Second-Order Linear Equations with Constant Coefficients
Try an exponential and the equation becomes a quadratic in the exponent. Real, repeated and complex roots give three shapes of general solution.
What a learner can do afterwards
- Write and solve the characteristic equation
- Give the general solution in each of the three root cases
- Convert a complex-root solution into a decaying sine and cosine form
1 · Read
Constant coefficient equations give way to one guess: try y equals e to the r x and see which r work. Substitution turns the differential equation into the characteristic polynomial, a plain quadratic whose roots decide the shape of the solution. The general solution combines two independent pieces with constants C1 and C2, and your starting position plus starting velocity fix those constants. Newton is why you care: springs, circuits, and pendulums all lead here.
Two distinct real roots give two exponentials, side by side. A repeated root adds a factor of x, because the second piece must stay independent of the first. A complex pair becomes an oscillation wrapped in an exponential, which you rewrite with sine and cosine. Take y double prime plus 4 y prime plus 4 y equals 0: the polynomial is r squared plus 4 r plus 4, which is r plus 2 squared, so r is minus 2 twice and the answer is C1 plus C2 times x, all times e to the minus 2 x.
Distinct real roots r1 and r2 give y = C1 e^(r1 x) + C2 e^(r2 x). A repeated root r gives y = (C1 + C2 x) e^(rx). A complex pair a ± bi gives, by Euler's formula, e^(ax)(cos bx + i sin bx), and the real solutions are y = e^(ax)(C1 cos bx + C2 sin bx). The discriminant of the characteristic quadratic picks the case: positive, zero or negative. When a is negative the oscillation decays.
Fit the constants. Solve y'' - y' - 6y = 0 with y(0) = 1 and y'(0) = 8. The characteristic equation r² - r - 6 = 0 factorises as (r - 3)(r + 2) = 0, so y = C1 e^(3x) + C2 e^(-2x). Position: y(0) = C1 + C2 = 1. Velocity: y' = 3C1 e^(3x) - 2C2 e^(-2x), so y'(0) = 3C1 - 2C2 = 8. Solving, C1 = 2 and C2 = -1, so y = 2e^(3x) - e^(-2x). Check the template before fitting: the wrong template can never meet both conditions.
Guess the exponential, read the roots, pick the matching template, then fit the two constants.
Readings 'One guess cracks the equation', 'Why second order, and why two conditions' and 'The three templates with examples' are original to this dataset. All 6 extra questions are original to this dataset.
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14 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.