Section 4.1 Homogeneous Linear Equations
Objectives

To understand that a secondorder linear differential equation with constant coefficients is an equation of the form
\begin{equation*} a x'' + bx' + cx = 0. \end{equation*}and can be solved by examining the roots of the characteristic polynomial \(ar^2 + br + c = 0\text{.}\)

To understand that simple harmonic oscillator can be modeled by the equation
\begin{equation*} m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + k x = 0, \end{equation*}where \(m \gt 0\text{,}\) \(k \gt 0\text{,}\) and \(b \geq 0\text{.}\)
A differential equation of the form
is called a secondorder linear differential equation. We will first consider the case
where \(a\text{,}\) \(b\text{,}\) and \(c\) are constants and \(a \neq 0\text{.}\) An equation of this form is said to be homogeneous with constant coefficients. We already know how to solve such equations since we can rewrite them as a system of firstorder linear equations. Thus, we can find the general solution of a homogeneous secondorder linear differential equation with constant coefficients by computing the eigenvalues and eigenvectors of the matrix of the corresponding system.
Subsection 4.1.1 RLC Circuits
Recall the RC circuits that we studied earlier (see SectionÂ 1.3). Such circuits contained a voltage source, a capacitor, and a resistor. A battery or generator is an example of a voltage source, and a toaster or an electric stove is an example of something that might provide a resistance in a circuit. Capacitors store an electrical charge and are used in electronic flashes for cameras. We will now add an inductor such as a solenoid, a coil that generates a magnetic field. Inductor applications include transformers, power supplies, televisions, and radios. Our new circuit is called an RLC circuit (FigureÂ 4.1.1).
Current, \(I(t)\text{,}\) is the rate at which a charge flows through this circuit and is measured in amperes or amps. We assign a direction to the current, and a current flowing in the opposite direction will be given negative values. The impressed voltage, \(E(t)\text{,}\) is measured in volts, the resistance \(R\) is measured in ohms, and the capacitance \(C\) is measured in farads. The charge on the capacitor \(Q(t)\) at time \(t\) is measured in coulombs. Inductance on the coil, \(L\text{,}\) is measured in henrys.
The following laws from physics govern how our circuit behaves.
\(I = \dfrac{dQ}{dt}\text{.}\)
The voltage drop across a resistor is \(IR\) (Ohm's Law).
The voltage drop across a capacitor is \(Q/C\text{.}\)
The voltage drop across an inductor is \(L (dI/dt)\text{.}\)
In a closed circuit the impressed voltage is equal to the sum of the voltage drops in the rest of the circuit (Kirchhoff's Second Law).
Applying Kirchhoff's Second Law to our circuit, we have the differential equation
or
Differentiating both sides of (4.1.1), we have
For example, we might consider an RLC circuit with \(R = 1\text{,}\) \(L = 1\text{,}\) and \(C = 1\text{.}\) At \(t = 0\) when both \(I(0) =0\) and \(I'(0) = Q(0) = 0\text{,}\) the impressed voltage on the circuit is given by \(E(t) = \sin(t)\text{.}\) Our equation becomes
This is an example of a secondorder linear differential equation.
Subsection 4.1.2 SecondOrder Linear Equations
Suppose that we have a homogeneous secondorder linear differential equation with constant coefficients,
The goal of this section is to be able to solve all such equations. However, we did a great deal of work finding unique solutions to systems of firstorder linear systems equations in ChapterÂ 3. Our efforts are now rewarded. Since each secondorder homogeneous system with constant coefficients can be rewritten as a firstorder linear system, we are guaranteed the existence and uniqueness of solutions. Indeed, we can rewrite (4.1.2) as a system of firstorder linear equations,
and then find the general solution by computing the eigenvalues and eigenvectors of the matrix of the corresponding system.
Example 4.1.2.
