Complex roots · Differential Equations

Intuition

When the discriminant is negative the roots are a complex pair, and their exponentials are complex-valued. Euler's formula rewrites them, and their real and imaginary parts are two real solutions: an exponential times a cosine, and the same exponential times a sine. The imaginary part of the root is a frequency, and the real part decides whether the oscillation grows, holds or dies.

A playground swing left to itself. It goes back and forth, and friction makes each swing a little smaller than the last: an oscillation inside a shrinking envelope. The imaginary part of the root is how fast it swings; the real part is how fast the envelope shrinks.

plane diagram: an-oscillation-inside-an-envelope

A solution of $y''+\tfrac12y'+\tfrac{65}{16}y=0$ , whose roots are $-\tfrac14\pm2i$ : it is $e^{-x/4}\cos2x$ , an oscillation of frequency $2$ inside the envelope $\pm e^{-x/4}$ set by the real part.

Real solutions from a complex pair

If $ar^{2}+br+c=0$ has roots $\alpha\pm i\beta$ with , the general real solution of is . Euler's formula, taken here as known, gives , and because the equation has real coefficients the real and imaginary parts of a complex solution are real solutions.

r=\alpha\pm i\beta\ \Longrightarrow\ y=e^{\alpha x}\bigl(C_{1}\cos\beta x+C_{2}\sin\beta x\bigr)

Reading an oscillation

$y''+2y'+5y=0$ : , so .

Real and imaginary parts of a complex solution are solutions

L[u+iv]=0,\ u,v\ \text{real}\ \Longrightarrow\ L[u]=0,\ L[v]=0

The operator is linear and its coefficients are real, so applied to u plus i v it gives L of u plus i times L of v, with L of u and L of v both real functions. A complex number is zero only when its real and imaginary parts are both zero. So each part solves the equation on its own.

Proof steps

\begin{aligned}e^{(\alpha+i\beta)x}&=e^{\alpha x}\cos\beta x+i\,e^{\alpha x}\sin\beta x\\&=u+iv\end{aligned}

Euler's formula splits the complex solution into a real part and an imaginary part.

L[u+iv]=L[u]+i\,L[v]

L is linear with real coefficients, so L of u and L of v are real functions.

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

The complex exponential solves the equation, because its rate is a root.

L[u]=0,\qquad L[v]=0

A complex number is zero only if both its parts are, so both parts are real solutions.

Applications

Practice

A Complex Pair Gives Sines and Cosines

For roots $\alpha\pm i\beta$ , the real part becomes an exponential envelope and the imaginary part a frequency.

y=e^{\alpha x}\bigl(C_{1}\cos\beta x+C_{2}\sin\beta x\bigr)

Try it

What is the general solution of $y''+2y'+5y=0$ ?