Section 4.1 Homogeneous Linear Equations
ΒΆSubsection 4.1.1 RLC Circuits
ΒΆRecall 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 4.1.1).- I=dQdt.
- The voltage drop across a resistor is IR (Ohm's Law).
- The voltage drop across a capacitor is Q/C.
- The voltage drop across an inductor is L(dI/dt).
- 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).
Subsection 4.1.2 Second-Order Linear Equations
ΒΆSuppose that we have a homogeneous second-order linear differential equation with constant coefficients,Example 4.1.2.
Solutions of a linear system xβ²=Ax often include terms of the form ert. 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 ert, we can substitute this expression into the left-hand side of (4.1.3) to obtain
Since ert is never zero, we find that (r+5)(rβ2)=0 or r=β5 or 2. Thus, we have two solutions
By the Principle of Superposition,
is a solution to xβ³+3xβ²β10x=0.
Indeed, this is the general solution of our second-order equation since we have a one-to-one correspondence between the solutions of
and the system
The matrix associated with this system
has characteristic polynomial Ξ»2+3Ξ»β10. The eigenvalues of A are Ξ»1=β5 and Ξ»2=2 with eigenvectors v1=(1,β5) and v2=(1,2), respectively. Consequently, the solution to our system is
which agrees with (4.1.4).
Example 4.1.3.
Now let us solve the initial value problem
Again, we will assume that our solution has the form x(t)=ert. Substituting this function into our differential equation, we find that
As in Example 4.1.2, r2+4r+5=0; 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 x1(t) and x2(t) are solutions to our differential equation. Indeed, since x(t)=x1(t)+ix2(t) is a solution,
Since the real part and the imaginary part of x(t) must both be zero, we can conclude that axβ³1+bxβ²1+cx1=0 and axβ³2+bxβ²2+cx2=0. Therefore, the general solution to our equation is
To apply our initial conditions x(0)=1 and xβ²(0)=1, we first calculate
Thus,
and c1=1 and c2=3. Hence, the solution to our initial value problem is
Example 4.1.4.
Consider the equation
If we choose eΞ»t as our guess, we find
Thus, Ξ»=β1 and we have a solution x1(t)=eβt.
In order to find a general solution to xβ³+2xβ²+x=0, we must find a second solution that is not a multiple of x1(t)=eβt. Since we already know that cx1(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)x1(t) is a solution to xβ³+2xβ²+x=0. Indeed, if
then
and
Consequently,
and vβ³=0. Therefore, v=c1t+c2. Letting c1=1 and c2=0, we can assume that v(t)=t, and the second solution to our equation is x=teβt. Hence, the general solution to xβ³+2xβ²+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 first-order linear system.
Subsection 4.1.3 Classifying Harmonic Oscillators
ΒΆRecall from Subsection 1.1.3 that we can model harmonic motion using the equationExample 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)=xβ²(t) is the velocity of the oscillator (Figure 4.1.6). Examining the phase plane of the undamped oscillator, we find that the period of the oscillations is given by 2Ο/Ο=2Ο/3β2.094 (Figure 4.1.7).


If the damping value of b is small when compared to 4mk, then b2β4mk<0 and the roots of (4.1.8) will be complex. Furthermore, the real part of each root, βb/2m, is always negative. In such a situation, we say that the oscillator is under-damped.
If the damping value of b is large , then b2β4mk>0, and we obtain distinct real negative roots for (4.1.8). The oscillator is over-damped.
Finally, we say that the oscillator is critically-damped if b2β4mk=0.
Example 4.1.8. 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. The characteristic equation of the differential equation is Ξ»2+0.4Ξ»+1.04=0, which has roots Ξ»=β0.2Β±i. 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, 2Ο/Ο=2Οβ6.283, does not change; however, the amplitude slowly decreases (Figure 4.1.9 and Figure 4.1.10).


Example 4.1.11. An Over-Damped Oscillator.
We can expect a different type of behavior in the case of an over-damped 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 (Figure 4.1.12 and Figure 4.1.13).


Example 4.1.14. A Critically-Damped Oscillator.
As we increase the damping, the oscillations will cease to occur for some value of b. This will happen when b2β4mk=0. 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 (Figure 4.1.15 and Figure 4.1.16), oscillations will commence as soon as we start to reduce the damping constant b=4.


