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Section4.1Homogeneous Linear Equations

Write an introduction.

Subsection4.1.1

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, radios. Our new circuit is called an RLC circuit (Figure 4.1).

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Figure4.1An 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)\label{equation-secondorder01-RLC}\tag{4.1} \end{equation} or \begin{equation*} LQ'' + RQ' + \frac{1}{C} Q = E(t). \end{equation*} Differentiating both sides of (4.1), we have \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.

Subsection4.1.2Second-Order Linear Equations

An differential equation of the form \begin{equation*} a(t) x'' + b(t) x' + c(t) x = g(t) \end{equation*} is called a second-order linear differential equation. We will first consider the case \begin{equation} a x'' + b x' + cx = 0,\label{equation-secondorder01-constant-coef}\tag{4.2} \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. We already know how to solve such equations since we can rewrite them as a system of first-order linear equations, \begin{align*} x' & = y\\ y' & = -\frac{c}{a} x - \frac{b}{a} y. \end{align*} Thus, we can find the general solution of a homogeneous second-order linear differential equation with constant coefficients by computing the eigenvalues and eigenvectors of the matrix of the corresponding system.

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 (4.2) take the same form.

Example4.2

Consider the equation \begin{equation} x'' + 7 x' + 10 x = 0.\label{equation-secondorder01-guess}\tag{4.3} \end{equation} If we assume that a solution is of the form \(e^{rt}\text{,}\) we can substitute this expression into the left-hand side of (4.3) to obtain \begin{align*} \frac{d^2}{dt^2} e^{rt} + \frac{d}{dt} 7e^{rt} + 10 e^{rt} & = r^2 e^{rt} + 7 r e^{rt} + 10 e^{rt}\\ & = (r^2 + 7r + 10) e^{rt}\\ & = (r + 5)(r+ 2) e^{rt}. \end{align*} 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 \begin{equation*} x_1(t) = e^{-5t} \mbox{ and } x_2(t) = e^{-2t}. \end{equation*} By the Principle of Superposition, \begin{equation} x(t) = c_1 x_1(t) + c_2 x_2(t) = c_1 e^{-5t} + c_2 e^{-2t}\label{equation-secondorder01-guess-soln}\tag{4.4} \end{equation} is a solution to \(x'' + 7x' + 10 x = 0\text{.}\)

Indeed, this is the general solution of our second-order equation since we have a one-to-one correspondence between the solutions of \begin{equation*} x'' + 7x' + 10 x = 0 \end{equation*} and the system \begin{align*} x' & = y\\ y' & = - 10x - 7 y. \end{align*} The matrix associated with this system \begin{equation*} A = \begin{pmatrix} 0 \amp 1 \\ -10 \amp -7 \end{pmatrix} \end{equation*} has eigenvalues \(\lambda_1 = -5\) 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 \begin{equation*} \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = c_1 e^{-5t} \begin{pmatrix} 1 \\ -5 \end{pmatrix} + c_2 e^{-2t} \begin{pmatrix} 1 \\ -2 \end{pmatrix}, \end{equation*} which agrees with (4.4).

In general, suppose that \begin{equation*} a x'' + b x' + c x = 0, \end{equation*} where \(a \neq 0\text{.}\) Applying the startegy in Example 4.2, we can find the general solution for this equation by finding the roots of the quadratic polynomial \(a \lambda^2 + b \lambda + c\text{,}\) \begin{equation*} \lambda = - \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \end{equation*} If \(b^2 - 4ac \gt 0\text{,}\) we have real roots \begin{equation*} \lambda_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad \text{and} \quad \lambda_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}, \end{equation*} and the solution to our second-order differential equation is \begin{equation} x(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t},\label{equation-secondorder01-real-soln}\tag{4.5} \end{equation} where \(c_1\) and \(c_2\) are arbitrary constants.