Solutions of a linear system \({\mathbf x}' = A {\mathbf x}\) often include terms of the form \(e^{r t}\text{.}\) It makes sense that solutions to equation (4.1.2) take the same form. Consider the equation
If we assume that a solution is of the form \(e^{rt}\text{,}\) we can substitute this expression into the lefthand side of (4.1.3) to obtain
Since \(e^{rt}\) is never zero, we find that \((r + 5)(r  2) = 0\) or \(r = 5\) or \(2\text{.}\) Thus, we have two solutions
By the Principle of Superposition,
is a solution to \(x'' + 3x'  10 x = 0\text{.}\)
Indeed, this is the general solution of our secondorder equation since we have a onetoone correspondence between the solutions of
and the system
The matrix associated with this system
has characteristic polynomial \(\lambda^2 + 3 \lambda  10\text{.}\) The eigenvalues of \(A\) are \(\lambda_1 = 5\) and \(\lambda_2 = 2\) with eigenvectors \(\mathbf v_1 = (1, 5)\) and \(\mathbf v_2 = (1, 2)\text{,}\) respectively. Consequently, the solution to our system is
which agrees withÂ (4.1.4).
In general, suppose that
where \(a \neq 0\text{.}\) Applying the strategy in ExampleÂ 4.1.2, we can find the general solution for this equation by finding the roots of the quadratic polynomial \(a \lambda^2 + b \lambda + c\text{,}\)
If \(b^2  4ac \gt 0\text{,}\) we have real roots
and the solution to our secondorder differential equation is
where \(c_1\) and \(c_2\) are arbitrary constants.
To prove that equationÂ (4.1.5) is indeed the general solution to the secondorder equation \(a x'' + b x' + c x = 0\text{,}\) we can study the equivalent system of linear equations. If we let \(y = x'\text{,}\) the corresponding linear system is \(\mathbf x' = A \mathbf x\text{,}\) where
The characteristic polynomial of \(A\) is
The roots of \(p(\lambda)\) are the same as the roots of \(a\lambda^2 + b\lambda + c\text{.}\)
If \(b^2  4ac \gt 0\text{,}\) we have real roots
We can find eigenvectors
for \(\lambda_1\) and \(\lambda_2\text{,}\) respectively. Thus, the general solution to the system of differential equations \(\mathbf x' = A \mathbf x\) is
which agrees withÂ (4.1.5).
Example 4.1.3.
Now let us solve the initial value problem
Again, we will assume that our solution has the form \(x(t) = e^{rt}\text{.}\) Substituting this function into our differential equation, we find that
As in ExampleÂ 4.1.2, \(r^2 + 4r + 5 = 0\text{;}\) however, the roots of this polynomial are complex,
Using Euler's formula, we can find a complex solution
The real and imaginary parts of our solution are
respectively. We claim that both \(x_1(t)\) and \(x_2(t)\) are solutions to our differential equation. Indeed, since \(x(t) = x_1(t) + i x_2(t)\) is a solution,
Since the real part and the imaginary part of \(x(t)\) must both be zero, we can conclude that \(ax_1'' + bx_1' + cx_1 = 0\) and \(ax_2'' + bx_2' + cx_2 = 0\text{.}\) Therefore, the general solution to our equation is
To apply our initial conditions \(x(0) = 1\) and \(x'(0) = 1\text{,}\) we first calculate
Thus,
and \(c_1 =1\) and \(c_2= 3\text{.}\) Hence, the solution to our initial value problem is
As before, the corresponding linear sytem is \(\mathbf x' = A \mathbf x\text{,}\) where
If \(b^2  4ac \lt 0\text{,}\) the eigenvalue of \(A\) are \(\lambda = \alpha + i \beta\) and \(\overline{\lambda} = \alpha  i \beta\text{,}\) where
The vector \(\mathbf v = (1, \alpha + i \beta)\) is an eigenvector for \(\lambda\text{.}\) Thus, a solution to our system of differential equations is
Taking the real and imaginary parts of \(x(t)\text{,}\) we obtain two real solutions to the system, \(x_1(t) = e^{\alpha t} \cos \beta t\) and \(x_2(t) = e^{\alpha t} \sin \beta t\text{.}\) Therefore, the general solution to \(ax_1'' + bx_1' + cx_1 = 0\) is
Given a secondorder linear differential equation with constant coefficients, \(ax'' + bx' + cx = 0\text{,}\) our strategy has been to solve the characteristic equation \(a \lambda^2 + b \lambda + c = 0\) to obtain two linearly independent solutions. We have covered the case where this equation has two distinct real solutions as well as when there are complex solutions, but what if there is only a single real solution \(\lambda = b/2a\text{?}\)
Example 4.1.4.
