right) \\ \\ &= v''(t)e^{-2t} - 4v'(t)e^{-2t} + 4v(t)e^{-2t} + 4v'(t)e^{-2t} - 8v(t)e^{-2t} + 4v(t)e^{-2t} \\ \\ &= \left(v''(t) - 4v'(t) + 4v(t) + 4v'(t) - 8v(t) + 4v(t)\right)e^{-2t} \\ \\ &= v''(t)e^{-2t} = 0 \\ \end{align*}$$
Since $e^{-2t}$ is never equal to zero we have $v''(t) = 0$. We can solve this by integrating twice

$$\begin{align*} v''(t) &= 0 \\ \\ v'(t) &= \displaystyle\int v''(t)\,dt = \displaystyle\int 0\,dt = c_1 \\ \\ v(t) &= \displaystyle\int v'(t)\,dt = \displaystyle\int c_1\,dt = c_1t + c_2. \\ \end{align*}$$
Therefore our solutions can be written

$$y(t) = v(t)e^{-2t} = \left(c_1t + c_2\right) e^{-2t} = c_1te^{-2t} + c_2e^{-2t}.$$
If $e^{-2t}$ and $te^{-2t}$ are linearly independent then we have our set of fundamental solutions. The Wronskian is given by

$$\begin{align*} W\left[e^{-2t},\,te^{-2t}\right](t) &= \left|\begin{array}{cc} e^{-2t} & te^{-2t} \\ -2e^{-2t} & e^{-2t} - 2te^{-2t} \end{array}\right| \\ \\ &= e^{-2t}\left(e^{-2t} - 2te^{-2t}\right) - \left(-2e^{-2t}\cdot te^{-2t}\right) \\ \\ &= e^{-4t} - 2te^{-4t} + 2te^{-4t} = e^{-4t} \neq 0. \\ \end{align*}$$
Since $e^{-4t}\neq 0$ for any $t\in\mathbb{R}$ we have that the two functions $e^{-2t}$ and $te^{-2t}$ are linearly independent for all real number $t$. Hence we indeed found a linearly independent solution and our fundamental set of solutions is

$$\left[ e^{-2t}, te^{-2t} \right]$$
and every solution of our differential equation may be written

$$y(t) = c_1e^{-2t} + c_2te^{-2t}.$$

3.4.5 Operator Method

With this method we will still use a form of the characteristic polynomial and solve two first-order differential equations.

$$\begin{align*} y'' + 4y' + 4 &= 0 \\ \\ (d^2 + 4d + 4)[y] &= 0 \\ \\ (d + 2)(d + 2)[y] &= 0 \end{align*}$$
Now define $\ v(t) = (d + 2)[y]\ $ so that our differential equation becomes

$$(d + 2)[v] = 0.$$
Solve this 1st-order equation.

$$\begin{align*} v' + 2v &= 0 \\ \\ \dfrac{dv}{dt} &= -2v \\ \\ \dfrac{1}{v}\dfrac{dv}{dt} &= -2 \\ \\ \log(v) &= -2t + c_1 \\ \\ v(t) &= c_1e^{-2t} \end{align*}$$
Replacing $\ v(t)\ $ with $\ (d + 2)[y]\ $ gives us

$$\begin{align*} (d + 2)[y] &= c_1e^{-2t} \\ \\ y' + 2y &= c_1e^{-2t} \\ \\ \mu(t) &= e^{\int 2\,dt} = e^{2t}\qquad\text{integrating factor}\\ \\ e^{2t}y' + 2e^{2t}y &= c_1 \\ \\ \left[e^{2t}y\right]' &= c_1 \\ \\ e^{2t}y &= \displaystyle\int c_1\,dt = c_1t + c_2 \\ \\ y(t) &= c_1te^{-2t} + c_2e^{-2t} \end{align*}$$
We again have the general solution of our differential equation with perhaps fewer steps and like d'Alembert's method can be used for nonhomogeneous equations later in this chapter.

