← Back to Math 055

4.3 Homogeneous Linear Equations with Constant Coefficients: Class Notes

4.3a

Consider the differential equation $y″-5y\prime +6y=0$.

Tip: Why try an exponential? Functions such as $x^3$, $\ln x$, $\tan x$, and $\sqrt{x}$ change form when differentiated, so the terms of the equation can't cancel. Every derivative of $e^{mx}$ is a multiple of $e^{mx}$.

Start with $y=e^{mx}$. Then $y\prime =m{e}^{mx}$ and $y″=m^2{e}^{mx}$. Substitute:

$$\begin{array}{cc}\hfill {\displaystyle m^2{e}^{mx}-5m{e}^{mx}+6e^{mx}}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (m^2-5m+6)e^{mx}}& {\displaystyle =0.}\hfill \end{array}$$

Since $e^{mx}\ne 0$,

$$m^2-5m+6=0\text{(auxiliary equation)}$$

$$\begin{array}{cc}\hfill {\displaystyle (m-2)(m-3)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle m}& {\displaystyle =2,3}\hfill \end{array}$$

So $y_1=e^{2x}$ and $y_2=e^{3x}$ are solutions, and the general solution is

Definition

The auxiliary equation of the second-order linear differential equation $ay″+by\prime +cy=0$ is $a{m}^2+bm+c=0$.

The auxiliary equation has solutions

$${\displaystyle m=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.}$$

This leads to 3 cases.

Case 1: $m_1\ne m_2$, both real. Then

$$y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}$$

Case 2: $m_1=m_2$, real (repeated roots). If $b^2-4ac=0$, then ${\displaystyle m_1=-\frac{b}{2a}}$ and

$${\displaystyle y_1=e^{-\frac{b}{2a}x}.}$$

Find $y_2$ by reduction of order, ${\displaystyle y_2=y_1\int \frac{e^{-\int Pdx}}{y_1^2}dx}$. In standard form,

$${\displaystyle ay″+by\prime +cy=0\Rightarrow y″+\frac{b}{a}y\prime +\frac{c}{a}y=0,}$$

so ${\displaystyle e^{-\int Pdx}=e^{-\frac{b}{a}x}}$, which is the same as ${\displaystyle y_1^2=e^{-\frac{b}{a}x}}$:

$${\displaystyle y_2=y_1\int \frac{e^{-\frac{b}{a}x}}{e^{-\frac{b}{a}x}}dx=x{y}_1.}$$

Then

$$y=c_1{e}^{m_{1}x}+c_{2}x{e}^{m_{1}x}$$

Case 3: $b^2-4ac<0$, so $m_1=\alpha +\beta i$, $m_2=\alpha -\beta i$. Then, for all $c_1,c_2$,

$$\begin{array}{cc}\hfill {\displaystyle y}& {\displaystyle =c_1{e}^{(\alpha +\beta i)x}+c_2{e}^{(\alpha -\beta i)x}}\hfill \\ \hfill {\displaystyle y}& {\displaystyle =e^{\alpha x}(c_1{e}^{\beta ix}+c_2{e}^{-\beta ix}).}\hfill \end{array}$$

To rewrite this with real functions, use Euler's formula:

$$e^{i\theta }=\cos \theta +i\sin \theta$$

Tip: Side note: if $\theta =\pi$, then $e^{i\pi }=-1$, so $e^{i\pi }+1=0$.

$$\begin{array}{cc}\hfill {\displaystyle e^{\beta ix}}& {\displaystyle =\cos \beta x+i\sin \beta x}\hfill \\ \hfill {\displaystyle e^{-\beta ix}}& {\displaystyle =\cos \beta x-i\sin \beta x}\hfill \end{array}$$

Adding the two lines gives $e^{\beta ix}+e^{-\beta ix}=2\cos \beta x$, so

$${\displaystyle \cos \beta x=\frac{e^{\beta ix}+e^{-\beta ix}}{2}.}$$

Likewise,

$${\displaystyle \sin \beta x=\frac{e^{\beta ix}-e^{-\beta ix}}{2i}.}$$

Strategic choices for $c_1,c_2$ above (for example, ${\displaystyle c_1=c_2=\frac{1}{2}}$ gives $e^{\alpha x}\cos \beta x$) give the following general solution. If $m=\alpha \pm \beta i$, then

$$y=e^{\alpha x}(c_1\cos \beta x+c_2\sin \beta x)$$

4.3b

Example

Find the general solution of the given second-order differential equation.