Consider the equation
If we choose \(e^{\lambda t}\) as our guess, we find
Thus, \(\lambda = 1\) and we have a solution \(x_1(t) = e^{t}\text{.}\)
In order to find a general solution to \(x'' + 2x' + x = 0\text{,}\) we must find a second solution that is not a multiple of \(x_1(t) = e^{t}\text{.}\) Since we already know that \(c x_1(t)\) is a solution to our differential equation, we will try to generalize this observation by replacing \(c\) with a nonconstant function \(v(t)\) and then try to determine \(v(t)\) so that \(v(t) x_1(t)\) is a solution to \(x'' + 2x' + x = 0\text{.}\) Indeed, if
then
and
Consequently,
and \(v'' = 0\text{.}\) Therefore, \(v = c_1 t + c_2\text{.}\) Letting \(c_1 = 1\) and \(c_2 = 0\text{,}\) we can assume that \(v(t) = t\text{,}\) and the second solution to our equation is \(x = t e^{t}\text{.}\) Hence, the general solution to \(x'' + 2x' + x = 0\) is
We leave it as an exercise to show that our solution agrees with the solution that we would obtain from solving the equivalent firstorder linear system.
The technique that we have used in ExampleÂ 4.1.4 is called reduction of order. We leave it as an exercise to show that this technique works in general. That is, given a secondorder linear differential equation
such that \(b^2  4ac = 0\text{,}\) then the general solution is given by
where \(\lambda = b/2a\text{.}\)
Activity 4.1.1. Solving SecondOrder Homogeneous Linear Differential Equations.
Solve each of the following initial value problems.
(a)
(b)
(c)
(d)
(e)
Subsection 4.1.3 Classifying Harmonic Oscillators
Recall from SubsectionÂ 1.1.3 that we can model harmonic motion using the equation
In the case of a springmass system, \(m\) is the oscillating mass, \(b\) is the damping coefficient, and \(k\) is the spring constant. It is important to remember that both \(m\) and \(k\) are positive constants and \(b \geq 0\text{.}\)
If \(b = 0\text{,}\) then the oscillator is undamped In this case,
The characteristic equation is
and we have eigenvalues \(\lambda = \pm i \sqrt{k/m}\text{.}\) Hence, the complex solution to our undamped oscillator is
where \(\omega = \sqrt{k/m}\text{.}\) Since both the real and imaginary parts of the complex solution are also solutions to (4.1.6), the general solution to the undamped harmonic oscillator is
giving us the position of the mass at time \(t\text{.}\) Now, of course, it is easy to determine the velocity of the mass at time \(t\) to be
Example 4.1.5.
Suppose that an undamped harmonic oscillator is modeled by the initial value problem
We can quickly determine the solution of this initial value problem to be
where \(v(t) = x'(t)\) is the velocity of the oscillator (FigureÂ 4.1.6). Examining the phase plane of the undamped oscillator, we find that the period of the oscillations is given by \(2 \pi / \omega = 2 \pi / 3 \approx 2.094\) (FigureÂ 4.1.7).
If we add damping to the oscillator, the equation becomes
where \(b \gt 0\text{.}\) The charactersitic equation of (4.1.7) is
which has roots
There are three possible types of types of motion for the oscillator depending on the nature of the roots of (4.1.8).
If the damping value of \(b\) is small when compared to \(4mk\text{,}\) then \(b^2  4mk \lt 0\) and the roots of (4.1.8) will be complex. Furthermore, the real part of each root, \(b/ 2m\text{,}\) is always negative. In such a situation, we say that the oscillator is underdamped.
