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Math 555: Differential Equations

3.1 Homogeneous Equations with Constant Coefficients

3.1.1 Second-Order Linear Differential Equations

In this chapter, we consider linear second-order differential equations. Linear second-order equations have a wide range of applications to mechanical and electrical systems, which will be explored later. The most intriguing thing about higher-order linear differential equations (of which second-order is the simplest) is that they can be explored pretty deeply with straightforward mathematics.

Second-order ordinary differential equations are typically of the form

$$ \dfrac{d^2 y}{dt^2} = f\left(t,y,\dfrac{dy}{dt}\right), $$
where $f$ is given. Typically, $t$ is used for the independent variable since time is often the independent quantity in physics applications. The letter $y$ is usually the dependent variable, but other choices such as $u$ for displacement in mechanical oscillation problems or $I$ for current in an electrical circuit are common. The second-order equation is linear if $f$ has the form

$$ f\left(t,y,\dfrac{dy}{dt}\right) = g(t) - p(t)\dfrac{dy}{dt} - q(t)y, $$
where $g$, $p$, and $q$ depend on $t$ but not $y$. Second-order linear equations are often written in the standard form

$$ y'' + p(t)y' + q(t)y = g(t), $$
where the primes are differentiations with respect to $t$ in the usual sense. The equation

$$ P(t)y'' + Q(t)y' + R(t)y = G(t) $$
is also linear and may be converted to standard form by dividing everything by $P(t)$ if $P(t)\neq 0$,

$$ p(t) = \dfrac{Q(t)}{P(t)},\quad q(t) = \dfrac{R(t)}{P(t)},\quad g(t) = \dfrac{G(t)}{P(t)}. $$
An equation is nonlinear if it fails to be reducible to one of the above forms. Nonlinear equations are much more difficult to solve and so will be set aside for now while we delve into linear second-order equations of various types.

3.1.2 Homogeneous Equations with Constant Coefficients

A second-order differential equation is called homogeneous if the function $g(t)$ or $G(t)$ is equal to $0$ for all $t$. An equation is nonhomogeneous if these functions are nonzero. They are sometimes referred to as forcing functions, since they represent an external force being applied to the physical system modeled by the differential equation. In this section, we will examine the homogeneous equation

$$ P(t)y'' + Q(t)y' + R(t)y = 0, $$
where the functions $P$, $Q$, and $R$ will be constant functions $a$, $b$, and $c$

$$ ay'' + by' + cy = 0 $$
with initial conditions given by

$$ y(t_0) = y_0,\qquad y'(t_0) = y_0'. $$
The initial conditions stated are for the functions $y$ and $y'$ evaluated at $t_0$. The number of initial conditions for a differential equation is equal to the order of the ODE, so we need two for a second-order equation.

The homogeneous linear second-order equation with constant coefficients described here can be solved fairly easily using elementary functions, so we continue by demonstrating this with an example.

Example 3.1.1

Suppose we have the initial value problem

$$ y'' - y = 0,\qquad y(0) = 1,\ y'(0) = 2. $$
This equation is in the same form as $ay'' + by' + cy = 0$ with $a=1$, $b=0$, and $c = -1$. However, what it is really asking is what function or functions are equal to its second derivative? After thinking for a second, $y_1(t) = e^t$ should come to mind because it is equal to its first derivative (and hence is equal to its second derivative). Also, $y_2(t) = e^{-t}$ is a solution since two derivatives handle the minus sign.

These aren't the only functions that solve this ODE. Due to the linearity of the equation, any constant times either of these functions such as $7e^t$ or $-3e^{-t}$ will also satisfy the equation. Additionally, and again from linearity, any sum of solutions is also a solution, in particular a linear combination of $y_1$ and $y_2$

$$ y(t) = c_1 y_1(t) + c_2 y_2(t) $$
is a general solution to the homogeneous equation. Applying the initial conditions will specify constants $c_1$ and $c_2$, yielding a solution to the initial value problem.

In our example,
$$ y(t) = c_1 e^t + c_2 e^{-t} $$
and
$$ y'(t) = c_1 e^t - c_2 e^{-t}. $$
Applying the initial conditions $y(0) = 1$, and $y'(0) = 2$ to these yields the system of equations

$$ \begin{align*} 1 &= c_1 + c_2 \\ 2 &= c_1 - c_2 \end{align*} $$
which are easily solved using the elimination method from algebra: $c_1 = \frac{3}{2}$ and $c_2 = -\frac{1}{2}$.

Hence, the solution to the initial value problem is

$$ y(t) = \frac{3}{2}e^t - \frac{1}{2}e^{-t}. $$

3.1.3 General Approach to Finding Solutions

After having worked through our first example, let's analyze the results and see how this may be applied to all ODEs of the form

$$ ay'' + by' + cy = 0, $$
where $a$, $b$, and $c$ are arbitrary real constants. To begin, we note that in our previous example, both of the solutions were of the form $e^{rt}$ with $r = 1$ and $r = -1$. So, what happens if we guess that our solution $y = e^{rt}$ for some $r$? We plug that into the ODE and see that

$$ \begin{align*} a(e^{rt})'' + b(e^{rt})' + ce^{rt} &= 0 \\ \\ ar^2 e^{rt} + br e^{rt} + ce^{rt} &= 0 \\ \\ \left(ar^2 + br + c\right) e^{rt} &= 0, \end{align*} $$
which is satisfied only when $ ar^2 + br + c = 0$ since $e^{rt} > 0$ for all $t$.