3.4.6 Reduction of Order

In reduction of order we solve the characteristic equation as in d'Alembert's method so that we have one of our two fundamental solutions. The we use Abel's theorem to find another linearly independent solution. If we call the second linearly independent solution $y_2(t)$ then we can substitute $\ y_1(t)=e^{-2t}\ $ into Abel's formula to get

$$\begin{align*} W\left[e^{-2t},\ y_2(t)\right](t) &= \left|\begin{array}{cc} e^{-2t} & y_2(t) \\ -2e^{-2t} & y_2'(t) \end{array}\right| \\ \\ &= e^{-2t}y_2'(t) + 2e^{-2t}y_2(t) \\ \\ &= c_1\,e^{\int 4\,dt} = c_1e^{4t} \\ \\ e^{-2t}y_2'(t) + 2e^{-2t}y_2(t) &= c_1e^{4t} \\ \\ y_2'(t) + 2y_2(t) &= c_1e^{2t} \\ \\ \mu(t) &= e^{\int 2\,dt} = e^{2t}\qquad\text{integrating factor} \\ \\ e^{2t}y_2'(t) + 2e^{2t}y_2(t) &= c_1 \\ \\ \left[ e^{2t}y_2(t)\right]' &= c_1 \\ \\ e^{2t}y_2(t) &= \displaystyle\int c_1\,dt = c_1t + c_2 \\ \\ y_2(t) &= c_1te^{-2t} + c_2e^{-2t} \end{align*}$$
Here we see that the first fundamental solution is part of the linear combination that makes up the second fundamental solution. We can use either

$$y_2(t) = c_1te^{-2t} + c_2e^{-2t}$$
or we can simple use the "new" part of the linear combination and take

$$y_2(t) = c_1te^{-2t}.$$
This give us our fundamental set of solutions $\ te^{-2t}\ $ and $\ c_2e^{-2t}\ $. Again our general solution to our differential equation is

$$y(t) = c_1te^{-2t} + c_2e^{-2t}.$$

Example 4

Find the solution of the initial value problem

$$y'' - y' + \frac{y}{4} = 0,\qquad y(0)=2,\ y'(0) = \frac{1}{3}$$

The characteristic equation is

$$d^2 - d + \frac{1}{4} = \left(d - \frac{1}{2}\right)^2 = 0.$$
Thus if we use what we learned about repeated roots our general solution is of the form

$$y(t) = c_1e^{t/2} + c_2te^{t/2}.$$
Using our initial conditions we have

$$\begin{align*} 2 = y(0) &= c_1e^0 + c_2(0)e^0 = c_1 \\ \\ y(t) &= 2e^{t/2} + c_2te^{t/2} \\ \\ y'(t) &= e^{t/2} + c_2e^{t/2} + \frac{c_2}{2}te^{t/2} \\ \\ \frac{1}{3} = y'(0) &= e^0 + c_2e^0 + \frac{c_2}{2}(0)e^0 = 1 + c_2 \\ \\ c_2 &= \frac{1}{3} - 1 = -\frac{2}{3} \\ \\ y(t) &= 2e^{-t/2} - \frac{2}{3}te^{-t/2} \end{align*}$$

Exercise 1

Find the solution to the initial value problem. Sketch the graph of the solution and describe its behavior for increasing $t$.
$$9y'' - 12y' + 4y = 0,\qquad y(0) = 2,\ y'(0) = -1$$

View Solution

In characteristic equation is

$$\begin{align*} 9d^2 - 12d + 4 &= 0 \\ \\ d &= \dfrac{12\,\pm\sqrt{144 - 4(4)9}}{18} = \dfrac{2}{3} \\ \\ \end{align*}$$
This give us the general solution
$$y(t) = c_1e^{2t/3} + c_2te^{2t/3}$$
Using our initial conditions
$$\begin{align*} 2 = y(0) &= c_1 + 0 \\ \\ y(t) &= 2e^{2t/3} + c_2te^{2t/3} \\ \\ y'(t) &= \frac{4}{3}e^{2t/3} + c_2e^{2t/3} + \frac{2c_2t}{3}e^{2t/3} \\ \\ -1 = y'(0) &= \frac{4}{3} + c_2 + 0 \\ \\ c_2 &= -\dfrac{7}{3} \\ \\ y(t) &= 2e^{2t/3} - \dfrac{7t}{3}e^{2t/3} \end{align*}$$

$$\lim_{t\rightarrow\infty} 2e^{2t/3} - \dfrac{7t}{3}e^{2t/3} = -\infty$$

Exercise 2

Find the solution to the initial value problem. Sketch the graph of the solution and describe its behavior for increasing $t$.
$$ 4y'' + 4y' + y = 0,\qquad y(0) = 1,\ y'(0) = 2 $$

Check Your Work

$$ y(t) = e^{-\frac{t}{2}} + \dfrac{5}{2}te^{-\frac{t}{2}}$$

Video Solution

Example 5

Given that $\ y_1(t) = t^{-1}\ $ is a solution of

$$2t^2y'' + 3ty' - y = 0\qquad t > 0,$$
find a fundamental set of solutions.

