When the characteristic quadratic has two different real roots, each gives an exponential solution, and the general solution is a combination of the two with a constant in front of each. The two constants are fitted from the starting value and the starting rate, and the lab checks the fitted solution by substitution. Whether the motion dies away, runs off or settles is decided by the signs of the roots, before any constant is found.
Core ideas
Three ideas: the general solution has both exponentials and a constant on each, the constants come from a two-by-two fit, and the roots' signs forecast the motion.
A root r of a·r² + b·r + c = 0 makes e^(rt) a solution. If there are two roots, C₁·e^(r₁t) + C₂·e^(r₂t) is a solution for every C₁ and C₂, since combinations of solutions are solutions. The constants are what make it general.
At t = 0 the exponentials are both 1, so y(0) = C₁ + C₂, and the rate is y′(0) = r₁·C₁ + r₂·C₂. Two equations in two unknowns have exactly one solution whenever r₁ ≠ r₂, because the equations then disagree about the proportions of C₁ and C₂.
A negative root gives a term that shrinks to zero, a positive root gives a term that grows without limit, and a root of zero gives a constant. If both roots are negative every solution dies away; if one is positive, a solution grows unless the constant on that term is exactly zero.
The material
The form, the fit, the exact check, and a case where a growing root is present and the solution decays anyway.
The three kinds of discriminant give three shapes of answer. A positive discriminant gives two distinct real roots r₁ and r₂, and the answer is a combination of the two exponentials. The word “general” matters: the form carries both constants, and a particular motion is picked out by giving them values.
If a·r² + b·r + c = 0 has two different real roots r₁ and r₂, then every solution of a·y″ + b·y′ + c·y = 0 has the form y = C₁·e^(r₁t) + C₂·e^(r₂t), and every such function is a solution.
That every such function is a solution is the combination rule from “The Second-Order Equation” applied to two exponentials, and the lab checks it case by case. That there are no other solutions is the part this course states and does not prove in full; “Superposition and the Wronskian” shows why two constants always suffice to fit a starting value and rate, and that is the part you can use.
Take y″ + 3y′ + 2y = 0. The roots are −1 and −2, so y = C₁·e^(−t) + C₂·e^(−2t). At t = 0 the exponentials are 1, so y(0) = C₁ + C₂. The rate is y′ = −C₁·e^(−t) − 2·C₂·e^(−2t), so y′(0) = −C₁ − 2·C₂. With y(0) = 1 and y′(0) = 0 the equations are C₁ + C₂ = 1 and −C₁ − 2·C₂ = 0.
Checking by substitution is what makes the answer trustworthy. From y = 2·e^(−t) − e^(−2t) the rates are y′ = −2·e^(−t) + 2·e^(−2t) and y″ = 2·e^(−t) − 4·e^(−2t). Then y″ + 3y′ + 2y has the e^(−t) terms 2 − 6 + 4 = 0 and the e^(−2t) terms −4 + 6 − 2 = 0. The lab prints the same residual, 0, in its status line, and also checks that both starting values hold.
With roots −1 and −2 every solution is a sum of two terms that shrink, so every solution tends to 0. With roots 2 and 3, as in y″ − 5y′ + 6y = 0, both terms grow. With roots −1 and 1, as in y″ − y = 0, one term shrinks and one grows, and the growing one wins as soon as its constant is not zero.
For y″ − y = 0 with y(0) = 1 and y′(0) = −1, the equations are C₁ + C₂ = 1 and C₁ − C₂ = −1. Adding gives 2·C₁ = 0, so C₁ = 0 and C₂ = 1, and the solution is y = e^(−t), which decays.
Change the starting rate to y′(0) = −99/100 and C₁ = 1/200 is not zero, so the term in e^t is present and, though it starts tiny, it grows past everything. A decaying solution here is balanced on one exact value.
That balance is why the signs forecast the motion only up to a constant being exactly zero. For a starting value and rate chosen from measured data that never happens, so the safe forecast is that a positive root means growth.
Interactive
Roots, the general solution and the constants, exactly
Method
Five moves. The fit comes last, and the check after it.
Solve a·r² + b·r + c = 0. Confirm the discriminant is positive, so that there are two different real roots.
y = C₁·e^(r₁t) + C₂·e^(r₂t), with a constant in front of each exponential. The lab puts the larger root first, so C₁ goes with the larger root.
C₁ + C₂ = y(0) and r₁·C₁ + r₂·C₂ = y′(0). The second comes from the rate of the general solution, taken at t = 0.
Eliminate one constant. Keep the answers as fractions.
Differentiate twice and put the result into the equation. The residual is 0 or something was mis-copied.
Two real roots, a two-by-two fit, and a substitution as the check.
The second equation is the rate of the general solution at t = 0: y′ = −C₁·e^(−t) − 2·C₂·e^(−2t). A wrong sign there is the most common slip, and the check at the end catches it.
Both roots are negative, so the solution decays to zero; the term −e^(−2t) goes faster, and the 2·e^(−t) is left.
You should be able to write C₁·e^(r₁t) + C₂·e^(r₂t), set up and solve the two equations for the constants from y(0) and y′(0), verify the result by substituting it, and say which solutions decay, grow or settle.
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