If the damping value of \(b\) is large , then \(b^2  4mk \gt 0\text{,}\) and we obtain distinct real negative roots for (4.1.8). The oscillator is overdamped.
Finally, we say that the oscillator is criticallydamped if \(b^2  4mk = 0\text{.}\)
Example 4.1.8. An UnderDamped Oscillator.
Suppose that an oscillator is modeled by the initial value problem
Notice that the damping \(b = 0.4\) is very small compared with the spring constant \(k = 1.04\text{.}\) The characteristic equation of the differential equation is \(\lambda^2 + 0.4 \lambda + 1.04 = 0\text{,}\) which has roots \(\lambda = 0.2 \pm i\text{.}\) Therefore, the complex solution must be
and the general solution must be
Applying the initial conditions, our solution becomes
Notice that the period of the oscillations, \(2 \pi / \omega = 2 \pi \approx 6.283\text{,}\) does not change; however, the amplitude slowly decreases (FigureÂ 4.1.9 and FigureÂ 4.1.10).
Example 4.1.11. An OverDamped Oscillator.
We can expect a different type of behavior in the case of an overdamped oscillator. For example,
The characteristic equation of this initial value problem is
hence, we have the general solution
Applying the initial conditions, our solution is
Notice that the damping is too strong for any oscillations to occur (FigureÂ 4.1.12 and FigureÂ 4.1.13).
Example 4.1.14. A CriticallyDamped Oscillator.
As we increase the damping, the oscillations will cease to occur for some value of \(b\text{.}\) This will happen when \(b^2  4mk = 0\text{.}\) the At this point we have critical damping. Consider the system
The general solution to this initial value problem is
The solution to the initial value problem is
Although we see that no oscillations for this oscillator (FigureÂ 4.1.15 and FigureÂ 4.1.16), oscillations will commence as soon as we start to reduce the damping constant \(b = 4\text{.}\)
Subsection 4.1.4 Important Lessons

A secondorder linear differential equation with constant coefficients is an equation of the form
\begin{equation*} a x'' + bx' + cx = 0. \end{equation*}We can guess the solution to this equation. Since we can rewrite this equation as a system of firstorder linear differential equations, we can determine the general solution to \(a x'' + bx' + cx = 0\text{.}\)

Suppose that
\begin{equation*} a x'' + b x' + c x = 0, \end{equation*}where \(a \neq 0\) and \(b^2  4ac \gt 0\text{.}\) If the roots of \(ar^2 + br + c\) are \(r_1\) and \(r_2\text{,}\) the general solution to this differential equation is
\begin{equation*} x(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}. \end{equation*} 
If \(b^2  4ac \lt 0\text{,}\) the differential equation
\begin{equation*} a x'' + b x' + c x = 0 \end{equation*}has a general solution
\begin{equation*} x(t) = c_1 e^{\alpha t} \cos \beta t + c_2 e^{\alpha t} \sin \beta t, \end{equation*}where \(\alpha \pm i \beta\) are the roots of \(a r^2 + br + c = 0\text{.}\)

If \(b^2  4ac = 0\text{,}\) the differential equation
\begin{equation*} a x'' + b x' + c x = 0 \end{equation*}has a general solution
\begin{equation*} x(t) = c_1 e^{ bt/2a} + c_2 t e^{bt/2a}. \end{equation*} 
A simple harmonic oscillator can be modeled by the equation
\begin{equation*} m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + k x = 0, \end{equation*}where \(m \gt 0\text{,}\) \(k \gt 0\text{,}\) and \(b \geq 0\text{.}\) There are three possible types of motion for the oscillator depending on the sign of \(b^2  4mk\text{.}\)
If \(b^2  4mk \lt 0\text{,}\) the oscillator is underdamped.
If \(b^2  4mk \gt 0\text{,}\) the oscillator is overdamped.
If \(b^2  4mk = 0\text{,}\) the oscillator is critically damped.
Reading Questions 4.1.5 Reading Questions
1.
What is the characteristic equation of \(ax'' + bx' + cx = 0\text{?}\)
2.