Using d'Alembert's method we let $y(t) = v(t)y_1(t)$ and

$$\begin{align*} y(t) &= v(t)t^{-1} \\ \\ y'(t) &= v'(t)t^{-1} - v(t)t^{-2} \\ \\ y''(t) &= v''(t)t^{-1} - 2v'(t)t^{-2} + 2v(t)t^{-3} \\ \\ 2t^2y'' + 3ty' - y &= 2t^2\left(v''(t)t^{-1} - 2v'(t)t^{-2} + 2v(t)t^{-3}\right) + 3t\left(v'(t)t^{-1} - v(t)t^{-2}\right) - v(t)t^{-1} \\ \\ &= 2tv''(t) - 4v'(t) + 4v(t)t^{-1} + 3v'(t) - 3v(t)t^{-1} - v(t)t^{-1} \\ \\ &= 2tv''(t) - v'(t) = 0 \\ \\ \end{align*}$$
Notice that this 2nd-order differential equation has only derivative terms so we will use another form of reduction of order by substituting $\ w(t) = y'(t)\ $.

$$\begin{align*} 2tw'(t) - w(t) &= 0 \\ \\ 2t\dfrac{dw}{dt} &= w \\ \\ \dfrac{1}{w}\dfrac{dw}{dt} &= \dfrac{1}{2t} \\ \\ \log|w| &= \dfrac{1}{2}\log|t| + c_1 = \log\left|t^{1/2}\right| + c_1 \\ \\ w(t) &= c_1t^{1/2} \\ \\ y'(t) &= c_1t^{1/2} \\ \\ y(t) &= \displaystyle\int c_1t^{1/2}\,dt = \dfrac{2c_1}{3}t^{3/2} + c_2 \end{align*}$$

$$y(t) = v(t)t^{-1} = \left(\dfrac{2c_1}{3}t^{3/2} + c_2\right)t^{-1}$$
$$y(t) = \dfrac{2c_1}{3}t^{1/2} + c_2t^{-1}.$$
From this we see that the other linearly independent solution is

$$y_2(t) = t^{1/2}.$$
This give us a the following set of fundamental solutions

$$\left[ t^{-1},\ t^{1/2}\ \right].$$
We know they are linearly independent because the Wronskian

$$\begin{align*} W\left[t^{-1},\ t^{1/2}\right](t) &= \left|\begin{array}{cc} t^{-1} & t^{1/2} \\ -t^{-2} & \frac{1}{2}t^{-1/2} \end{array}\right| \\ \\ &= \dfrac{1}{2}t^{-3/2} + t^{-3/2} = \dfrac{3}{2}t^{-3/2} \neq 0 \end{align*}$$
when $t > 0$.

Exercise 3

Use d'Alembert's method or use Abel's formula to find a second linearly independent of the equation

$$t^2y'' - 4ty' + 6y = 0$$
when $y_1(t) = t^2$ is a solution.

View Solution

In standard form our differential equation becomes
$$y'' - \dfrac{4}{t}y' + \dfrac{6}{t^2}y = 0.$$
We are guaranteed a unique solution only if $\ t > 0\ $ or $\ t < 0\ $. We will chose $\ t > 0\ $. Using Abel's formula instead of d'Alembert's method like Example 5 we have that the Wronskian
$$\begin{align*} W\left[t^2,\ y_2(t)\right](t) &= \left|\begin{array}{cc} t^2 & y_2(t) \\ 2t & y_2'(t) \end{array}\right| \\ \\ &= t^2y_2'(t) - 2ty_2(t) \\ \\ &= c_1\text{exp}\left(-\displaystyle\int -\frac{4}{t}\,dt\right) \\ \\ &= c_1e^{4\log(t)} = c_1e^{\log\left(t^4\right)}= c_1t^4 \\ \\ t^2y_2'(t) - 2ty_2(t) &= c_1t^4 \\ \\ y_2' - \dfrac{2}{t}y_2 &= c_1t^2 \\ \mu(t) &= e^{\int-\frac{2}{t}\,dt} = e^{-2\log(t)} = e^{\log(t^{-2})} = t^{-2} \\ \\ t^{-2}y_2' - 2t^{-2}y_2 &= c_1 \\ \\ \left[ t^{-2}y_2 \right]' &= c_1 \\ \\ t^{-2}y_2 &= \displaystyle\int c_1\,dt = c_1t + c_2 \\ \\ y_2(t) &= c_1t^3 + c_2t^2 \\ \end{align*}$$
The other fundamental solution is $y_2(t) = t^3$.