Second-Order Constant-Coefficient Equations, differential equations lesson

A mass on a spring with friction and the current in a simple circuit both lead to equations such as $y'' + 2y' + 5y = 0$ , where $t$ is time and $y' = dy/dt$ . Because the coefficients are fixed numbers, an exponential guess turns the differential equation into a quadratic equation. A real quadratic has two distinct real roots, one repeated root or a pair of complex roots, and each case gives its own pair of basic solutions. This lesson works through all three and shows how complex roots still lead to real solutions and real initial-value answers.

Key terms

Constant-coefficient homogeneous equation

Definition A second-order constant-coefficient homogeneous equation is

y'' + a\,y' + b\,y = 0,

where $y$ is an unknown function of $t$ , $y' = dy/dt$ , and $a$ and $b$ are real constants. An equation $a_2y'' + a_1y' + a_0y = 0$ with real constants and $a_2 \ne 0$ takes this form after division by $a_2$ , with $a = a_1/a_2$ and $b = a_0/a_2$ . Constant coefficients are continuous everywhere, so all of $\mathbb{R}$ is a regular interval. Every initial-value problem $y(t_0) = k_0$ , $y'(t_0) = k_1$ therefore has exactly one solution, defined for all real $t$ , and once a fundamental set $y_1$ , $y_2$ is known, every solution is $C_1y_1 + C_2y_2$ . The characteristic equation below finds such a set.

In plain words The coefficients are fixed numbers rather than functions of time, and nothing is left over on the right side. Every solution then lives for all time, and two independent solutions describe them all.

Example The equation $2y'' + 4y' + 10y = 0$ becomes $y'' + 2y' + 5y = 0$ after division by $2$ , so $a = 2$ and $b = 5$ . By contrast, $y'' + ty = 0$ has a coefficient that changes with $t$ , and the method of this lesson does not apply to it. An exponential trial $e^{rt}$ there gives $(r^2 + t)\,e^{rt}$ , which is not zero for all $t$ for any constant $r$ . The equation $y'' + 2y' + 5y = \cos t$ is not homogeneous. Its solutions are the homogeneous family found here plus one particular solution, which still has to be found, for example by undetermined coefficients.

Characteristic equation

Definition Write $L[y] = y'' + ay' + by$ . For every constant $r$ , real or complex,

L[e^{rt}] = (r^2 + ar + b)\,e^{rt}.

The polynomial $r^2 + ar + b$ is the characteristic polynomial, and

r^2 + ar + b = 0

is the characteristic equation. Since $e^{rt}$ is never zero, the exponential $e^{rt}$ solves $L[y] = 0$ exactly when $r$ is a root. The discriminant

\Delta = a^2 - 4b

decides the case. If $\Delta > 0$ there are two distinct real roots, if $\Delta = 0$ there is one repeated real root, and if $\Delta < 0$ there is a pair of complex conjugate roots.

In plain words An exponential keeps its shape under differentiation, and each derivative only multiplies it by $r$ . So the differential equation reduces to a quadratic equation for the rate $r$ .

Example For $y'' - y' - 2y = 0$ the characteristic equation $r^2 - r - 2 = 0$ factors as $(r - 2)(r + 1) = 0$ , and the discriminant is $1 + 8 = 9 > 0$ . For $y'' + 4y' + 4y = 0$ the discriminant is $16 - 16 = 0$ , and for $y'' + 2y' + 5y = 0$ it is $4 - 20 = -16 < 0$ . These three equations illustrate the three cases below.

Distinct real roots

Definition If $\Delta > 0$ , the characteristic equation has two real roots $r_1 \ne r_2$ , namely

r_{1,2} = \frac{-a \pm \sqrt{\Delta}}{2}.

The solutions $e^{r_1t}$ and $e^{r_2t}$ form a fundamental set on $\mathbb{R}$ , because their Wronskian

W = (r_2 - r_1)\,e^{(r_1 + r_2)t}

is never zero. Every solution is

y = C_1e^{r_1t} + C_2e^{r_2t}.

In plain words Two different rates give two different exponentials, and together they build every solution. A positive root contributes a growing term and a negative root a decaying one, and the initial data decide how much of each is present.

Example For $y'' - y' - 2y = 0$ the roots are $2$ and $-1$ , so every solution is $y = C_1e^{2t} + C_2e^{-t}$ , and the Wronskian is $-3e^{t}$ . The data $y(0) = 1$ and $y'(0) = -1$ give $C_1 + C_2 = 1$ and $2C_1 - C_2 = -1$ , so $C_1 = 0$ , $C_2 = 1$ and $y = e^{-t}$ , which decays to $0$ . The data $y(0) = 1$ and $y'(0) = 0$ give $3C_1 = 1$ instead, and

y = \tfrac{1}{3}e^{2t} + \tfrac{2}{3}e^{-t},

which grows without bound as $t \to \infty$ because its coefficient of $e^{2t}$ is not zero.