Describe the possible types of damping of a harmonic oscillator?
Exercises 4.1.6 Exercises
Finding General Solutions.
Find the general solution for each equation in Exercise GroupÂ 4.1.6.1â€“10.
1.
\(\dfrac{d^2 y}{dx^2}  y = 0\)
2.
\(x''  2x'  8x = 0\)
3.
\(y'' + 5y = 0\)
4.
\(\dfrac{d^2 x}{dt^2} + 6 \dfrac{dx}{dt} + 5x = 0\)
5.
\(x''  10x' + 25x = 0\)
6.
\(\dfrac{d^2 y}{dx^2}  2\dfrac{dy}{dx} + 4y = 0\)
7.
\(y''  8y' + 4y = 0\)
8.
\(\dfrac{d^2 x}{dx^2} + 3\dfrac{dx}{dt}  10x = 0\)
9.
\(\dfrac{d^2 Q}{dt^2}  4 \dfrac{dQ}{dt} + 9Q = 0\)
10.
\(\dfrac{d^2 y}{dt^2} + 6 \dfrac{dy}{dt} + 9y = 0\)
Solving Initial Value Problems.
Solve the initial value problems in Exercise GroupÂ 4.1.6.11â€“20.
11.
\(\dfrac{d^2 y}{dx^2}  y = 0\text{,}\) \(y(0) = 1\text{,}\) \(y'(0) = 0\)
12.
\(x''  2x'  8x = 0\text{,}\) \(x(0) = 1\text{,}\) \(x'(0) = 2\)
13.
\(y'' + 5y = 0\text{,}\) \(y(0) = 1\text{,}\) \(y'(0) = 1\)
14.
\(\dfrac{d^2 x}{dt^2} + 6 \dfrac{dx}{dt} + 5x = 0\text{,}\) \(x(0) = 2\text{,}\) \(x'(0) = 1\)
15.
\(x''  10x' + 25x = 0\text{,}\) \(x(0) = 1\text{,}\) \(x'(0) = 0\)
16.
\(\dfrac{d^2 y}{dx^2}  2\dfrac{dy}{dx} + 4y = 0\text{,}\) \(x(0) = 1\text{,}\) \(x'(0) = 1\)
17.
\(y''  8y' + 4y = 0\text{,}\) \(y(0) = 1\text{,}\) \(y'(0) = 2\)
18.
\(\dfrac{d^2 x}{dx^2} + 3\dfrac{dx}{dt}  10x = 0\text{,}\) \(x(0) = 1\text{,}\) \(x'(0) = 2\)
19.
\(\dfrac{d^2 Q}{dt^2}  4 \dfrac{dQ}{dt} + 9Q = 0\text{,}\) \(Q(0) = 1\text{,}\) \(Q'(0) = 2\)
20.
\(\dfrac{d^2 y}{dt^2} + 6 \dfrac{dy}{dt} + 9y = 0\text{,}\) \(y(0) = 0\text{,}\) \(y'(0) = 0\)
Harmonic Oscillators.
Consider the harmonic oscillators with mass \(m\text{,}\) damping coeeficient \(b\text{,}\) and spring constant \(k\) in Exercise GroupÂ 4.1.6.21â€“28.
Write the secondorder initial value problem corresponding for the harmonic oscillator.
Classify the oscillator as undamped, underdamped, overdamped, or critically damped.
Solve the initial value problem.
Sketch the \(x(t)\) and \(v(t)\)graphs of the solution of the initial value problem.
Sketch the phase portrait of the initial value problem.
21.
\(m = 1\text{,}\) \(b = 1\) \(k = 1\text{,}\) \(x(0) = 1\text{,}\) \(v(0) = 0\)
22.
\(m = 1\text{,}\) \(b = 2\) \(k = 3\text{,}\) \(x(0) = 3\text{,}\) \(v(0) = 4\)
23.
\(m = 1\text{,}\) \(b = 5\) \(k = 3\text{,}\) \(x(0) = 2\text{,}\) \(v(0) = 3\)
24.
