Educational tool; results are estimates for learning, verify with your textbook or instructor.
This integrating factor calculator is a first order linear differential equation solver for y' + P(x) y = Q(x). The coefficient of y' is already 1. P(x) is a constant or a polynomial. Q(x) is a polynomial, a polynomial times an exponential, or a polynomial times a sine and a cosine of the same frequency. The steps name P and Q, build μ(x) = e^(∫ P dx), multiply through, integrate, and write the general solution. An optional condition y(x₀) = y₀ fixes the constant.
The boxes open on y' + 2y = 4x with y(0) = 1. The particular solution is 2x − 1 + 2 e^(−2x). Leave both condition boxes blank when you only want the family with C still in it.
Separation is a different first order method. A product of a function of x and a function of y is on the separable walkthrough. Constant-coefficient second order equations use a characteristic polynomial, not an integrating factor. That solver is the second order differential equation calculator. The map around both is differential equations.
How the integrating factor works
Start from standard form, y' + P(x) y = Q(x). If a textbook writes a(x) y' + b(x) y = g(x), divide by a(x) first, on an interval where a(x) is not zero. This page does not do that division. P and Q are the functions after it.
Choose any antiderivative of P. The page uses the one with constant term 0. Then μ(x) = e raised to that antiderivative. The product rule says (μ y)' = μ y' + μ' y. Because μ' = μ P, the left side of the multiplied equation is exactly (μ y)'. Integrate both sides. Divide by μ, which is never zero. The constant of integration becomes the C in y = (1/μ) (∫ μ Q dx + C).
For a constant P = p, μ(x) = e^(p x). A polynomial Q, a polynomial times e^(kx), and a polynomial times A sin(bx) + B cos(bx) all integrate against that factor by a finite reduction. The page writes the coefficients. For a nonconstant polynomial P, a polynomial Q sometimes has a polynomial particular solution, and the homogeneous piece is still C e^(−∫ P dx). The page checks every power before it trusts that polynomial. If a power disagrees, it does not print a formula.
How to read the result
The large line is the general solution, or the particular solution when C came out as an exact fraction. Under it, the steps follow the order above: standard form, integrating factor, multiplied equation, integral, general solution, then the initial condition if you supplied one.
Fractions stay exact. A decimal such as 0.5 is kept as 1/2. If the initial time makes an exponential, a sine, or a cosine that is not 0, 1, or −1, the page still writes an exact expression for C when the rest of the value is rational, and it prints a decimal on its own line, labelled as a decimal approximation.
The curve is the particular solution when a condition was entered. The marked point is (x₀, y₀). With no condition, the curve uses the sample choice C = 1, and the legend says so. It is one member of the family. A Runge-Kutta plot is the numerical table, not a closed form. The legend says that too.
When the integral is not elementary
Some integrating factors do not have an elementary antiderivative against Q. The example to keep is y' + x y = 1. Here P(x) = x, so μ(x) = e^(x^2/2). With y(0) = 0 the exact initial-value statement is still an integral, y(x) = e^(−x^2/2) times the integral from 0 to x of e^(t^2/2) dt. That integral is not a finite combination of polynomials, exponentials, and sines and cosines. Printing one would be a wrong answer.
The preset "Not elementary" asks for that initial condition, the endpoint x = 1, and 10 steps. Classical fourth-order Runge-Kutta then reports a value near 0.724778 at x = 1. The later digits move if you shorten the step. Every row is labelled as a numerical approximation. Without an initial condition the page stops after the integrating factor and asks for y(x₀), because a table has to start somewhere.
Three worked examples
The three solutions below are written on the page, so you can read them with scripts turned off. The form opens on the first one. Each check substitutes the solution back into the differential equation.
Polynomial forcing. y' + 2y = 4x, y(0) = 1.
P(x) = 2 and Q(x) = 4x.
∫ P dx = 2x, taking the antiderivative with no added constant.
