Every solution of y′ + p·y = q is one particular solution y_p plus a solution y_h of the homogeneous equation y′ + p·y = 0. Adding is allowed because the left side is linear in y, so the sum gives q + 0; and nothing is missed, because the difference of two solutions solves the homogeneous equation. The constant in y_h is fitted to the starting value after the two parts are added, never before.
Core ideas
Three ideas: the two equations, why adding their solutions is legitimate, and why one particular solution is enough.
The homogeneous solution y_h solves y′ + p·y = 0, the equation with the forcing removed, and carries the constant C. The particular solution y_p solves the forced equation y′ + p·y = q and carries no constant. Any one particular solution will do; the lessons that follow find them by guessing a form and matching coefficients.
Substitute y_p + y_h: the left side is (y_p + y_h)′ + p·(y_p + y_h) = (y_p′ + p·y_p) + (y_h′ + p·y_h) = q + 0. That works for any p and any q, and it fails for nonlinear equations: the sum of two solutions of y′ = y² is not a solution.
If y is any solution at all, then y − y_p solves the homogeneous equation, because q − q = 0. So y − y_p = C/μ for some C, and every solution has the form y_p + C/μ. A different y_p changes only the constant that the starting value will fix.
The material
The two parts, the proof that their sum solves the equation and that nothing else does, how a particular solution is found by a guess, and the order in which the constant is fitted.
For y′ + p(t)·y = q(t), a particular solution y_p is any one function that satisfies it, and a homogeneous solution y_h is any function that satisfies y′ + p(t)·y = 0.
The general solution is y = y_p + y_h, where y_h = C/μ carries one arbitrary constant.
If y_p solves y′ + p·y = q and y_h solves y′ + p·y = 0, then y_p + y_h solves y′ + p·y = q. And if y is any solution, then y − y_p solves the homogeneous equation.
The derivative of a sum is the sum of the derivatives, so (y_p + y_h)′ + p·(y_p + y_h) = (y_p′ + p·y_p) + (y_h′ + p·y_h) = q + 0.
For the second claim, (y − y_p)′ + p·(y − y_p) = (y′ + p·y) − (y_p′ + p·y_p) = q − q = 0. Both steps use only that y and y′ appear once and to the first power.
□
Take y′ + 3y = 2t. The homogeneous equation is y′ + 3y = 0, with factor μ = e^(3t) from the last lesson, so y_h = C·e^(−3t). A particular solution has to be found, and a guess whose form is borrowed from the forcing is the standard way.
The forcing 2t is a polynomial of degree one, so try a polynomial of degree one, y_p = A·t + B. Substituting: A + 3·(A·t + B) = 2t, which is 3A·t + (A + 3B) = 2t. Matching the coefficients of t and of 1 gives 3A = 2 and A + 3B = 0, so A = 2/3 and B = −2/9.
The check is the last line, and it is exact. The guess carried a constant term B even though the forcing 2t has none, for the reason the next lesson makes into a rule: with A·t alone the equations would be 3A = 2 and A = 0, which no A satisfies. The derivative of A·t is a constant, and something in the guess has to cancel it. “Polynomial Forcing and Undetermined Coefficients” turns the guess into a method.
For y′ + 2y = 6 try a constant, y_p = A. Then 0 + 2A = 6 gives A = 3, which is the steady state from “Constant Coefficients and the Steady State”.
The gap (y₀ − 3)·e^(−2t) found there is exactly the homogeneous part, C·e^(−2t) with C = y₀ − 3. That lesson was this one for a constant q.
Add the two parts first: y = (2/3)·t − 2/9 + C·e^(−3t). Only then put in the starting value, because the starting value is a statement about the whole solution. With y(0) = 1: −2/9 + C = 1, so C = 11/9. The −2/9 is the value of the particular part at t = 0, and it has to be subtracted before C is read off.
For y′ + 2y = e^t with y(0) = 0 try y_p = A·e^t. Then A·e^t + 2A·e^t = e^t gives 3A = 1, so y_p = (1/3)·e^t.
The solution is (1/3)·e^t + C·e^(−2t), and y(0) = 1/3 + C = 0 gives C = −1/3. The lab shows the three curves in its parts view: the homogeneous part, the particular part, and their sum.
Because the two parts are separate, the lab can draw them separately. In the parts view the homogeneous curve decays on its own, the particular curve is the forced response the starting value cannot change, and the solution is their sum. A different particular solution would give a different split and the same sum.
Interactive
μ, y_h, y_p and the constant, exactly
Method
Solve the easy equation, find one solution of the forced one, add, and fit the constant last.
Set q = 0 and use y_h = C/μ. For p = 3 that is C·e^(−3t).
Guess a form with the same shape as q, with unknown constants, substitute, and match coefficients. Any single solution will do.
Substitute y_p alone into y′ + p·y and read q. If it does not give q, the guess was wrong, however neat the constants look.
y = y_p + C/μ. Put in y(t₀), subtract the particular part's value at t₀, and divide by 1/μ(t₀).
The finished y must give the starting value at t₀ and must satisfy the original equation. Both are one line.
The homogeneous equation has the factor e^(3t), so its solution is known at once. The particular solution is a degree-one polynomial.
The order matters. The constant 11/9 is not 1, and the difference is exactly the 2/9 that the particular part contributes at t = 0. Reading C off the homogeneous part alone would have set it to 1 and made y(0) equal to 7/9.
For a rehearsal, take the third preset, y′ + 2y = e^t with y(0) = 0, and write down y_h, y_p and C before pressing the button. The particular solution is a multiple of e^t; find the multiple by substitution, then compute the constant from the sum.
You should be able to solve the homogeneous equation, find one particular solution by a guess and matched coefficients, show that the sum solves the forced equation and that the difference of two solutions solves the homogeneous one, and compute the constant from a starting value after the two parts are added.
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