\(m = 1\text{,}\) \(b = 0\) \(k = 25\text{,}\) \(x(0) = 2\text{,}\) \(v(0) = 0\)
25.
\(m = 2\text{,}\) \(b = 3\) \(k = 5\text{,}\) \(x(0) = 2\text{,}\) \(v(0) = 1\)
26.
\(m = 4\text{,}\) \(b = 4\) \(k = 1\text{,}\) \(x(0) = 2\text{,}\) \(v(0) = 1\)
27.
\(m = 3\text{,}\) \(b = 4\) \(k = 1\text{,}\) \(x(0) = 2\text{,}\) \(v(0) = 1\)
28.
\(m = 8\text{,}\) \(b = 4\) \(k = 1\text{,}\) \(x(0) = 2\text{,}\) \(v(0) = 1\)
Oscillations of a Hanging Mass.
In Exercise GroupÂ 4.1.6.29â€“31, we will consider the motion of a mass \(m\) hanging at the end of a vertical spring as in FigureÂ 4.1.17. The mass stretches the spring in a downward (positive) direction by length \(L\text{.}\) There are two forces acting on the point where the mass is attached to the springâ€”the force exerted by the spring and gravity. The force of gravity, the weight of the mass, acts downward with a magnitude of \(mg\text{,}\) where \(g\) is the acceleration due to gravity. On the other hand, the force of the spring acts upward and is given by \(kL\text{,}\) where \(L\) is the length of the spring.^{â€‰1â€‰} When the mass is hanging in equilibrium the force of gravity and the force of the spring balance each other out; that is,
We would like to investigate the motion of the mass if is initially displaced or acted on by an external force. Let \(x(t)\) be the displacement of the mass from its equilibrium position, where a downward displacement is positive. The force acting on the mass are the weight of the mass \(mg\) and the force exerted by the spring, which is the total elongation of the spring, or
By Newton's second law of motion,
Furthermore, it is possible to add a damping term, \(bx'\text{,}\) or even an external force \(F(t)\) to obtain
Since \(mg  kL = 0\text{,}\) we obtain the familiar equation
29.
Suppose that a mass of \(100\) grams stretches a spring \(2\) centimeters.
Determine the spring constant \(k\text{.}\)
If the mass is displaced an additional \(4\) centimeters and released, write an initial value problem that will model the motion of the oscillating mass.
Solve the initial value problem.
30.
Suppose that a mass of \(1\) kilogram stretches a spring \(5\) centimeters.
Determine the spring constant \(k\text{.}\)
If the mass is displaced an additional \(5\) centimeters and released, write an initial value problem that will model the motion of the oscillating mass.
Suppose the the springmass system is suspended in a fluid that exerts a resistance of \(0.25\) kilograms when the mass has a velocity of \(2\) centimeters per second. Modify the intialvalue problem that you wrote in (b) to take this fact into account.
Solve the initial value problem.
31.
Suppose that a mass weighing 4 lbs stretches a spring 3 inches.
If \(g = 32 \text{ft/sec}^2\text{,}\) determine \(m\text{.}\)
Determine the spring constant \(k\text{.}\)
If the mass is displaced an additional 6 inches and released, write an initial value problem that will model the motion of the oscillating mass.
Solve the initial value problem.
Pay careful attention to units.
32.