μ(x) = e^(2x).
Multiply through.
d/dx [ e^(2x) y ] = 4x e^(2x).
Integrate. For ∫ x^n e^(a x) dx with a = 2, the antiderivative is e^(2x) times a polynomial.
The coefficient of x is 4/2 = 2.
The constant term is (0 − 1·2)/2 = −1.
∫ 4x e^(2x) dx = e^(2x) (2x − 1), up to a constant.
Divide by μ.
y = 2x − 1 + C e^(−2x).
Initial condition y(0) = 1.
−1 + C = 1, so C = 2.
y = 2x − 1 + 2 e^(−2x).
Check.
y' = 2 − 4 e^(−2x).
y' + 2y = 2 − 4 e^(−2x) + 4x − 2 + 4 e^(−2x) = 4x.
y(0) = −1 + 2 = 1.
Sine forcing. y' + y = sin(x), y(0) = 1.
P(x) = 1 and Q(x) = sin(x).
μ(x) = e^x.
d/dx [ e^x y ] = e^x sin(x).
∫ e^(a x) sin(bx) dx with a = 1 and b = 1 is e^x (sin(x) − cos(x)) / 2.
Divide by e^x.
y = (1/2) sin(x) − (1/2) cos(x) + C e^(−x).
y(0) = 1 gives −1/2 + C = 1, so C = 3/2.
y = (1/2) sin(x) + (−1/2) cos(x) + (3/2) e^(−x).
Check at x = 0.
(1/2)·0 + (−1/2)·1 + 3/2 = 1.
The derivative of the particular piece at 0, plus the derivative of (3/2) e^(−x) at 0, plus y(0), returns sin(0) = 0.
Integrating factor e^(x^2). y' + 2x y = 2x, y(0) = 3.
P(x) = 2x and Q(x) = 2x.
∫ P dx = x^2.
μ(x) = e^(x^2).
d/dx [ e^(x^2) y ] = 2x e^(x^2).
The right side is the derivative of e^(x^2). A polynomial particular solution is the constant 1, because 0 + 2x·1 = 2x.
y = 1 + C e^(−x^2).
y(0) = 3 gives 1 + C = 3, so C = 2.
y = 1 + 2 e^(−x^2).
Check.
y' = −4x e^(−x^2).
y' + 2x y = −4x e^(−x^2) + 2x + 4x e^(−x^2) = 2x.
y(0) = 1 + 2 = 3.
Common mistakes
The first mistake is skipping the division that puts the equation in standard form. If the book says 2y' + 4y = 8x, then P is 2 and Q is 4x, not 4 and 8x. This page assumes you already divided by the coefficient of y'.
The second mistake is keeping a constant inside the exponent of μ and then treating that constant as part of the answer. It multiplies every term and cancels. The arbitrary constant belongs with the integral of μ Q, as C, not inside μ.
The third mistake is integrating μ Q and forgetting to multiply back by 1/μ. The product μ y equals that integral. y itself is the integral divided by μ.
The fourth mistake is reaching for a characteristic equation. That method solves a y'' + b y' + c y = 0, a second order equation with constant coefficients. It is the second order calculator. A first order linear equation has an integrating factor even when P depends on x.
The fifth mistake is inventing an elementary antiderivative for e^(x^2/2). The preset "Not elementary" exists so the page can refuse that guess and show a labelled numerical table instead.
Related reading
Place the equation on the differential equations map. Separation, which is a different method, is the separable walkthrough. The short definitions are differential equation and initial value problem. A course shaped path that arrives at this chapter is the AP Calculus AB bridge.
After this first order chapter, constant coefficients of a second derivative are the second order differential equation calculator. Two candidate solutions from that page can be tested on the Wronskian calculator. A zero at one point is not, by itself, a proof of dependence. The same initial value problem can also be started from a transform. Once you have Y(s), the Laplace transform calculator inverts it.
Last reviewed: 28 October 2026.
