The Auxiliary Equation

Try an exponential

A second-order linear equation with constant coefficients has the form $a\frac{d^{2}y}{d{x}^2}+b\frac{dy}{dx}+cy=0$a d²y/dx² + b 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 $y=e^{mx}$y = e^(mx) has exactly that property: $\frac{dy}{dx}=m{e}^{mx}$dy/dx = m e^(mx) and $\frac{d^{2}y}{d{x}^2}=m^2{e}^{mx}$d²y/dx² = m² e^(mx). So try $y=e^{mx}$y = e^(mx) and look for the values of m that work.

The auxiliary equation

Substitute: $a{m}^2{e}^{mx}+bm{e}^{mx}+c{e}^{mx}=0$am² e^(mx) + bm e^(mx) + c e^(mx) = 0. Every term has the factor $e^{mx}$e^(mx), so this is $\left(a{m}^2+bm+c\right)e^{mx}=0$(am² + bm + c)e^(mx) = 0.

An exponential is never zero, so the bracket must be: $a{m}^2+bm+c=0$am² + 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 $\frac{d^{2}y}{d{x}^2}-5\frac{dy}{dx}+6y=0$d²y/dx² − 5 dy/dx + 6y = 0 the auxiliary equation is $m^2-5m+6=0$m² − 5m + 6 = 0, which factors as (m − 2)(m − 3) = 0. So $e^{2x}$e^(2x) and $e^{3x}$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 $e^{2x}$e^(2x) and $e^{3x}$e^(3x) each give 0, then $A{e}^{2x}+B{e}^{3x}$A e^(2x) + B e^(3x) gives 0 for any constants A and B.

The general solution is $y=A{e}^{2x}+B{e}^{3x}$y = A e^(2x) + B e^(3x). A second-order equation needs two constants, and here they are. In general, two real roots $m_1$m₁ and $m_2$m₂ give $y=A{e}^{m_{1}x}+B{e}^{m_{2}x}$y = A e^(m₁x) + B e^(m₂x).

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

For $\frac{d^{2}y}{d{x}^2}+3\frac{dy}{dx}+2y=0$d²y/dx² + 3 dy/dx + 2y = 0 the auxiliary equation is $m^2+3m+2=\left(m+1\right)\left(m+2\right)=0$m² + 3m + 2 = (m + 1)(m + 2) = 0, with roots −1 and −2. So $y=A{e}^{-x}+B{e}^{-2x}$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).

$y=2e^{-x}-e^{-2x}$y = 2e^(−x) − e^(−2x), the solution of $\frac{d^{2}y}{d{x}^2}+3\frac{dy}{dx}+2y=0$d²y/dx² + 3 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 $b^2-4ac=0$b² − 4ac = 0 the auxiliary equation has only one root, $m=-\frac{b}{2a}$m = −b/2a, and $e^{mx}$e^(mx) gives only one constant. A second solution is $x{e}^{mx}$x e^(mx).

To see why, differentiate: $\frac{dy}{dx}=\left(1+mx\right)e^{mx}$dy/dx = (1 + mx)e^(mx) and $\frac{d^{2}y}{d{x}^2}=\left(2m+m^{2}x\right)e^{mx}$d²y/dx² = (2m + m²x)e^(mx). Substituting, the left side is $\left[x\left(a{m}^2+bm+c\right)+\left(2am+b\right)\right]e^{mx}$[x(am² + bm + c) + (2am + b)]e^(mx). The first bracket is 0 because m is a root, and 2am + b is 0 because $m=-\frac{b}{2a}$m = −b/2a. So $x{e}^{mx}$x e^(mx) is a solution exactly when the root repeats.

The general solution is $y=\left(A+Bx\right)e^{mx}$y = (A + B x)e^(mx). For $\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}+y=0$d²y/dx² + 2 dy/dx + y = 0 the auxiliary equation is ${\left(m+1\right)}^2=0$(m + 1)² = 0, so m = −1 twice and $y=\left(A+Bx\right)e^{-x}$y = (A + B x)e^(−x). The start y = 1, $\frac{dy}{dx}=0$dy/dx = 0 gives A = 1 and B − A = 0, so $y=\left(1+x\right)e^{-x}$y = (1 + x)e^(−x).

The gold curve is $y=\left(1+x\right)e^{-x}$y = (1 + x)e^(−x), the solution of $\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}+y=0$d²y/dx² + 2 dy/dx + y = 0 from the same start. The dashed curve is $e^{-x}$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 $b^2-4ac<0$b² − 4ac < 0 the roots are a complex pair, $p\pm qi$p ± qi. The exponential still works: $e^{\left(p+qi\right)x}=e^{px}{e}^{iqx}$e^((p + qi)x) = e^(px)e^(iqx), and $e^{iqx}=\cos qx+i\sin qx$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 $e^{px}\cos qx$e^(px) cos qx and $e^{px}\sin qx$e^(px) sin qx. So $y=e^{px}\left(A\cos qx+B\sin qx\right)$y = e^(px)(A cos qx + B sin qx).

For $\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}+5y=0$d²y/dx² + 2 dy/dx + 5y = 0, $b^2-4ac=4-20=-16$b² − 4ac = 4 − 20 = −16, so $m=-1\pm 2i$m = −1 ± 2i and p = −1, q = 2 and $y=e^{-x}\left(A\cos 2x+B\sin 2x\right)$y = e^(−x)(A cos 2x + B sin 2x). From y = 1 and $\frac{dy}{dx}=0$dy/dx = 0 at x = 0, A = 1 and −A + 2B = 0, so B = ½.

The gold curve is $y=e^{-x}\left(\cos 2x+\frac{1}{2}\sin 2x\right)$y = e^(−x)(cos 2x + ½ sin 2x). It swings across the x-axis, first at $x\approx 1.02$x ≈ 1.02, and reaches −0.21 at $x=\frac{\pi }{2}$x = π/2. The dashed curves are $\pm \frac{1}{2}\sqrt{5}{e}^{-x}$±½√5 e^(−x): the cosine and sine together never exceed $\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}$√(1 + ¼) = √5/2, so the swing stays between them and shrinks as they do.

The sign of p

In $e^{px}\left(A\cos qx+B\sin qx\right)$e^(px)(A cos qx + B sin qx) 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 $b^2-4ac$b² − 4ac decides the shape of every solution. Positive gives two exponentials, zero gives $\left(A+Bx\right)e^{mx}$(A + B x)e^(mx), and negative gives an oscillation.

The usual mistakes

Changing a sign on the way to the auxiliary equation. $\frac{d^{2}y}{d{x}^2}-5\frac{dy}{dx}+6y=0$d²y/dx² − 5 dy/dx + 6y = 0 gives $m^2-5m+6=0$m² − 5m + 6 = 0, with roots 2 and 3, so the solution has $e^{2x}$e^(2x) and $e^{3x}$e^(3x), not $e^{-2x}$e^(−2x) and $e^{-3x}$e^(−3x).

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

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

Forgetting the $e^{px}$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 $\lambda$λ, because there m counts minutes: ${\lambda }^2+0.4\lambda +0.03=0$λ² + 0.4λ + 0.03 = 0 has roots −0.1 and −0.3.