Let \(a x'' + b x' + cx = 0\text{,}\) where \(a \neq 0\) and \(b^2  4ac = 0\text{.}\)
Show that \(x_1(t) = e^{bt/2a}\) is a solution to \(a x'' + b x' + cx = 0\text{.}\)

Assume that
\begin{equation*} y = v(t) x_1(t) = v(t) e^{bt/2a} \end{equation*}is a solution to \(a x'' + b x' + cx = 0\) and show that \(v(t) = c_1 + c_2 t\text{.}\) Thus,
\begin{equation*} x(t) = c_1 e^{bt/2a} + c_2 t e^{bt/2a} \end{equation*}is a general solution for \(a x'' + b x' + cx = 0\text{.}\)

Observe that
\begin{align*} a x_1'' + b_1' + cx_1 & = a \left(\frac{b}{2a}\right)^2e^{bt/2a} + b \left( \frac{b}{2a} \right) e^{bt/2a} + c e^{bt/2a}\\ & = e^{bt/2a} \left( \frac{b^2}{4a}  \frac{b^2}{2a} + c \right)\\ & = e^{bt/2a} \left( \frac{b^2 + 4ac}{4a} \right)\\ & = 0. \end{align*} 
If \(y = v(t) x_1(t) = v(t) e^{bt/2a}\) is a solution to our differential equation, then
\begin{align*} a y'' + b y' + cy & = a (v''x_1 + 2 v'x_1' + vx_1'' ) + b(v' x_1 + v x_1') + cv x_1\\ & = a v''x_1 + 2a v'x_1' + bv' x_1 + v(a x_1'' +b x_1' + c x_1)\\ & = a v'' e^{bt/2a} + \left[2a \left( \frac{b}{2a} \right) e^{bt/2a} + b e^{bt/2a} \right] v'\\ & = a v'' e^{bt/2a}\\ & = 0. \end{align*}Since \(a \neq 0\text{,}\) we know that \(v'' = 0\text{.}\) Hence, \(v(t) = c_1 + c_2 t\text{.}\)
33.
Reduction of Order. Suppose that \(x_1(t)\) is a solution (not identically zero) to the equation

Assume that \(x(t) = v(t) x_1(t)\) is a solution to \(x'' + p(t) x' + q(t) x = 0\) and derive the equation
\begin{equation} x_1 v'' +(2x_1' + px_1)v' = 0.\tag{4.1.9} \end{equation} Let \(u = v'\) and show that (4.1.9) is a firstorder linear differential equation in \(u\text{.}\)

Show that \(x_1(t) = 1/t\) is a solution to
\begin{equation} 2 t^2 x'' + 3t x'  x = 0 \tag{4.1.10} \end{equation}for \(t \gt 0\) and find a second linearly independent solution to (4.1.10).
 \begin{align*} x'' + px' + qx & = (v''x_1 + 2 v' x_1' + vx_1'') + p(v'x_1 + vx_1') + q(vx_1)\\ & = x_1 v'' + 2v' x_1' + p x_1 v' + v (x_1'' + p x_1' + q x_1)\\ & = x_1 v'' +(2x_1' + px_1)v' \\ & = 0. \end{align*}
If \(u = v'\text{,}\) then \(x_1 u' +(2x_1' + px_1)u= 0\text{.}\)

If \(x_1(t) = 1/t\text{,}\) then
\begin{equation*} 2 t^2 x_1'' + 3t x_1'  x_1 = 2 t^2 \left(\frac{2}{t^3}\right) + 3t \left(\frac{1}{t^2}\right)  \frac{1}{t} = 0. \end{equation*}If we assume that \(x = v/t\) is a second solution, then
\begin{equation*} 2 t^2 x'' + 3t x'  x = 2tv''  v' = 0. \end{equation*}If we let \(u = v'\text{,}\) then a solution of \(2tu'  u = 0\) is \(u = \sqrt{t}\) and \(v = \int \sqrt{t} \, dt = 2 t^{3/2} / 3\text{.}\) Therefore, the second solution to our equation is
\begin{equation*} x = \frac{v}{t} = \frac{2}{3} \sqrt{t}. \end{equation*}
34.
Let \(a x'' + b x' + cx = 0\text{,}\) where \(a \neq 0\) and \(b^2  4ac = 0\text{.}\)
Show that \(x_1(t) = e^{bt/2a}\) is a solution to \(a x'' + b x' + cx = 0\text{.}\)

Assume that
\begin{equation*} y = v(t) x_1(t) = v(t) e^{bt/2a} \end{equation*}is a solution to \(a x'' + b x' + cx = 0\) and show that \(v(t) = c_1 + c_2 t\text{.}\) Thus,
\begin{equation*} x(t) = c_1 e^{bt/2a} + c_2 t e^{bt/2a} \end{equation*}is a general solution for \(a x'' + b x' + cx = 0\text{.}\)
35.