This equation is called the characteristic equation for the ODE and determines with values of $r$ yield solutions. In this case, the characteristic equation is a quadratic polynomial whose roots may be real and distinct (covered in this section), complex conjugates (covered in 3.3 ), or real and repeated (covered in 3.4 ). In the case of real and distinct roots, finding a general solution is very easy:

  1. Find the characteristic equation.
  2. Determine the roots of the characteristic equation $r_1$ and $r_2$.
  3. $y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}$.
  4. Apply the initial conditions to the general solution (only if you are solving an initial value problem).

Example 3.1.2

Find the solution to the initial value problem

$$ y'' - y' - 6y = 0,\qquad y(0) = 2,\ y'(0) = -1. $$
This has the characteristic equation

$$ r^2 - r - 6 = (r-3)(r+2) = 0, $$
and so $r_1 = 3$ and $r_2 = -2$. Which implies that the general solution is

$$ y(t) = c_1 e^{3t} + c_2 e^{-2t}. $$

To solve the initial value problem, we use the general solution for $y$ and its derivative

$$ y'(t) = 3c_1 e^{3t} - 2c_2 e^{-2t}, $$
and plug in the initial values $y(0) = 2$ and $y'(0) = -1$. This again yields a system of equations

$$ \begin{align*} c_1 + c_2 &= 2 \\ 3c_1 - 2c_2 &= -1, \end{align*} $$
which we solve to find $c_1 = \frac{3}{5}$ and $c_2 = \frac{7}{5}$ (see below).

The solution

$$ y(t) = \dfrac{3}{5} e^{3t} + \dfrac{7}{5} e^{-2t} $$
has the following graph Solution to initial value problem for y''-y'-6y=0, y(0)=2, y'(0) = -1

3.1.4 Solving the Initial Value Problem: Cramer's Rule

For the previous two example, applying the initial conditions required that we solve a system of two equations and two unknowns for the constants $c_1$ and $c_2$. There are many ways to do this, including the elimination method mentioned above. That method will always work, but here we will cover another method called Cramer's rule that also allows us to solve this $2\times 2$ system of equations. To begin, we take our previous example

$$ \begin{aligned} c_1 + c_2 &= 2 \\ 3c_1 - 2c_2 &= -1 \end{aligned} $$
and rewrite it as a matrix equation

$$ \begin{bmatrix} 1 & 1 \\ 3 & -2 \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}.$$
To make the description of Cramer's rule easier, let's give the elements of this equation names

$$ \underbrace{\begin{bmatrix} 1 & 1 \\ 3 & -2 \end{bmatrix}}_A \underbrace{\begin{bmatrix} c_1 \\ c_2 \end{bmatrix}}_\mathbf{c} = \underbrace{\begin{bmatrix} 2 \\ -1 \end{bmatrix}}_\mathbf{b}$$
so we have the equation $A\mathbf{c} = \mathbf{b}$.

Cramer's rule allows us to manipulate this matrix equation in a way to yield explicit formulas for the unknown quantities $c_1$ and $c_2$. The procedure involves replacing a column of $A$ with $\mathbf{b}$, and then computing a ratio of determinants. To find the first unknown $c_1$, we replace the first column of $A$ with $\mathbf{b}$ and find the determinant of this matrix divided by the determinant of $A$. The second unknown $c_2$ is found by the same formula, except we replace the second column of $A$ in the numerator.

$$ \begin{align*} c_1 = \dfrac{\begin{vmatrix} \color{#307fe2}{2} & 1 \\ \color{#307fe2}{-1} & -2 \end{vmatrix}}{\begin{vmatrix} 1 & 1 \\ 3 & -2 \end{vmatrix}} &= \dfrac{2(-2) - 1(-1)}{1(-2) - 3(1)} = \frac{3}{5} \\ \\ c_2 = \dfrac{\begin{vmatrix} 1 & \color{#307fe2}{2} \\ 3 & \color{#307fe2}{-1} \end{vmatrix}}{\begin{vmatrix} 1 & 1 \\ 3 & -2 \end{vmatrix}} &= \dfrac{1(-1) - 2(3)}{1(-2) - 3(1)} = \frac{7}{5} \end{align*} $$

The vertical bars surrounding a matrix indicate computing a determinant. The determinant of a $2\times 2$ matrix is given by

$$ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc. $$

This procedure is a quick way to apply initial conditions and is the basis for a technique we will use later in the chapter, so please become familiar with it. Here is a great video from 3Blue1Brown that describes the geometry underlying this technique.

3.1.5 Video Examples

Example 3.1.3

Solve the initial value problem

$$ 4y'' + 8y' + 3y = 0;\qquad y(0)=3,\ y'(0) = -1. $$

Check Your Work
$$ y(t) = \dfrac{7}{2}e^{-\frac{t}{2}} - \dfrac{1}{2}e^{-\frac{3t}{2}} $$
Solution to initial value problem for y''-y'-5y=0, y(0)=2, y'(0) = -1

Video Solution

Example 3.1.4

Solve the initial value problem

$$ 2y'' + y' - 4y = 0;\qquad y(0)=0,\ y'(0) = 1. $$

Check Your Work
$$ y(t) = \dfrac{2}{\sqrt{33}}e^{\frac{t}{4}\left(-1 + \sqrt{33}\right)} - \dfrac{2}{\sqrt{33}}e^{\frac{t}{4}\left(-1 - \sqrt{33}\right)} $$
Solution to initial value problem for 2y'' + y' - 4y = 0; y(0)=0, y'(0) = 1

Video Solution

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