A constant-coefficient initial value problem is solved in four moves: transform both sides with the derivative rules, solve the resulting algebraic equation for Y(s), take Y apart into table forms by partial fractions, and read the table backwards to get y(t). The initial values enter at the first move, so no constants are fitted at the end. The lab runs the whole chain in exact fractions and substitutes the answer back into the equation.
Core ideas
Three ideas: the equation turns into algebra, the initial values arrive with it, and the inverse is the table read from right to left.
Transform every term of the equation. The derivative rules turn y′ and y″ into multiples of Y plus terms that are already known. What remains is (a·s² + b·s + c)·Y = something, where the left coefficient is the characteristic polynomial of the equation, and the right side is made of the initial values and the transform of the forcing.
The corrections in the derivative rules are the initial values, so the right side already contains them. Dividing by the characteristic polynomial gives Y(s) as one fraction, and its inverse is the solution to this initial value problem, with no constants left to fit.
A fraction like (s + 3)/((s + 1)(s + 2)) is not in the table. Taking it apart into A/(s + 1) + B/(s + 2) gives two entries that are, and the table read from right to left returns A·e^(−t) + B·e^(−2t). When the denominator has an irreducible quadratic, complete the square: (s + 1)² + 4 is the form the sine and cosine entries have.
The material
The transformed equation in general, a first-order case in full, and a second-order case with real roots and one with complex roots.
Take the equation a·y″ + b·y′ + c·y = q(t) with starting values y(0) and y′(0). Transform each term with the rules of the previous lesson and collect everything that multiplies Y.
The polynomial a·s² + b·s + c multiplying Y is the characteristic polynomial of the equation, with s in place of r. That is the first connection to the earlier method, and the next lessons make it precise. For a first-order equation y′ + a·y = q the same steps give (s + a)·Y = y(0) + ℒ[q].
Solve y′ + 2y = 6 with y(0) = 0. The transform of the constant 6 is 6/s, so (s + 2)·Y = 0 + 6/s, and Y = 6/(s·(s + 2)).
Partial fractions: 6/(s·(s + 2)) = A/s + B/(s + 2). At s = 0 the factor s + 2 is 2, so A = 6/2 = 3. At s = −2 the factor s is −2, so B = 6/(−2) = −3. Then y = 3 − 3e^(−2t), which starts at 0 and settles at 3.
For y″ + 3y′ + 2y = 0 with y(0) = 1 and y′(0) = 0 the right side is s·1 + 0 + 3·1 = s + 3, and the left coefficient is s² + 3s + 2 = (s + 1)(s + 2). The forcing is zero, so nothing else contributes.
To find A, multiply both sides by s + 1 and set s = −1: A = (−1 + 3)/(−1 + 2) = 2. To find B, multiply by s + 2 and set s = −2: B = (−2 + 3)/(−2 + 1) = −1. So Y = 2/(s + 1) − 1/(s + 2), and reading the table backwards gives y = 2e^(−t) − e^(−2t).
This is the answer to the real-roots example in “Real Distinct Roots” in Second-Order Linear Equations, found there by fitting two constants from a 2×2 system. Here the starting values went in at the first line and the constants came out of two substitutions.
For y″ + 2y′ + 5y = 0 with y(0) = 0 and y′(0) = 2 the right side is s·0 + 2 + 2·0 = 2 and the left coefficient is s² + 2s + 5, which has no real roots. Complete the square: s² + 2s + 5 = (s + 1)² + 4. Then Y = 2/((s + 1)² + 4).
The sine entry is b/(s² + b²), and replacing s by s + 1 multiplies the signal by e^(−t), as in the table's last entries. With b = 2 the numerator is already the 2 the entry needs, so y = e^(−t)·sin(2t). The lab checks this by transforming the answer back and reading its residual in the original equation.
Two limits of the lab, stated where they matter. A linear factor of the denominator must have a rational root, so s² − 2, whose roots are ±√2, is refused with the factor named. An irreducible quadratic must have a rational frequency, so s² + 2, whose inverse would carry cos(√2·t), is refused the same way rather than printed with a surd inside a cosine. The mathematics is still fine in both cases; the lab is limited.
Interactive
Transforms, partial fractions and solutions, exactly
Method
Transform, solve, decompose, invert, check. The check is the step that finds a slip in any of the others.
Replace each derivative by its rule, with the starting values written in. Transform the forcing from the table.
Move every term that contains Y to the left and factor it out. Divide by the characteristic polynomial. Y is now a single fraction in s.
Factor the denominator. Write one term for each linear factor, and complete the square for a quadratic that does not factor. Find the coefficients by substituting the roots.
Replace each fraction by the signal that has it as its transform, and add the signals.
Put the answer into the original equation and its starting values. The residual must be zero, and the answer must start where the problem says.
The first preset, solved by hand and then read against the lab.
Check the starting values: y(0) = 2 − 1 = 1 and y′(0) = −2 + 2 = 0. Check the equation: y″ + 3y′ + 2y has the e^(−t) coefficients 2 − 6 + 4 = 0 and the e^(−2t) coefficients −4 + 6 − 2 = 0. The lab reports the same residual.
Notice what was not done. No general solution with two constants was written and no pair of equations for them was solved. The starting values were in the first line.
You should be able to write the transformed equation with the starting values inside it, solve for Y(s), decompose it by substituting the roots, read the table backwards, and verify the answer by its two starting values and by substitution into the equation.
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