To prove that equation (4.5) is indeed the general solution to the second-order 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 \begin{equation*} A = \begin{pmatrix} 0 \amp 1 \\ -c/a \amp -b/a \end{pmatrix}. \end{equation*} The characteristic polynomial of \(A\) is \begin{equation*} p(\lambda) = \det(A - \lambda I) = \lambda^2 + \frac{b}{a} \lambda + \frac{c}{a}. \end{equation*} 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 \begin{equation*} \lambda_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad \text{and} \quad \lambda_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}. \end{equation*} We can find eigenvectors \begin{align*} \mathbf v_1 \amp = \begin{pmatrix} 1 \\ (-b + \sqrt{b^2 - 4ac}\,)/2a \end{pmatrix} = \begin{pmatrix} 1 \\ \lambda_1 \end{pmatrix}\\ \mathbf v_1 \amp = \begin{pmatrix} 1 \\ (-b - \sqrt{b^2 - 4ac}\,)/2a \end{pmatrix} = \begin{pmatrix} 1 \\ \lambda_2 \end{pmatrix} \end{align*} for \(\lambda_1\) and \(\lambda_2\text{,}\) respectively. Thus, the general solution to the system of differential equations \(\mathbf x' = A \mathbf x\) is \begin{equation*} \mathbf x(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = \begin{pmatrix} x(t) \\ x'(t) \end{pmatrix} = c_1 e^{\lambda_1 t} \begin{pmatrix} 1 \\ \lambda_1 \end{pmatrix} + c_2 e^{\lambda_2 t} \begin{pmatrix} 1 \\ \lambda_2 \end{pmatrix}, \end{equation*} which agrees with (4.5).

Example4.3

Now let us solve the initial value problem \begin{align*} x'' + 4 x' + 5 x & = 0\\ x(0) & = 1\\ x'(0) & = 1. \end{align*} 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 \begin{equation*} 0 = x'' + 4 x' + 5 x = r^2 e^{rt} + 4r e^{rt} + 5 e^{rt} = (r^2 + 4r + 5) e^{rt}. \end{equation*} As with the real case, \(r^2 + 4r + 5 = 0\) or \begin{equation*} r = \frac{-4 \pm \sqrt{-4}}{2} = - 2 \pm i. \end{equation*} Thus, we can find a solution \begin{equation*} x(t) = e^{(-2 + i)t} = e^{-2t} e^{it} = e^{-2t} (\cos t + i \sin t). \end{equation*} The real and imaginary parts of our solution are \begin{align*} x_1(t) \amp = e^{-2t} \cos t\\ x_2(t) \amp = e^{-2t} \sin t, \end{align*} 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, \begin{align*} 0 \amp = ax'' + bx' + cx\\ \amp = a(x_1 + i x_2)'' + b(x_1 + i x_2)' + c(x_1 + i x_2)\\ \amp = (ax_1'' + bx_1' + cx_1) + i(ax_2'' + bx_2' + cx_2). \end{align*} Since the real part and the imaginiary 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 \begin{equation*} x(t) = c_1 e^{-2t} \cos t + c_2 e^{-2t} \sin t. \end{equation*} Applying our initial conditions, we can quickly calculate \(c_1 =1\) and \(c_2= 3\text{.}\) Hence, the solution to our initial value problem is \begin{equation*} x(t) = e^{-2t} \cos t + 3 e^{-2t} \sin t. \end{equation*}

As before, the corresponding linear sytem is \(\mathbf x' = A \mathbf x\text{,}\) where \begin{equation*} A = \begin{pmatrix} 0 \amp 1 \\ -c/a \amp -b/a \end{pmatrix}. \end{equation*} If \(b^2 - 4ac \lt 0\text{,}\) the eigenvalue of \(A\) are \(\lambda = \alpha + i \beta\) and \(\overline{\lambda} = \alpha - i \beta\text{,}\) where \begin{equation*} \alpha = - \frac{b}{2a} \quad \text{and} \quad \beta = \frac{\sqrt{4ac - b^2}}{2a}. \end{equation*} The vector \(\mathbf v = (1, \alpha + i \beta)\) is an eigenvector for \(\lambda\text{.}\) Thus, a solution to our system of differential equations is \begin{align*} \mathbf x(t) \amp = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}\\ \amp = \begin{pmatrix} x(t) \\ x'(t) \end{pmatrix}\\ \amp = e^{(\alpha + i \beta)t} \begin{pmatrix} 1 \\ \alpha + i \beta \end{pmatrix}\\ \amp = e^{\alpha t} (\cos \beta t + i \sin \beta t) \begin{pmatrix} 1 \\ \alpha + i \beta \end{pmatrix}\\ \amp = e^{\alpha t} \begin{pmatrix} \cos \beta t \\ \alpha \cos \beta t - \beta \sin \beta t \end{pmatrix} + i e^{\alpha t} \begin{pmatrix} \sin \beta t \\ \alpha \sin \beta t + \beta \cos \beta t \end{pmatrix}. \end{align*} 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 \begin{equation*} x(t) = c_1 e^{\alpha t} \cos \beta t + c_2 e^{\alpha t} \sin \beta t. \end{equation*}

