is called a second-order linear differential equation. We will first consider the case
\begin{equation*}
a x'' + b x' + cx = 0,
\end{equation*}
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. In this section, we will develop a technique to solve differential equations of this form.
Subsection2.1.1RLC Circuits
Recall the RC circuits that we studied earlier (see Section 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 (Figure 2.1.1).
an electrical circuit containing a voltage source, and resistor, a capacitor, and an inductor
Figure2.1.1.An RLC Circuit
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
\begin{equation}
L \frac{dI}{dt} + RI + \frac{1}{C} Q = E(t)\tag{2.1.1}
\end{equation}
\begin{equation*}
L I'' + RI' + \frac{1}{C} I = E'(t).
\end{equation*}
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
\begin{equation*}
I'' + I' + I = E'(t) = \cos t.
\end{equation*}
This is an example of a second-order linear differential equation.
Subsection2.1.2Solving Second-Order Linear Equations
Suppose that we have a homogeneous second-order linear differential equation with constant coefficients,
\begin{equation}
a x'' + b x' + cx = 0.\tag{2.1.2}
\end{equation}
The goal of this section is to be able to solve all such equations. To build some intuition, we start by considering the homogeneous first-order linear differential equation
Our work in Chapter 1 suggests that this equation has a solution of the form \(x(t) = e^{\lambda t}\) for some constant \(\lambda\text{.}\) We can identify \(\lambda\) by substuting our guess for the solution into (2.1.3) and solving algebraically. Since \(\frac{d}{dt}e^{\lambda t} = \lambda e^{\lambda t}\text{,}\) this substitution yields the equation
The term \(e^{\lambda t}\) is always non-zero, and therefore we can safely divide both sides of the equation above by \(e^{\lambda t}\) to reach
\begin{equation*}
a\lambda + b = 0,
\end{equation*}
which has solution \(\lambda = -\frac{b}{a}\text{.}\) Thus (2.1.3) has \(x(t) = e^{-\frac{b}{a}t}\) as a solution.
A similar strategy works for homogeneous second-order linear differential equations. We “guess” that (2.1.2) has a solution of the form \(x=e^{\lambda t}\) and substitute our guess into the left-hand side of the equation. In detail,
\begin{align*}
a\left(\frac{d^2}{dt^2} e^{\lambda t}\right) + b\left(\frac{d}{dt}e^{\lambda t}\right) + c e^{\lambda t} & = a\lambda^{2} e^{\lambda t} + b \lambda e^{\lambda t} + c e^{\lambda t}\\
& = (a\lambda^{2} + b\lambda + c) e^{\lambda t}
\end{align*}
Dividing out the non-zero term \(e^{\lambda t}\) as before, we see that \(\lambda\) must satisfy
\begin{equation}
a\lambda^{2} + b\lambda + c = 0.\tag{2.1.4}
\end{equation}
We call (2.1.4) the characteristic equation for \(ax'' + bx' + cx = 0\text{.}\) Its roots are
has characteristic equation \(\lambda^{2} + 3\lambda - 10 = 0\text{.}\) Rather than using the quadratic formula, we can simply factor the left-hand side as \((\lambda + 5)(\lambda-2)\) to deduce that the roots are \(\lambda_{1} =-5\) and \(\lambda_{2} = 2\text{.}\) The roots correspond to a pair of solutions to (2.1.5):
The characteristic equation (2.1.4) also has distinct roots when \(b^{2} - 4ac \lt 0\text{,}\) although now the roots are complex numbers. Even so, we can make sense of the corresponding solutions using tools from Section 1.8. In this case, it is customary to write \(\lambda = \alpha \pm i\beta\text{,}\) where
Using the facts that \(\cos(-\beta t)=\cos(\beta t)\) and \(\sin(-\beta t) = -\sin(\beta t)\text{,}\) a similar computation on \(x_{2}(t)\) results in the expression
To get real solutions, we can take the real part of \(x_{1}(t)\) and (negative) the imaginary part of \(x_{2}(t)\text{.}\) Going forward, we will use the resulting real functions as the canonical solutions to (2.1.2) when \(b^{2} - 4ac \lt 0\text{,}\) re-defining \(x_{1}(t)\) and \(x_{2}(t)\) as
Note that complex roots of the characteristic equation induce oscillatory behavior in the solutions of the differential equation due to the presence of the \(\sin\) and \(\cos\) terms.
