Whatever kind of root the equation has, the starting value and the starting rate give two linear equations for the two constants, because at t = 0 every exponential is 1, cos is 1 and sin is 0. The equations differ only in a few entries, and they are solved by elimination with exact fractions. When the roots are irrational the constants would be surds, and the lab declines to fit them and says so.
Core ideas
Three ideas: each starting number is one equation, the three kinds of root change only a few entries of the pair, and a dash from the lab has a specific meaning.
The general solution is a formula in t, C₁ and C₂. Putting t = 0 in it gives y(0), and putting t = 0 in its rate gives y′(0). Each is a linear expression in C₁ and C₂, so two starting numbers give two linear equations in two unknowns.
That is why the pair is short. For two real roots the first equation is C₁ + C₂ = y(0); for a repeated root it is C₁ = y(0); for a complex pair it is again C₁ = y(0). The second equation carries the roots, because the rate of e^(rt) at 0 is r.
When the roots contain a square root the constants do too, and the lab fits only rational constants. It prints a dash and gives the reason in its status line. The two equations still have an exact solution; it is a number the lab does not print.
The material
The pair for each kind of root, a general solution of the real case by elimination, a complex fit, and the case the lab declines.
Every earlier lesson on this course ended by fitting two constants, and each did it in one line. This lesson shows that the lines are one computation. The unknowns are C₁ and C₂, the data are the starting value y(0) and the starting rate y′(0), and the equations come from putting t = 0 in the general solution and in its rate.
Read the second line of each pair as “the rate at the start”. For two exponentials it adds the roots in proportion to the constants. For the repeated root, the product rule on (C₁ + C₂·t)·e^(rt) at t = 0 gives r·C₁ + C₂. For the complex pair, the product rule on the exponential and the cosine and sine at t = 0 gives α·C₁ + β·C₂.
Multiply the first equation by r₂ and subtract it from the second, then multiply the first by r₁ and subtract the second from it. The unknowns separate.
The factor r₁ − r₂ is why the roots have to differ. If they are different it is not zero and may be divided out, so there is exactly one pair C₁, C₂ for every starting value and rate. If they are equal the two exponentials are one function, and the repeated-root form takes over.
For y″ + 3y′ + 2y = 0 the roots are r₁ = −1 and r₂ = −2, so r₁ − r₂ = 1. With y(0) = 0 and y′(0) = 1, the first formula gives C₁ = (1 − 0)/1 = 1 and the second C₂ = (0 − 1)/1 = −1.
The solution is e^(−t) − e^(−2t). It starts at 0 and is positive afterwards, with a starting rate of 1: the rate at the start is not the rate of the number 0.
For y″ + 2y′ + 5y = 0 the roots are −1 ± 2i, so the pair is C₁ = y(0) and −C₁ + 2·C₂ = y′(0). Take y(0) = 2 and y′(0) = 0. Then C₁ = 2 and −2 + 2·C₂ = 0, so C₂ = 1, and the solution is e^(−t)·(2·cos(2t) + sin(2t)). Its rate at 0 is zero because two effects cancel exactly: the exponential pulls it down by 2 and the sine pushes it up by 2. The rate is not zero because y(0) happens to be a constant.
For y″ − y′ − y = 0 the roots are (1 ± √5)/2. The two equations are exactly as above, and they have a solution; by hand it is C₁ = (5 − √5)/10 and C₂ = (5 + √5)/10 for y(0) = 1 and y′(0) = 0, and the first equation checks, since the two add to 1. The constants contain √5, though, and the lab fits rational constants only. It prints a dash in the constants tile and the solution tile and says in its status line that the roots are irrational.
That is a limit of the arithmetic. The lab differentiates and substitutes solutions whose exponents and constants are fractions, which it can do without rounding; a surd in an exponent or a constant would have to be rounded, and then the residual would be a rounded number instead of a zero. A complex pair whose β is a surd is declined for the same reason, as in y″ + 3y = 0 with the roots ±√3·i, and the status line names the irrational frequency. The method is the same for any roots, and a solution with irrational constants is a perfectly good one.
Interactive
Roots, the general solution and the constants, exactly
Method
Five moves, the same for every kind of root.
Two exponentials, (C₁ + C₂·t)·e^(rt), or e^(αt) with a cosine and sine. The lab names the kind.
That gives the first equation. The exponentials become 1, the cosine 1 and the sine 0.
Differentiate once, by the product rule where t or an exponential multiplies something. This gives the second equation, which equals y′(0).
Keep the fractions exact. Check that the number you divide by is not zero.
Put t = 0 into the answer and into its rate, and read the status line of the lab for the residual.
A complex pair, the pair of equations, and the answer checked at the start.
The fourth line is the rate of the general solution at t = 0. A wrong sign on the −C₁ shows up in the last line as a starting rate that is not 0.
The lab prints the same solution expanded, as 2·e^(−t)·cos(2t) + e^(−t)·sin(2t). The status line also reports the residual of the whole solution, which is exactly 0; the check on the last line covers the starting values, and the residual covers the equation.
You should be able to set t = 0 in the general solution and in its rate, solve the resulting pair exactly, check both starting values, and say why irrational roots make the lab decline.
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