The Auxiliary Equation

Try an exponential

A second-order linear equation with constant coefficients has the form $a \frac{d^{2}y}{dx^{2}} + b \frac{dy}{dx} + c y = 0$, where a, b and c are numbers. A solution y must combine with its first and second derivatives to give 0, so its derivatives should be multiples of y itself.

The exponential $y = e^{mx}$ has exactly that property: $\frac{dy}{dx} = m e^{mx}$ and $\frac{d^{2}y}{dx^{2}} = m^{2} e^{mx}$. So try $y = e^{mx}$ and look for the values of m that work.

The auxiliary equation

Substitute: $am^{2} e^{mx} + bm e^{mx} + c e^{mx} = 0$. Every term has the factor $e^{mx}$, so this is $\left(am^{2} + bm + c\right)e^{mx} = 0$.

An exponential is never zero, so the bracket must be: $am^{2} + bm + c = 0$. This quadratic is the auxiliary equation. The coefficients carry straight across, each with its own sign, and the differential equation has become algebra.

For $\frac{d^{2}y}{dx^{2}} - 5 \frac{dy}{dx} + 6y = 0$ the auxiliary equation is $m^{2} - 5m + 6 = 0$, which factors as (m − 2)(m − 3) = 0. So $e^{2x}$ and $e^{3x}$ are both solutions. Check the roots: 4 − 10 + 6 = 0 and 9 − 15 + 6 = 0.

Two real roots

The equation is linear: putting a sum of two functions into the left side gives the sum of what each gives on its own. So if $e^{2x}$ and $e^{3x}$ each give 0, then $A e^{2x} + B e^{3x}$ gives 0 for any constants A and B.

The general solution is $y = A e^{2x} + B e^{3x}$. A second-order equation needs two constants, and here they are. In general, two real roots $m_{1}$ and $m_{2}$ give $y = A e^{m_{1}x} + B e^{m_{2}x}$.

Two known values fix the constants. If y = 0 and $\frac{dy}{dx} = 1$ when x = 0, then A + B = 0 and 2A + 3B = 1, so A = −1 and B = 1, and $y = e^{3x} - e^{2x}$.

Three equations, one start

The three kinds of root are easiest to compare on three equations solved from the same start, y = 1 and $\frac{dy}{dx} = 0$ at x = 0.

For $\frac{d^{2}y}{dx^{2}} + 3 \frac{dy}{dx} + 2y = 0$ the auxiliary equation is $m^{2} + 3m + 2 = \left(m + 1\right)\left(m + 2\right) = 0$, with roots −1 and −2. So $y = A e^{-x} + B e^{-2x}$. The start gives A + B = 1 and −A − 2B = 0, so A = 2 and B = −1: $y = 2e^{-x} - e^{-2x}$.

$y = 2e^{-x} - e^{-2x}$, the solution of $\frac{d^{2}y}{dx^{2}} + 3 \frac{dy}{dx} + 2y = 0$ that starts at 1 with gradient 0. Both roots are negative, so both exponentials shrink, and y falls to 0.60 at x = 1 and 0.25 at x = 2 without ever crossing the x-axis.

A repeated root

When $b^{2} - 4ac = 0$ the auxiliary equation has only one root, $m = - \frac{b}{2a}$, and $e^{mx}$ gives only one constant. A second solution is $x e^{mx}$.

To see why, differentiate: $\frac{dy}{dx} = \left(1 + mx\right)e^{mx}$ and $\frac{d^{2}y}{dx^{2}} = \left(2m + m^{2}x\right)e^{mx}$. Substituting, the left side is $\left[x\left(am^{2} + bm + c\right) + \left(2am + b\right)\right]e^{mx}$. The first bracket is 0 because m is a root, and 2am + b is 0 because $m = - \frac{b}{2a}$. So $x e^{mx}$ is a solution exactly when the root repeats.