From these complex solutions, using the techniques of Subsection 1.8.4, from these solutions, we can get the following equivalent set of real solutions:
There is one more case to consider: when \(b^{2}-4ac=0\text{.}\) In this case, the charactersitic equation has a repeated real root at \(\lambda = -\frac{b}{2a}\text{,}\) which appears to give only one solution \(x_{1}(t) = e^{-\frac{b}{2a}t}.\) However, as demonstrated in the next example, it is possible to generate a second solution using \(x_{1}(t)\text{.}\)
Example2.1.4.
Consider the equation
\begin{equation*}
x'' + 2x' + x = 0.
\end{equation*}
Thus, there is a repeated real root \(\lambda = -1\) and we have a solution \(x_1(t) = e^{-t}\text{.}\)
We seek a second solution that is not a multiple of \(x_1(t) = e^{-t}\text{.}\) We will “guess” that a solution to \(x'' + 2x' + x = 0\) of the form \(v(t) x_1(t)\) exists for some unknown function \(v(t)\) and attempt to solve for \(v(t)\text{.}\) Indeed, if
and \(v'' = 0\text{.}\) Therefore, \(v = c_1 t + c_2\text{.}\) Letting \(c_1 = 1\) and \(c_2 = 0\) gives \(v(t) = t\text{,}\) and a second solution to our equation is given by
\begin{equation*}
x(t) = t e^{-t}\text{.}
\end{equation*}
Note that we could just as well have chosen, say, \(c_{1} = 0\) and \(c_{2}=1\text{,}\) or \(c_{1} = 1\) and \(c_{2}=1\text{.}\) The former, however, results in \(x(t)=e^{-t}\text{,}\) which is the solution we already knew. The latter choice gives \(x(t) = te^{-t} + e^{-t}\text{.}\) While this solution is distinct from the one we started with, it still contains information from the original solution in the second term. The assignment \(c_{1}=1\text{,}\)\(c_{2}=0\) is the simplest choice of constants which gives another solution that is, in a sense, independent from the first.
The technique that we have used in Example 2.1.4 is called reduction of order. In general, given a second-order linear differential equation
\begin{equation*}
a x'' + bx' + cx = 0
\end{equation*}
such that \(b^2 - 4ac = 0\text{,}\) we have a pair of solutions
using the roots of the characteristic equation \(a\lambda^{2} + b \lambda + c = 0.\) In this section, we will show that these solutions can in fact be used to describe all solutions to the differential equation.
A key idea in this direction is the principle of superposition, which states that scalar multiples and sums of solutions to (2.1.2) are again solutions to the same equation.
This fact can be established using basic properties of the derivative. For instance, suppose that \(x(t)\) is a solution to \(ax'' + bx' + cx = 0\text{,}\) and let \(A\) be an arbitrary real constant. Recalling that for any differentiable function \(f\text{,}\) we have \(\frac{d}{dt}Af(t)=A\frac{d}{dt}f(t)\text{,}\) we may compute
where in the second-to-last line we use the fact that \(x(t)\) is a solution to (2.1.2), that is, \(ax'' + bx' + cx = 0\text{.}\) So, \(Ax(t)\) is also a solution.
Similarly, since \(\frac{d}{dt}(f(t) + g(t)) = \frac{d}{dt}f(t) + \frac{d}{dt}g(t)\) for any differentiable \(f\) and \(g\text{,}\) if \(x_{1}(t)\) and \(x_{2}(t)\) are solutions to (2.1.2), then
So \(x_{1}(t) + x_{2}(t)\) is also a solution to (2.1.2). In conjunction, the preceding computations yield the following theorem.
Theorem2.1.5.Principle of Superposition.
