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. We already know how to solve such equations since we can rewrite them as a system of first-order linear equations. 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.

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).

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}

\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.

Suppose that we have a homogeneous second-order linear differential equation with constant coefficients,

\begin{equation}
a x'' + b x' + cx = 0.\label{equation-secondorder01-constant-coef}\tag{4.2}
\end{equation}

Since we can rewrite this equation as a system of first-order linear equations,

\begin{align*}
x' & = y\\
y' & = -\frac{c}{a} x - \frac{b}{a} y.
\end{align*}

we can find the general solution 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.

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*}
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 \(a \lambda^2 + b \lambda + c\text{,}\)

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

for \(\lambda_1\) and \(\lambda_2\text{,}\) respectively. Thus, the general solution to the system of differential equations \(\mathbf x' = A \mathbf x\) is

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

\begin{equation*}
x(t) = c_1 e^{-2t} \cos t + c_2 e^{-2t} \sin t.
\end{equation*}

To apply our initial conditions \(x(0) = 1\) and \(x'(0) = 1\text{,}\) we first calculate

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*}

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

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

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

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.

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

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*}