If two functions solve a·y″ + b·y′ + c·y = 0 then so does every combination of them, which is why the family has constants at all. Whether the combinations can fit any starting value and rate is decided by one number, the Wronskian of the two solutions at t = 0: when it is not zero, every start can be fitted, and when it is zero, some cannot.
Core ideas
Three ideas: why combinations of solutions are solutions, why two solutions may still not be enough, and the number that tells the difference.
The equation is linear with zero on the right: y″, y′ and y each appear alone and to the first power. So the left side of a sum is the sum of the left sides, and a constant multiple is carried through. If the left side is zero for y₁ and for y₂ it is zero for C₁·y₁ + C₂·y₂.
The functions e^(−t) and 2·e^(−t) both solve y″ + 3y′ + 2y = 0, yet their combinations are all multiples of e^(−t), a family with one real constant in it. Its starting rate is always −1 times its starting value, so a start with y(0) = 1 and y′(0) = 0 is out of reach.
The pair of equations for C₁ and C₂ has the coefficients y₁(0), y₂(0) and y₁′(0), y₂′(0). Their determinant is y₁(0)·y₂′(0) − y₁′(0)·y₂(0), which is the Wronskian W = y₁·y₂′ − y₁′·y₂ evaluated at 0. Not zero means exactly one fit for every start.
The material
The superposition rule and its proof, a pair of solutions that is not enough, the Wronskian, and the four lab presets.
If y₁ and y₂ solve a·y″ + b·y′ + c·y = 0, then C₁·y₁ + C₂·y₂ solves it for every pair of constants C₁ and C₂.
The rate of a sum is the sum of the rates, and a constant factor comes out of a rate, so for y = C₁·y₁ + C₂·y₂ the combination a·y″ + b·y′ + c·y splits into C₁·(a·y₁″ + b·y₁′ + c·y₁) + C₂·(a·y₂″ + b·y₂′ + c·y₂).
Each bracket is zero because y₁ and y₂ are solutions, so the whole is C₁·0 + C₂·0 = 0. The proof used two things: that the equation is linear, and that its right side is zero.
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The second condition matters. The equation y″ + y = 1 is linear, and y = 1 solves it, since y″ = 0 and 0 + 1 = 1. The double y = 2 does not: 0 + 2 = 2, not 1. Superposition belongs to equations whose right side is zero.
Take y₁ = e^(−t) and y₂ = 2·e^(−t) for y″ + 3y′ + 2y = 0. Both are solutions, and so is every combination C₁·y₁ + C₂·y₂ = (C₁ + 2·C₂)·e^(−t). The starting value is y(0) = C₁ + 2·C₂ and the starting rate is y′(0) = −C₁ − 2·C₂, which is minus the first. The start y(0) = 1, y′(0) = 0 would need C₁ + 2·C₂ to be 1 and 0 together.
For two functions y₁ and y₂, the Wronskian is W(t) = y₁(t)·y₂′(t) − y₁′(t)·y₂(t). Its value at t = 0 is the determinant of the coefficients of the pair of equations that fit C₁ and C₂.
For the pair above, y₁(0) = 1, y₂(0) = 2, y₁′(0) = −1 and y₂′(0) = −2. The determinant is 1·(−2) − 2·(−1) = 0, and the Wronskian is zero for every t, because the second function is a multiple of the first.
If y₁ and y₂ solve a·y″ + b·y′ + c·y = 0 and W(0) ≠ 0, then for every y(0) and y′(0) there is exactly one pair C₁, C₂ with C₁·y₁ + C₂·y₂ starting there.
The equations are C₁·y₁(0) + C₂·y₂(0) = y(0) and C₁·y₁′(0) + C₂·y₂′(0) = y′(0). Multiply the first by y₂′(0) and the second by y₂(0) and subtract, and the C₂ terms cancel: W(0)·C₁ = y₂′(0)·y(0) − y₂(0)·y′(0).
In the same way W(0)·C₂ = y₁(0)·y′(0) − y₁′(0)·y(0). If W(0) is not zero both can be divided by it, which gives one pair of constants and no other. That the combination is then the solution with that start is the fact that two starting values fix one solution, which the course uses and does not prove.
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The lab's four presets give the pair of solutions that the roots determine and print W(0). For y″ + 3y′ + 2y = 0 the pair is e^(−t) and e^(−2t), and the worked example below computes it. For y″ + 4y = 0 the pair is cos(2t) and sin(2t), and W is 2 for every t. For y″ + 4y′ + 4y = 0 the pair is e^(−2t) and t·e^(−2t), and W = e^(−4t), which is 1 at 0.
For two exponentials W(0) is the second root minus the first, with the larger root first, and a repeated root would make it zero: that is why the repeated case needed the second function t·e^(rt). For a complex pair W(0) is β. The fourth preset, y″ − y′ − y = 0, has the roots (1 ± √5)/2, which differ by √5, and the lab prints −√5 as an exact surd, not a rounded number; the sign is the order of the two roots. In all three computed cases W(t) = W(0)·e^(−(b/a)·t), so a Wronskian that starts nonzero never reaches zero; this is stated as a pattern, not proved.
The sign of W(0) carries no information. Swapping the roles of y₁ and y₂ reverses it, and what the guarantee needs is only that it is not zero.
Interactive
Roots, the general solution and the constants, exactly
Method
Five moves. The determinant is the only new computation.
From the roots: two exponentials, or e^(rt) and t·e^(rt), or e^(αt) with cosine and with sine.
You need y₁′ and y₂′, and the product rule where a factor of t or an exponential is multiplied.
Keep the order: first function times the rate of the second, minus the rate of the first times the second.
Exponentials become 1, cosines 1, sines 0, and t becomes 0.
Not zero: every start can be fitted, by exactly one pair. Zero: find the relation between y(0) and y′(0) that the pair can reach.
The Wronskian of the two exponentials, and the determinant it equals.
The determinant of the pair of equations is −1, the same as W(0), so the pair has exactly one solution for every right side. Adding the two equations gives C₂ = −(y(0) + y′(0)), then C₁ = 2·y(0) + y′(0); for y(0) = 1 and y′(0) = 0 that is C₁ = 2, C₂ = −1.
The value is negative, and nothing turns on that: with the two functions in the other order it would be +1.
You should be able to show that a combination of solutions is a solution, compute y₁·y₂′ − y₁′·y₂ at t = 0 exactly, say that a nonzero value means every starting value and rate can be fitted, and give a pair for which it cannot.
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