Repeated root

Definition If $\Delta = 0$ , the only root is $r_0 = -a/2$ . The characteristic polynomial is then $(r - r_0)^2$ , so $a = -2r_0$ and $b = r_0^2$ . The exponential $e^{r_0t}$ gives only one solution, and a second one is $te^{r_0t}$ . Its derivatives are

\begin{aligned} (te^{r_0t})' &= (1 + r_0t)\,e^{r_0t}, \\[0.4em] (te^{r_0t})'' &= (2r_0 + r_0^2t)\,e^{r_0t}. \end{aligned}

Substituting them into the equation with $a = -2r_0$ and $b = r_0^2$ , the constant terms give $2r_0 - 2r_0 = 0$ and the terms with $t$ give $r_0^2 - 2r_0^2 + r_0^2 = 0$ . The Wronskian $W(e^{r_0t}, te^{r_0t}) = e^{2r_0t}$ is never zero, so every solution is

y = (C_1 + C_2t)\,e^{r_0t}.

In plain words When the two rates merge into one, the exponential alone supplies only one solution. Multiplying it by $t$ gives the missing second one. For a negative root the factor $t$ changes the early shape of the curve, but the exponential still drives every solution to zero as $t \to \infty$ .

Example For $y'' + 4y' + 4y = 0$ the characteristic polynomial is $(r + 2)^2$ , so $r_0 = -2$ and $y = (C_1 + C_2t)\,e^{-2t}$ . Writing $e^{-2t}$ twice would not be enough. The pair $e^{-2t}$ , $e^{-2t}$ has Wronskian $0$ , and its combinations $Ce^{-2t}$ only reach data with $y'(0) = -2y(0)$ . For the data $y(0) = 2$ and $y'(0) = -1$ , the derivative $y' = (C_2 - 2C_1 - 2C_2t)\,e^{-2t}$ gives $C_1 = 2$ and $C_2 - 2C_1 = -1$ , so $C_2 = 3$ and

y = (2 + 3t)\,e^{-2t}.

Complex conjugate roots

Definition If $\Delta < 0$ , the roots are $r = \alpha \pm i\beta$ with

\begin{gathered} \alpha = -\frac{a}{2}, \\[0.4em] \beta = \frac{\sqrt{4b - a^2}}{2} > 0. \end{gathered}

They are complex conjugates of each other because $a$ and $b$ are real. By Euler’s formula the complex solution $e^{(\alpha + i\beta)t}$ equals $e^{\alpha t}(\cos \beta t + i\sin \beta t)$ , and its real and imaginary parts

\begin{gathered} y_1 = e^{\alpha t}\cos \beta t, \\[0.4em] y_2 = e^{\alpha t}\sin \beta t \end{gathered}

are real solutions with Wronskian $\beta e^{2\alpha t} \ne 0$ . Every real solution is therefore

\begin{aligned} y &= C_1e^{\alpha t}\cos \beta t \\[0.2em] &\quad + C_2e^{\alpha t}\sin \beta t \end{aligned}

with real constants $C_1$ and $C_2$ , which is often written $y = e^{\alpha t}(C_1\cos \beta t + C_2\sin \beta t)$ .

In plain words Complex roots mean oscillation. The number $\beta$ is the angular frequency of the swings, and the real part $\alpha$ sets the envelope $e^{\alpha t}$ . The swings die out when $\alpha < 0$ , keep a constant size when $\alpha = 0$ and grow when $\alpha > 0$ .

Example For $y'' - 2y' + 10y = 0$ the roots of $r^2 - 2r + 10 = 0$ are $1 \pm 3i$ , so every real solution is $e^{t}(C_1\cos 3t + C_2\sin 3t)$ . It oscillates with angular frequency $3$ inside a growing envelope. For $y'' + 9y = 0$ the roots are $\pm 3i$ , so $\alpha = 0$ and the solutions $C_1\cos 3t + C_2\sin 3t$ keep a constant amplitude.