Consider the equation
Determine all values of \(\alpha\text{,}\) if any, for which all solutions tend toward zero as \(t \to \infty\text{.}\) Also, determine the values of \(\alpha\text{,}\) if any, for which all nonzero solutions become unbounded as \(t \to \infty\text{.}\)
36.
Solve each of the following initial value problems.
 \begin{align*} y''  2y' + 5y & = 0\\ y(\pi/2) & = 0\\ y'(\pi/2) & = 2 \end{align*}
 \begin{align*} 9y''  12y' + 4y & = 0\\ y(0) & = 2\\ y'(0) & = 1 \end{align*}
 \begin{align*} y'' + 8y' 9y & = 0\\ y(1) & = 1\\ y'(1) & = 0 \end{align*}
 \begin{align*} y'' + 2ay' + (a^2 + 1)y & = 0\\ y(0) & = 1\\ y'(0) & = 0 \end{align*}
37.
Reduction of Order. Suppose that \(x_1(t)\) is a solution (not identically zero) to the equation

Assume that \(x(t) = v(t) x_1(t)\) is a solution to \(x'' + p(t) x' + q(t) x = 0\) and derive the equation
\begin{equation} x_1 v'' +(2x_1' + px_1)v' = 0.\tag{4.1.11} \end{equation} Let \(u = v'\) and show that ((4.1.11)) is a firstorder linear differential equation in \(u\text{.}\)

Show that \(x_1(t) = 1/t\) is a solution to
\begin{equation} 2 t^2 x'' + 3t x'  x = 0\tag{4.1.12} \end{equation}for \(t \gt 0\) and find a second linearly independent solution to ((4.1.12)).
38.
Euler Equations. An important class of secondorder linear differential equations are equations of the form
where \(t \gt 0\) and \(\alpha\) and \(\beta\) are real constants. An equation of this form is called an Euler equation.

Show that the substitution \(x = \ln t\) transforms an Euler equation into an equation of constant coefficients. {\em Hint}: Show that
\begin{align*} \frac{dy}{dt} & = \frac{dx}{dt} \frac{dy}{dx}\\ \frac{d^2y}{dt^2} & = \left( \frac{dx}{dt} \right)^2 \frac{d^2y}{dx^2} + \frac{d^2x}{dt^2} \frac{dy}{dx}. \end{align*} 
Solve the equation
\begin{equation*} t^2 y'' + 4ty' + 2y = 0. \end{equation*}
Subsection 4.1.7 Solving Second Order Linear Equations with Sage
Second order homogeneous linear differential equations with constant coefficeints can be solved sybolically using Sage. For example,
can be solved using the following Sage commands.
We can even solve initial value problems such as
Of course, we can add a forcing term to our initial value problem,
Exercises 4.1.8 Exercises
Solve each of the following differential equations using Sage.
1.
\(\dfrac{d^2 y}{dx^2}  y = 0\)
2.
\(\dfrac{d^2 y}{dx^2}  y = 0\text{,}\) \(y(0) = 1\text{,}\) \(y'(0) = 0\)
3.
\(\dfrac{d^2 y}{dx^2}  y = \cos 2t\)
4.
\(\dfrac{d^2 y}{dx^2}  y = \cos 2t\text{,}\) \(y(0) = 1\text{,}\) \(y'(0) = 0\)
5.
\(\dfrac{d^2 y}{dx^2}  y = \cos t\)
6.
\(\dfrac{d^2 y}{dx^2}  y = \cos t\text{,}\) \(y(0) = 1\text{,}\) \(y'(0) = 0\)
7.
\(x''  2x'  8x = 0\)
8.
\(x''  2x'  8x = 0\text{,}\) \(x(0) = 1\text{,}\) \(x'(0) = 2\)
9.
\(x''  2x'  8x = e^{2t}\)
10.
\(x''  2x'  8x = e^{2t} + e^{4t}\)