When the discriminant is negative the characteristic quadratic has a complex pair α ± βi, and the solutions of the equation are real functions all the same: an exponential e^(αt) multiplied by a cosine and a sine of βt. The real part α decides whether the oscillation shrinks, stays level or grows, and β decides how fast it turns. The lab fits the two constants and checks the result by substitution.
Core ideas
Three ideas: what a complex root hands over, why the answer is still a real function, and what the two parts of the root mean for the motion.
The quadratic formula gives r = (−b ± √(b² − 4ac))/(2a), and when the discriminant is negative the square root is β times i for a real β. The two roots are then α + βi and α − βi, with the real part α = −b/(2a) in both. Only two real numbers, α and β, are needed to write the solution.
Both e^(αt)·cos(βt) and e^(αt)·sin(βt) satisfy a·y″ + b·y′ + c·y = 0, and the lab checks each by substitution with no i in sight. The general solution has a constant on each. It is real whenever the constants are real.
The factor e^(αt) multiplies everything: it shrinks the oscillation when α is negative, leaves its size alone when α is zero and grows it when α is positive. The cosine and sine turn once every 2π/β units of t, so a larger β is a faster oscillation.
The material
The form of the answer, a proof that it works with algebra the reader has, a hand check, and the three shapes the real part can give.
Take y″ + 2y′ + 5y = 0. The discriminant is 4 − 20 = −16, so the roots are (−2 ± 4i)/2 = −1 ± 2i. There is no real number that solves the characteristic quadratic, and it is tempting to conclude that no real function solves the equation. The conclusion does not follow: the roots are numbers that satisfy a quadratic, and the solutions are functions of t built from them.
The reason a cosine and a sine appear is that an exponential with an imaginary exponent is a cosine plus i times a sine. This course does not use that identity, because the answer can be checked directly, which is the standard throughout: a candidate is a solution when its residual is zero.
If a·r² + b·r + c = 0 has the roots α ± βi with α and β real and β not zero, then e^(αt)·cos(βt) and e^(αt)·sin(βt) both solve a·y″ + b·y′ + c·y = 0, and the general solution is e^(αt)·(C₁·cos(βt) + C₂·sin(βt)).
Take y = e^(αt)·cos(βt). The product rule, with the chain rule supplying the factor β in front of the sine, gives y′ = e^(αt)·(α·cos(βt) − β·sin(βt)) and y″ = e^(αt)·((α² − β²)·cos(βt) − 2·α·β·sin(βt)). Then a·y″ + b·y′ + c·y equals e^(αt) times (a·(α² − β²) + b·α + c)·cos(βt) − β·(2a·α + b)·sin(βt).
The root pair has real part α = −b/(2a), so 2a·α + b = 0 and the sine term vanishes. It also has product α² + β² = c/a, since (α + βi)·(α − βi) = α² + β². With b = −2a·α the cosine bracket is a·α² − a·β² − 2a·α² + c = c − a·(α² + β²), which is 0.
The sine case is the same computation with the roles of the two terms exchanged. That these two functions are all the solutions is stated here and not proved; “Superposition and the Wronskian” shows why they can fit any starting value and rate.
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For y″ + 2y′ + 5y = 0 the roots give α = −1 and β = 2. Take the sine solution y = e^(−t)·sin(2t). Its rates, by the product rule, are below, and the cosine terms and the sine terms of the residual each cancel exactly.
The same equation has the envelope e^(−t) and turns at speed 2, so one full turn takes π units of t. The lab's three presets have three different real parts. The first, α = −1, shrinks. The second is y″ + 4y = 0, with α = 0: it neither shrinks nor grows and keeps its size for ever. The third has α = 1 and grows.
For y″ − 2y′ + 2y = 0 the discriminant is 4 − 8 = −4 and the roots are 1 ± i, so α = 1 and β = 1. The general solution is e^t·(C₁·cos t + C₂·sin t).
With y(0) = 1 and y′(0) = 0 the first condition gives C₁ = 1 and the second is α·C₁ + β·C₂ = 1 + C₂ = 0, so C₂ = −1. The solution turns once every 2π and its size multiplies by e each time t increases by 1.
That last sentence is the whole content of the real part. A spring that loses energy to friction is described by a solution with a negative α, and one that is driven harder at every swing has a positive α. The next course measures both.
Interactive
Roots, the general solution and the constants, exactly
Method
Five moves. The first three are about the root, and the last two are the fit and the check.
b² − 4ac < 0. Then there is no real root, and the roots come as a pair.
α = −b/(2a) is the real part and β is the positive number in front of i. For −1 ± 2i they are −1 and 2.
e^(αt)·(C₁·cos(βt) + C₂·sin(βt)). Keep the exponential factor; it is what the real part contributes.
C₁ = y(0), and α·C₁ + β·C₂ = y′(0) gives C₂. The division by β is the step most often skipped.
Differentiate twice by the product rule and read the residual. A missing factor of e^(αt) fails here at once.
A complex pair, the general real solution, and a two-condition fit.
The fourth line is the first condition, since cos 0 = 1 and sin 0 = 0. The fifth is the rate of the general solution at t = 0, which picks up α·C₁ from the exponential and β·C₂ from the sine.
The solution starts at 0, rises, and turns back as the factor e^(−t) shrinks the swing. The lab prints the residual 0 in its status line, so the answer is exact and not an approximation.
You should be able to compute a negative discriminant, read α and β from the roots, write e^(αt)·(C₁·cos(βt) + C₂·sin(βt)), fit C₁ and C₂ from y(0) and y′(0), and say whether the oscillation shrinks, stays level or grows.
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