If \(x_{1}(t)\) and \(x_{2}(t)\) are solutions to the differential equation \(ax'' + bx' + cx = 0\text{,}\) then for any constants \(c_{1}\) and \(c_{2}\text{,}\)\(c_{1}x_{1}(t) + c_{2}x_{2}(t)\) is also a solution.
Before continuing, it is worth noting that Theorem 2.1.5 is still true if the constant coefficients \(a\text{,}\)\(b\text{,}\) and \(c\) are replaced with variable coefficients \(a(t)\text{,}\)\(b(t)\text{,}\) and \(c(t)\) which depend on \(t\text{.}\) For our present purposes, however, we continue to focus solely on the constant coefficient case.
The expression \(c_{1}x_{1}(t) + c_{2}x_{2}(t)\) is called a linear combination of the functions \(x_{1}\) and \(x_{2}\text{.}\) Provided that \(x_{1}\) and \(x_{2}\) solve (2.1.2) and \(x_{2}\) is not simply a constant multiple of \(x_{1}\text{,}\) we can think of the linear combination \(c_{1}x_{1}(t) + c_{2}x_{2}(t)\) as a two-parameter family of solutions to (2.1.2).
Does such a family encompass all solutions? To make this question more tractable, we first state an existence and uniqueness result for second-order linear equations which is analogous to Theorem 1.6.1.
with initial conditions \(x(t_{0}) = x_{0}\) and \(x'(t_{0}) = v_{0}\text{.}\) If \(g(t)\) is continuous on some interval \((t_{1}, t_{2})\) containing \(t_{0}\text{,}\) then there exists a unique solution \(x(t)\) that is continuous on \((t_{1}, t_{2})\text{.}\)
In light of Theorem 2.1.6, to show that a two-parameter family of solutions \(c_{1}x_{1}(t) + c_{2}x_{2}(t)\) contains all solutions to the differential equation, it is enough to demonstrate that the constants \(c_{1}\) and \(c_{2}\) can be chosen to satisfy arbitrary initial conditions. That is, for any \(t_{0}\text{,}\)\(x_{0}\text{,}\) and \(v_{0}\text{,}\) there exists an assignment of \(c_{1}\) and \(c_{2}\) satisfying the system of equations
both solve the equation \(x'' + 3 x' - 10 x = 0\text{.}\) Taking an arbitrary linear combination of \(x_{1}\) and \(x_{2}\) then gives a two-parameter family of solutions
(There is an easy way to verify that \(x_{1}\) and \(x_{2}\) are not constant multiples of one another: just confirm that their ratio \(\frac{e^{-5t}}{e^{2t}} = e^{-7t}\) is not constant!)
Now, let \(t_{0}\text{,}\)\(x_{0}\text{,}\) and \(v_{0}\) be arbitrary. Theorem 2.1.6 gives that there is a unique solution \(x(t)\) to the differential equation satisfying \(x(t_{0}) = x_{0}\) and \(x'(t_{0}) = v_{0}\text{.}\) If we are able to choose \(c_{1}\) and \(c_{2}\) so that the system of equations
is the general solution to \(x'' + 3 x' - 10 x = 0\text{.}\) Here, we established this by explicitly solving a system of equations for \(c_{1}\) and \(c_{2}\text{.}\) Going forward, we will develop some theory which will allow us to reach the same conclusion with less explicit computation.
As we saw in Example 2.1.7, one can demonstrate that \(x(t) = c_{1}x_{1}(t) + c_{2}x_{2}(t)\) is a general solution by explicitly solving the system (2.1.6)-(2.1.7).
Solving the system (2.1.6)-(2.1.7) symbolically yields the following expressions for \(c_{1}\) and \(c_{2}\text{:}\)
This represents a valid (and unique) solution to the system if and only if the denominator which appears in the expressions for both \(c_{1}\) and \(c_{2}\) is not equal to \(0\text{.}\) We call this quantity the Wronskian of \(x_{1}\) and \(x_{2}\) and denote it by
Notice that \(W(x_{1},x_{2})\) is a function of \(t\text{.}\) Provided that \(W(x_{1},x_{2}) \neq 0\) for all values of \(t\text{,}\) by our computations above we may solve the system (2.1.6)-(2.1.7) for any values of \(t_{0}\text{,}\)\(x_{0}\text{,}\) and \(v_{0}\text{.}\) Thus we have the following criterion.
