The left side of y′ + p·y = q is almost the derivative of something and not quite. Multiply by the function μ whose rate is p times itself and it becomes exactly (μ·y)′, the product rule read backwards, so the equation turns into (μ·y)′ = μ·q and can be integrated. When p = a/t the factor is the power tᵃ, and the whole solution is a calculation in fractions.
Core ideas
Three ideas, in the order they are used: what the factor must do, what it therefore is, and what the equation becomes once it is applied.
The product rule says (μ·y)′ = μ·y′ + μ′·y. The left side of the equation is y′ + p·y. Multiply it by μ and it is μ·y′ + p·μ·y. These agree exactly when μ′ = p·μ, so the factor is whichever function satisfies that one condition.
For a constant p = a that function is e^(a·t). For p = a/t it is tᵃ, because the power rule gives (tᵃ)′ = a·t^(a − 1) = (a/t)·tᵃ. In general it is e^(∫p dt), where ∫p dt is one antiderivative of p; the exponent is the antiderivative of p, not p itself.
The equation becomes (μ·y)′ = μ·q. Integrate the right side and add a constant: μ·y = ∫μ·q dt + C. Then divide by μ, including the constant, and fit C to the starting value last.
The material
The condition on the factor, the check that proves it works, the two families of factor the lab handles, a full solution in fractions, and why the constant in the exponent does not matter.
As it stands the left side y′ + p·y is not the derivative of anything you can write down. y′ alone is the derivative of y, and the extra term p·y spoils it. A product μ·y has the derivative μ·y′ + μ′·y, which has the same two-term shape; the only question is whether a μ exists that makes the second term p·μ·y.
If μ′ = p·μ, then (μ·y)′ = μ·(y′ + p·y) for every differentiable y. So y′ + p·y = q implies (μ·y)′ = μ·q.
The product rule, verified exactly on polynomials in Rates of Change and the Derivative, gives (μ·y)′ = μ·y′ + μ′·y.
Replace μ′ by p·μ: μ·y′ + p·μ·y = μ·(y′ + p·y). If y′ + p·y = q the right side is μ·q. Only the product rule and the one condition on μ were used.
□
Two families of p need nothing else. For a constant p = a, μ = e^(a·t), since the rate of e^(a·t) is a times itself, a claim from the earlier courses. For p = a/t with a whole number a, μ = tᵃ, and the check is the power rule alone. It is exact, and the lab prints the result without a logarithm.
For the general formula μ = e^(∫p dt) the chain rule, verified on polynomials in Rates of Change and the Derivative and applied here to the exponential as a stated claim, gives μ′ = (∫p dt)′·e^(∫p dt) = p·μ, which is the condition. The exponent has to be an antiderivative of p. Using p itself fails at once: for the constant p = 3, e^p = e³ is a constant, its rate is 0, and p·μ = 3·e³ is not.
Which antiderivative does not matter. A different constant added to ∫p dt multiplies μ by a constant factor, such as 5 or e⁷, and that factor multiplies both sides of the equation and cancels. μ = 5t² works as well as μ = t², and the final y is the same.
Take y′ + 2y = e^t. Then μ = e^(2t), and (e^(2t)·y)′ = e^(2t)·e^t = e^(3t).
Integrate: e^(2t)·y = (1/3)·e^(3t) + C, so y = (1/3)·e^t + C·e^(−2t). The lab prints exactly this, with C left free because no starting value was given.
For p = a/t the same steps stay in fractions. With y′ + (2/t)·y = t² and y(1) = 1, the factor is t², (t²·y)′ = t⁴, t²·y = t⁵/5 + C, and y = t³/5 + C/t². Substituting to check: y′ = 3t²/5 − (2C)/t³ and (2/t)·y = 2t²/5 + (2C)/t³, which add to t² for every C.
The lab limits itself to the families above, and says so when it refuses. It takes a constant p, or a/t with a a whole number from −4 to 4; it will not integrate tᵃ·q when that integral has a ln t in it; and for p = a/t it asks for a starting time other than 0, since the equation is not defined at t = 0. These are limits of the lab, and the integrating factor itself has no such limit.
Interactive
μ, y_h, y_p and the constant, exactly
Method
Five moves: standard form, factor, multiply, integrate, divide. The check is a substitution, and it is cheap.
y′ + p(t)·y = q(t), with the signs. Everything else depends on p being right.
Find a function with μ′ = p·μ: e^(a·t) for p = a, tᵃ for p = a/t. Differentiate it and compare with p·μ before you use it.
Multiply the whole equation by μ, left and right. The left side is now (μ·y)′, so write it that way.
μ·y = ∫μ·q dt + C. If the integral is a power of t, the power rule gives it in fractions.
y = (∫μ·q dt + C)/μ, with C divided as well. Put the starting value in last, solve for C, and check by substituting into the original.
The equation is already in standard form, with p = 2/t and q = t². The factor is a power, so nothing leaves the rational numbers.
Every line is an exact fraction, and the answer is checked by substitution: y′ = 3t²/5 − (8/5)/t³, (2/t)·y = 2t²/5 + (8/5)/t³, and the sum is t². At t = 1 the solution is 1/5 + 4/5 = 1. Nothing here was rounded, because μ was a power of t and the integral of a power is a power.
For a rehearsal, work the third preset, y′ + y/t = 1 with y(1) = 2, before opening it. The factor is t; the integrated line is t·y = t²/2 + C; the starting value fixes C. Then compare the constant with the lab's.
You should be able to find a function with μ′ = p·μ and check it by differentiating, write the equation as (μ·y)′ = μ·q, integrate the right side with a constant, divide the constant by μ as well, fit it to a starting value last, and explain why e^(p) is not a factor.
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