$$y″-10y\prime +25y=0$$

Solution

$$\begin{array}{cc}\hfill {\displaystyle m^2-10m+25}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (m-5{)}^2}& {\displaystyle =0\Rightarrow m=5\text{(repeated)}}\hfill \end{array}$$

Repeated root (Case 2):

Example

$2y″-3y\prime +4y=0$

Solution

Auxiliary equation $2m^2-3m+4=0$. By the quadratic formula,

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle m}& {\displaystyle =\frac{3\pm \sqrt{9-32}}{4}}\hfill \\ \hfill {\displaystyle m}& {\displaystyle =\frac{3}{4}\pm \frac{\sqrt{23}}{4}i.}\hfill \end{array}}$$

Complex roots (Case 3) with ${\displaystyle \alpha =\frac{3}{4}}$ and ${\displaystyle \beta =\frac{\sqrt{23}}{4}}$:

Example

Find the general solution of the given higher-order differential equation.

$${\displaystyle \frac{d^{3}x}{d{t}^3}-\frac{d^{2}x}{d{t}^2}-4x=0}$$

Solution

We're looking for $x(t)$. The auxiliary equation is

$$m^3-m^2-4=0.$$

Its roots (found with wolframalpha.com) are

$${\displaystyle m=2,m=-\frac{1}{2}\pm \frac{\sqrt{7}}{2}i.}$$

Tip: To check by hand: $m=2$ works because $8-4-4=0$, and dividing by $m-2$ leaves $m^2+m+2$, whose roots are ${\displaystyle -\frac{1}{2}\pm \frac{\sqrt{7}}{2}i}$.

The real root gives $e^{2t}$, and the complex pair gives the Case 3 form:

Special 2nd-order differential equations

$y″+k^{2}y=0$ and $y″-k^{2}y=0$, $k\in ℝ$

First equation: $m^2+k^2=0$, so $m=\pm ki$ (Case 3 with $\alpha =0$):

$$y=c_1\cos kx+c_2\sin kx$$

Second equation: $m^2-k^2=0$, so $m=\pm k$:

$$y=c_1{e}^{kx}+c_2{e}^{-kx}$$

If ${\displaystyle c_1=c_2=\frac{1}{2}}$, then $y_1=\cosh kx$. If ${\displaystyle c_1=\frac{1}{2}}$, ${\displaystyle c_2=-\frac{1}{2}}$, then $y_2=\sinh kx$. Alternatively,

$$y=c_1\cosh kx+c_2\sinh kx$$

4.3c

Example

Find a homogeneous linear differential equation with constant coefficients whose general solution is given.

$$y=c_1{e}^{-4x}+c_2{e}^{-3x}$$

Solution

The exponents tell us the roots of the auxiliary equation:

$$\begin{array}{cc}\hfill {\displaystyle m=-4,-3}& {\displaystyle \Rightarrow (m+4)(m+3)=0}\hfill \\ \hfill {\displaystyle }& {\displaystyle \Rightarrow m^2+7m+12=0}\hfill \end{array}$$

Example

$y=c_1{e}^{10x}+c_{2}x{e}^{10x}$

Solution

The factor $x$ signals a repeated root: $m=10$ (repeated).

$$\begin{array}{cc}\hfill {\displaystyle (m-10{)}^2}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle m^2-20m+100}& {\displaystyle =0}\hfill \end{array}$$

Example

$y=c_1+c_2{e}^{2x}\cos 5x+c_3{e}^{2x}\sin 5x$

Solution

Three arbitrary constants, so this is a 3rd-order DE. The constant term is $y_1=e^{0x}$, so $m=0$; the other two terms give $m=2\pm 5i$.

$$\begin{array}{cc}\hfill {\displaystyle m(m-(2+5i))(m-(2-5i))}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle m(m-2-5i)(m-2+5i)}& {\displaystyle =0}\hfill \end{array}$$

Use the difference of squares $(A-B)(A+B)=A^2-B^2$ with $A=m-2$ and $B=5i$:

$$\begin{array}{cc}\hfill {\displaystyle m({(m-2)}^2-{(5i)}^2)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle m(m^2-4m+4+25)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle m^3-4m^2+29m}& {\displaystyle =0}\hfill \end{array}$$