Theorem2.1.8.
If \(x_{1}(t)\) and \(x_{2}(t)\) are solutions to the differential equation \(ax'' + bx' + cx = 0\) such that for all values of \(t\text{,}\)\(W(x_{1},x_{2}) \neq 0\text{,}\) then
is a general solution to \(ax'' + bx' + cx = 0\text{.}\)
In the example below, we revisit the differential equation from Example 2.1.7 in light of Theorem 2.1.8.
Example2.1.9.
Consider again the equation \(x'' + 3 x' - 10 x = 0\) from Example 2.1.7. We have a pair of solutions \(x_{1}(t)=e^{-5t}\) and \(x_{2}(t)=e^{2t}\text{.}\) We compute their Wronskian as
is a general solution to \(x'' + 3 x' - 10 x = 0,\) confirming our observations from Example 2.1.7.
We can now give an explicit general solution to the equation \(ax'' + bx' + cx = 0\text{.}\) This task combines our efforts from Subsection 2.1.2, where we constructed pairs of solutions using the characteristic equation, with the theory developed in this section.
Theorem2.1.10.General Solution to the Homogeneous Second-Order Equation.
Consider the equation \(ax'' + bx' + cx = 0\text{.}\)
If \(b^{2}-4ac \gt 0\text{,}\) a general solution is given by
\begin{equation*}
\alpha \pm \beta i
\end{equation*}
are the complex conjugate roots of the characteristic equataion.
Proof.
We prove the theorem only in the case \(b^{2} - 4ac \lt 0\) in detail, since this one is the most computationally taxing. The other two cases can be proved in the same way.
Suppose that \(b^{2} - 4ac \lt 0\text{,}\) and let \(\alpha\) and \(\beta\) be defined as in the theorem statement. We know from Subsection 2.1.2 that
are solutions to \(ax'' + bx' + cx = 0\text{,}\) so to show that \(x(t) = c_{1}x_{1}(t) + c_{2}x_{2}(t)\) is the general solution, by Theorem 2.1.8 it suffices to check that their Wronskian \(W(x_{1},x_{2})\) is non-zero for all \(t\text{.}\) We first do some preliminary computations: using the product rule,
Since \(e^{2\alpha t} \) is non-zero for all values of \(t\) and \(\beta \neq 0\text{,}\) we conclude that \(W(x_{1},x_{2})\) is non-zero for all values of \(t\text{.}\) Thus \(x(t) = c_{1}x_{1}(t) + c_{2}x_{2}(t)\) is the general solution.
With general solutions in hand, we can solve an initial value problem involving \(ax'' + bx' + cx = 0\) with initial conditions of the form \(x(0) = x_{0}\text{,}\)\(x'(0) = v_{0}\text{.}\) We demonstrate in the next example.
and so \(x(0) = 10\) imposes \(c_{1} = 10\text{.}\) To solve for \(c_{2}\text{,}\) we must differentiate \(x(t)\) and use \(x'(0) = -2\text{.}\) With some computation, we find
Recall from Subsection 1.1.3 that we can model harmonic motion using the equation
\begin{equation*}
m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + k x = 0.
\end{equation*}
In the case of a spring-mass 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,
\begin{equation}
m \frac{d^2 x}{dt^2} + k x = 0.\tag{2.1.8}
\end{equation}
The characteristic equation is
\begin{equation*}
m \lambda^2 + k = 0,
\end{equation*}
and we roots at \(\lambda = \pm i \sqrt{k/m}\text{.}\) Writing \(\omega = \sqrt{k/m}\) for simplicity, Theorem 2.1.10 gives the general solution
where \(v(t) = x'(t)\) is the velocity of the oscillator (Figure 2.1.13). Examining the phase plane of the undamped oscillator, we find that the period of the oscillations is given by \(2 \pi / \omega = 2 \pi / 3 \approx 2.094\) (Figure 2.1.14).