Subsection4.1.3Reduction of Order

Given a second-order 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 soluitons, but what if there is only a single real solution \(\lambda = -b/2a\text{.}\)

Example4.4

Consider the equation \begin{equation*} x'' + 2x' + x = 0. \end{equation*} If we choose \(e^{\lambda t}\) as our guess, we find \begin{align*} x'' + 2 x' + x & = \lambda^2 e^{\lambda t} + 2\lambda e^{\lambda t} + e^{\lambda t}\\ & = e^{\lambda t} (\lambda + 1)^2\\ & = 0. \end{align*} 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 \(x_1(t) = e^{-t}\text{.}\)

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 \begin{equation*} x(t) = v(t) x_1(t) = v(t) e^{-t}, \end{equation*} then \begin{equation*} x'(t) = v(t) x_1'(t) + v'(t) x_1(t) = -v(t) e^{-t} + v'(t) e^{-t} \end{equation*} and \begin{align*} x''(t) & = v''(t) x_1(t) + 2v'(t) x_1'(t) + v(t) x_1''(t)\\ & = v''(t) e^{-t} - 2 v'(t) e^{-t} + v(t) e^{-t}. \end{align*} Consequently, \begin{align*} x'' + 2 x' + x & = [v'' e^{-t} - 2 v' e^{-t} + v e^{-t}] + 2 [-v e^{-t} + v' e^{-t}] + [v e^{-t}]\\ & = e^{-t} v''\\ & = 0, \end{align*} 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 \begin{equation*} x(t) = c_1 e^{-t} + c_2 t e^{-t}. \end{equation*} 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.

The technique that we have used in Example 4.4 is called reduction of order. We leave it as an exercise to show that this technique works in general. That is, given a second-order linear differential equation \begin{equation*} a x'' + bx' + cx = 0 \end{equation*} such that \(b^2 - 4ac = 0\text{,}\) then the general solution is given by \begin{equation*} x(t) = c_1 e^{\lambda t} + c_2 t e^{\lambda t}, \end{equation*} where \(\lambda = -b/2a\text{.}\)

We did a great deal of work finding unique solutions to first-order linear systems of equations, \begin{align*} x' & = ax + by\\ y' & = cx + dy. \end{align*} Our efforts are now rewarded. Since each second-order homogeneous system with constant coefficients can be rewritten as a first-order linear system, we are guaranteed the existence and uniqueness of solutions.

Subsection4.1.4Important 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*} 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 \(a x'' + bx' + cx = 0\text{.}\)
  • 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*}

Subsection4.1.5Exercises

1

Let \(a x'' + b x' + cx = 0\text{,}\) where \(a \neq 0\) and \(b^2 - 4ac = 0\text{.}\)

  1. Show that \(x_1(t) = e^{-bt/2a}\) is a solution to \(a x'' + b x' + cx = 0\text{.}\)
  2. 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 \(a x'' + b x' + cx = 0\text{.}\)
2

Reduction of Order. Suppose that \(x_1(t)\) is a solution (not identically zero) to the equation \begin{equation*} x'' + p(t) x' + q(t) x = 0. \end{equation*}

  1. 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{4.6} \end{equation}
  2. Let \(u = v'\) and show that (4.6) is a first-order linear differential equation in \(u\text{.}\)
  3. 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{4.7} \end{equation} for \(t \gt 0\) and find a second linearly independent solution to (4.7).

Subsection4.1.6Project