Real-valued solutions

Definition Let $a$ and $b$ be real, and let $z = u + iv$ , with $u$ and $v$ real-valued, be a complex-valued solution of $L[y] = y'' + ay' + by = 0$ . Then $u$ and $v$ are real solutions, because

0 = L[z] = L[u] + i\,L[v]

and both $L[u]$ and $L[v]$ are real-valued, so both are zero. With complex roots the complex solutions are $c_1e^{(\alpha + i\beta)t} + c_2e^{(\alpha - i\beta)t}$ with complex constants $c_1$ and $c_2$ , and such a combination is real-valued exactly when $c_2 = \overline{c_1}$ . Choosing $c_1 = (C_1 - iC_2)/2$ and $c_2 = \overline{c_1}$ with real $C_1$ and $C_2$ gives the real form $e^{\alpha t}(C_1\cos \beta t + C_2\sin \beta t)$ , and real initial data always lead to real $C_1$ and $C_2$ . When $R = \sqrt{C_1^2 + C_2^2} > 0$ , the same solution can be written

y = R\,e^{\alpha t}\cos(\beta t - \varphi),

where the phase $\varphi$ satisfies $R\cos\varphi = C_1$ and $R\sin\varphi = C_2$ .

In plain words Complex numbers are a tool for finding the solutions, not part of the final answer. The real and imaginary parts of one complex solution already give two real solutions, and real starting values select a real curve.

Example Solve $y'' + 2y' + 5y = 0$ with $y(0) = 3$ and $y'(0) = 1$ . The roots of $r^2 + 2r + 5 = 0$ are $-1 \pm 2i$ , so every real solution has the form $y = e^{-t}(C_1\cos 2t + C_2\sin 2t)$ . The height gives $C_1 = 3$ . The product rule applied to the whole product gives

\begin{aligned} y' &= e^{-t}(2C_2 - C_1)\cos 2t \\[0.2em] &- e^{-t}(2C_1 + C_2)\sin 2t, \end{aligned}

so $y'(0) = 2C_2 - C_1 = 1$ and $C_2 = 2$ . The solution is

y = e^{-t}(3\cos 2t + 2\sin 2t),

an oscillation with angular frequency $2$ whose amplitude $\sqrt{13}\,e^{-t}$ decays. Dropping the derivative of $e^{-t}$ would give $2C_2 = 1$ and the wrong coefficient $C_2 = \tfrac{1}{2}$ .

Check yourself

Solve $y'' + y' - 6y = 0$ with $y(0) = 0$ and $y'(0) = 5$ .

Show the answer

The characteristic equation $r^2 + r - 6 = 0$ factors as $(r - 2)(r + 3) = 0$ , so $y = C_1e^{2t} + C_2e^{-3t}$ . The data give $C_1 + C_2 = 0$ and $2C_1 - 3C_2 = 5$ . Substituting $C_2 = -C_1$ gives $5C_1 = 5$ , so $C_1 = 1$ , $C_2 = -1$ and

y = e^{2t} - e^{-3t}.

For $y'' - 6y' + 9y = 0$ a classmate writes $y = C_1e^{3t} + C_2e^{3t}$ and tries to meet $y(0) = 1$ and $y'(0) = 2$ . Explain what goes wrong and find the correct solution.

Show the answer

The characteristic polynomial is $(r - 3)^2$ , so $3$ is a repeated root. The classmate’s family is only $Ce^{3t}$ with $C = C_1 + C_2$ , and every member has $y'(0) = 3y(0)$ . With $y(0) = 1$ that forces $y'(0) = 3$ rather than $2$ , so no member fits. The correct family is $y = (C_1 + C_2t)\,e^{3t}$ , with $y(0) = C_1$ and $y'(0) = 3C_1 + C_2$ . So $C_1 = 1$ , $C_2 = -1$ and

y = (1 - t)\,e^{3t}.

Its derivative $y' = (2 - 3t)\,e^{3t}$ gives $y'(0) = 2$ as required.

Solve $y'' + 4y' + 13y = 0$ with $y(0) = 1$ and $y'(0) = 0$ , and describe the behavior of the solution as $t \to \infty$ .

Show the answer

The roots of $r^2 + 4r + 13 = 0$ are $-2 \pm 3i$ , so $y = e^{-2t}(C_1\cos 3t + C_2\sin 3t)$ . The height gives $C_1 = 1$ . By the product rule $y'(0) = -2C_1 + 3C_2$ , so $-2 + 3C_2 = 0$ and $C_2 = \tfrac{2}{3}$ . The solution is

y = e^{-2t}\left(\cos 3t + \tfrac{2}{3}\sin 3t\right).

It oscillates with angular frequency $3$ inside the envelope $\pm\tfrac{\sqrt{13}}{3}e^{-2t}$ , so it tends to $0$ . Forgetting the derivative of $e^{-2t}$ would give $3C_2 = 0$ and the wrong answer $C_2 = 0$ .

Second-Order Constant-Coefficient Equations

Key terms

Constant-coefficient homogeneous equation

Characteristic equation

Distinct real roots

Repeated root

Complex conjugate roots

Real-valued solutions

Check yourself

How this lesson connects

Builds on

Used later in