The general solution is $y = \left(A + B x\right)e^{mx}$. For $\frac{d^{2}y}{dx^{2}} + 2 \frac{dy}{dx} + y = 0$ the auxiliary equation is $\left(m + 1\right)^{2} = 0$, so m = −1 twice and $y = \left(A + B x\right)e^{-x}$. The start y = 1, $\frac{dy}{dx} = 0$ gives A = 1 and B − A = 0, so $y = \left(1 + x\right)e^{-x}$.

The gold curve is $y = \left(1 + x\right)e^{-x}$, the solution of $\frac{d^{2}y}{dx^{2}} + 2 \frac{dy}{dx} + y = 0$ from the same start. The dashed curve is $e^{-x}$ alone, which starts at the right height but with gradient −1; the factor 1 + x supplies the second constant that makes the gradient 0 at the start.

Complex roots

When $b^{2} - 4ac < 0$ the roots are a complex pair, $p \pm qi$. The exponential still works: $e^{\left(p + qi\right)x} = e^{px}e^{iqx}$, and $e^{iqx} = \cos qx + i \sin qx$.

Adding and subtracting the solutions from the two roots removes the i, and leaves the two real solutions $e^{px} \cos qx$ and $e^{px} \sin qx$. So $y = e^{px}\left(A \cos qx + B \sin qx\right)$.

For $\frac{d^{2}y}{dx^{2}} + 2 \frac{dy}{dx} + 5y = 0$, $b^{2} - 4ac = 4 - 20 = -16$, so $m = -1 \pm 2i$ and p = −1, q = 2 and $y = e^{-x}\left(A \cos 2x + B \sin 2x\right)$. From y = 1 and $\frac{dy}{dx} = 0$ at x = 0, A = 1 and −A + 2B = 0, so B = ½.

The gold curve is $y = e^{-x}\left(\cos 2x + \frac{1}{2} \sin 2x\right)$. It swings across the x-axis, first at $x \approx 1.02$, and reaches −0.21 at $x = \frac{\pi}{2}$. The dashed curves are $\pm\frac{1}{2}\sqrt{5} e^{-x}$: the cosine and sine together never exceed $\sqrt{1 + \frac{1}{4}} = \frac{\sqrt{5}}{2}$, so the swing stays between them and shrinks as they do.

The sign of p

In $e^{px}\left(A \cos qx + B \sin qx\right)$ the number q sets how fast y swings and p sets the envelope. A negative p makes the swings shrink, p = 0 makes them keep the same size forever, and a positive p makes them grow.

So the discriminant $b^{2} - 4ac$ decides the shape of every solution. Positive gives two exponentials, zero gives $\left(A + B x\right)e^{mx}$, and negative gives an oscillation.

The usual mistakes

Changing a sign on the way to the auxiliary equation. $\frac{d^{2}y}{dx^{2}} - 5 \frac{dy}{dx} + 6y = 0$ gives $m^{2} - 5m + 6 = 0$, with roots 2 and 3, so the solution has $e^{2x}$ and $e^{3x}$, not $e^{-2x}$ and $e^{-3x}$.

Writing $A e^{mx} + B e^{mx}$ for a repeated root. That is $\left(A + B\right)e^{mx}$, which has only one constant. The second solution is $x e^{mx}$.

Putting the i into the solution, as in $e^{\left(p + qi\right)x}$ left as it is. The real solution is $e^{px}\left(A \cos qx + B \sin qx\right)$, with q the size of the imaginary part.

Forgetting the $e^{px}$ when p is not zero. Without it the solution would swing at a constant size, which only happens when the middle coefficient b is 0.

Two tanks of brine

In the application below, the salt in two connected tanks gives two first-order equations. Removing one variable leaves one second-order equation for the salt y in the second tank, with coefficients 1, 0.4 and 0.03. Its auxiliary equation is written with $\lambda$, because there m counts minutes: $\lambda^{2} + 0.4\lambda + 0.03 = 0$ has roots −0.1 and −0.3.