position and velocity of an undamped harmonic oscillator plotted agains time
Figure2.1.13.Position and velocity of an undamped harmonic oscillator
the velocity of a harmonic oscillator plotted against the position forming an ellipse in the position-velocity plane
Figure2.1.14.The phase plane of an undamped harmonic oscillator
If we add damping to the oscillator, the equation becomes
\begin{equation}
m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + k x = 0.\tag{2.1.9}
\end{equation}
where \(b \gt 0\text{.}\) The charactersitic equation of (2.1.9) is
\begin{equation*}
m\lambda^2 + b\lambda + k = 0\text{,}
\end{equation*}
There are three possible types of types of motion for the oscillator depending on the nature of the roots of (2.1.10).
If the damping value of \(b\) is small when compared to \(4mk\text{,}\) then \(b^2 - 4mk \lt 0\) and the roots of (2.1.10) 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 under-damped.
If the damping value of \(b\) is large , then \(b^2 - 4mk \gt 0\text{,}\) and we obtain distinct real negative roots for (2.1.10). The oscillator is over-damped.
Finally, we say that the oscillator is critically-damped if \(b^2 - 4mk = 0\text{.}\)
Example2.1.15.An Under-Damped 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 general solution is
\begin{equation*}
x(t) = c_1 e^{-0.2t} \cos t + c_2 e^{-0.2t} \sin t.
\end{equation*}
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 (Figure 2.1.16 and Figure 2.1.17).
velocity and position plotted against time with oscillations becoming smaller as time increases
Figure2.1.16.Position and velocity of an under-damped harmonic oscillator
the velocity of a harmonic oscillator plotted against the position spiraling towards the origin on the position-velocity plane
Figure2.1.17.The phase plane of an under-damped harmonic oscillator
Example2.1.18.An Over-Damped Oscillator.
We can expect a different type of behavior in the case of an over-damped oscillator. For example,
velocity and position plotted against time with the two curves approaching zero for large times
Figure2.1.19.Position and velocity of an over-damped harmonic oscillator
the velocity of a harmonic oscillator plotted against the position with the solution curve approaching the origin on the position-velocity plane
Figure2.1.20.The phase plane of an over-damped harmonic oscillator
Example2.1.21.A Critically-Damped 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
Although we see that no oscillations for this oscillator (Figure 2.1.22 and Figure 2.1.23), oscillations will commence as soon as we start to reduce the damping constant \(b = 4\text{.}\)
velocity and position plotted against time with the two curves approaching zero for large times
Figure2.1.22.Position and velocity of a critically damped harmonic oscillator
the phase plane of a critically damped harmonic oscillator with the solution curve approaching the origin
Figure2.1.23.The phase plane of a critically damped harmonic oscillator
Subsection2.1.5Important Lessons
A second-order linear differential equation with constant coefficients is an equation of the form
\begin{equation*}
a x'' + bx' + cx = 0.
\end{equation*}
The characteristic equation of \(a x'' + bx' + cx = 0\) is
\begin{equation*}
a\lambda^{2} + b\lambda + c = 0.
\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 under-damped.
If \(b^2 - 4mk \gt 0\text{,}\) the oscillator is over-damped.
If \(b^2 - 4mk = 0\text{,}\) the oscillator is critically damped.
Reading Questions2.1.6Reading Questions
1.
What is the characteristic equation of \(ax'' + bx' + cx = 0\text{?}\)
2.
Describe the possible types of damping of a harmonic oscillator.
In Exercise Group 2.1.7.29–31, we will consider the motion of a mass \(m\) hanging at the end of a vertical spring as in [cross-reference to target(s) "secondorder01-figure-hanging-mass" missing or not unique]. 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 spring—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. 1
When the mass is hanging in equilibrium the force of gravity and the force of the spring balance each other out; that is,
\begin{equation*}
mg - kL = 0.
\end{equation*}
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
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 spring-mass 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 intial-value 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.
Hint.
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 \(a x'' + b x' + cx = 0\text{.}\)
Hint.
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
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 \(a x'' + b x' + cx = 0\text{.}\)
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.