Subsection 4.1.4 Important Lessons
ΒΆ- A second-order linear differential equation with constant coefficients is an equation of the formaxβ³+bxβ²+cx=0.We can guess the solution to this equation. Since we can rewrite this equation as a system of first-order linear differential equations, we can determine the general solution to axβ³+bxβ²+cx=0.
- Suppose thataxβ³+bxβ²+cx=0,where aβ 0 and b2β4ac>0. If the roots of ar2+br+c are r1 and r2, the general solution to this differential equation isx(t)=c1er1t+c2er2t.
- If b2β4ac<0, the differential equationaxβ³+bxβ²+cx=0has a general solutionx(t)=c1eΞ±tcosΞ²t+c2eΞ±tsinΞ²t,where Ξ±Β±iΞ² are the roots of ar2+br+c=0.
- If b2β4ac=0, the differential equationaxβ³+bxβ²+cx=0has a general solutionx(t)=c1eβbt/2a+c2teβbt/2a.
-
A simple harmonic oscillator can be modeled by the equation
md2xdt2+bdxdt+kx=0,where m>0, k>0, and bβ₯0. There are three possible types of types of motion for the oscillator depending on the sign of b2β4mk.
If b2β4mk<0, the oscillator is under-damped.
If b2β4mk>0, the oscillator is over-damped.
If b2β4mk=0, the oscillator is critically damped.
Reading Questions 4.1.5 Reading Questions
ΒΆ1.
What is the characteristic equation of axβ³+bxβ²+cx=0?
2.
Describe the possible types of damping of a harmonic oscillator?
Exercises 4.1.6 Exercises
ΒΆFinding General Solutions
Solving Initial Value Problems
11.
d2ydx2βy=0, y(0)=1, yβ²(0)=0
12.
xβ³β2xβ²β8x=0, x(0)=1, xβ²(0)=2
13.
yβ³+5y=0, y(0)=1, yβ²(0)=1
14.
d2xdt2+6dxdt+5x=0, x(0)=2, xβ²(0)=β1
15.
xβ³β10xβ²+25x=0, x(0)=1, xβ²(0)=0
16.
d2ydx2β2dydx+4y=0, x(0)=β1, xβ²(0)=1
17.
yβ³β8yβ²+4y=0, y(0)=1, yβ²(0)=β2
18.
d2xdx2+3dxdtβ10x=0, x(0)=1, xβ²(0)=2
19.
d2Qdt2β4dQdt+9Q=0, Q(0)=β1, Qβ²(0)=2
20.
d2ydt2+6dydt+9y=0, y(0)=0, yβ²(0)=0
Harmonic Oscillators
Write the second-order initial-value problem corresponding for the harmonic oscillator.
Classify the oscillator as undamped, under-damped, over-damped, 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, b=1 k=1, x(0)=1, v(0)=0
22.
m=1, b=2 k=3, x(0)=β3, v(0)=4
23.
m=1, b=5 k=3, x(0)=2, v(0)=β3
24.
m=1, b=0 k=25, x(0)=2, v(0)=0
25.
m=2, b=3 k=5, x(0)=2, v(0)=β1
26.
m=4, 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
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 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.
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.\label{equation-exercise-secondorder01-reduction-of-order-1}\tag{4.1.9} \end{equation}
- Let u = v' and show that (4.1.9) is a first-order 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 \label{equation-exercise-secondorder01-reduction-of-order-2}\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*}
Subsection 4.1.7 Solving Second Order Linear Equations with Sage
ΒΆSecond order homogeneous linear differential equations with constant coefficeints can be solved sybolically using Sage. For example,xxxxxxxxxx
t = var('t')
x = function('x')(t)
DE = diff(x,t,2)+3*diff(x,t)+2*x == 0
desolve(DE, [x,t])
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t = var('t')
x = function('x')(t)
DE = diff(x,t,2)+3*diff(x,t)+2*x == 0
desolve(DE, [x,t], ics=[0, 1, 2])
xxxxxxxxxx
t = var('t')
x = function('x')(t)
DE = diff(x,t,2)+3*diff(x,t)+2*x == t*cos(2*t)
desolve(DE, [x,t], ics=[0, 1, 2])
Exercises 4.1.8 Exercises
ΒΆ1.
\dfrac{d^2 y}{dx^2} - y = 0
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β
2.
\dfrac{d^2 y}{dx^2} - y = 0\text{,} y(0) = 1\text{,} y'(0) = 0
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β
3.
\dfrac{d^2 y}{dx^2} - y = \cos 2t
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β
4.
\dfrac{d^2 y}{dx^2} - y = \cos 2t\text{,} y(0) = 1\text{,} y'(0) = 0
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β
5.
\dfrac{d^2 y}{dx^2} - y = \cos t
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β
6.
\dfrac{d^2 y}{dx^2} - y = \cos t\text{,} y(0) = 1\text{,} y'(0) = 0
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β
7.
x'' - 2x' - 8x = 0
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β
8.
x'' - 2x' - 8x = 0\text{,} x(0) = 1\text{,} x'(0) = 2
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β
9.
x'' - 2x' - 8x = e^{-2t}
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β
10.
x'' - 2x' - 8x = e^{-2t} + e^{4t}
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β