Integrating factor ($\mu (t)=\exp \left({\int }_{t_0}^{t}a(s)ds\right)$)

An integrating factor is a nonzero multiplier that makes a differential equation directly integrable. For a first-order linear differential equation, it gathers the unknown function and its derivative into one product rule expression. Suppose $a$ and $b$ are continuous functions on an interval containing $t_0$, and consider the initial value problemDefine $A(t)={\int }_{t_0}^{t}a(s)ds$ and $\mu (t)=e^{A(t)}$. The exact solution of a differential equation throughout that interval isThe definite integral fixes $\mu (t_0)=1$, so the initial condition appears without an extra constant to determine. If the equation originally has a coefficient multiplying $y^{\prime }$, first divide by that coefficient on an interval where it is nonzero; a zero of that coefficient needs separate attention. See OpenStax's first-order linear equations chapter for this normalization and the integrating factor procedure.

For a concrete balance, solve $y^{\prime }+2y=3$ with $y(0)=0$. Here $\mu =e^{2t}$, so multiplying the entire ordinary differential equation gives $(e^{2t}y{)}^{\prime }=3e^{2t}$. Accumulating from zero to $t$ givesThe computed values are $y(0.5)\approx 0.94818$, $y(1)\approx 1.29700$ and $y(2)\approx 1.47253$. Its derivative is $3e^{-2t}$, which verifies $y^{\prime }+2y=3$ exactly. Interpreting $3$ as a steady input and $2y$ as proportional removal explains the shape: initially all input increases $y$, then removal catches up. The constant particular solution $y_p=3/2$ balances input and removal, and the homogeneous solution $y_h=-\frac{3}{2}{e}^{-2t}$ is the transient response that corrects it to the initial condition. Because this autonomous differential equation has a time-independent right-hand side, $3/2$ is also an equilibrium point.

The first panel adds the particular solution and homogeneous solution; the second shows the same balance in rates, $3=y^{\prime }+2y$. This links the algebra to a physical accounting rule. Newton's law of cooling has this form with $a=k$ and $b=k{T}_a$, while an electrical circuit can have a varying input. In both cases the integrating factor systematically handles the input rather than requiring a guess for a particular solution.

Variable coefficients work just as directly. For $y^{\prime }+2ty=t$, $y(0)=0$, the integrating factor is $e^{t^2}$, andThus $y(0.5)\approx 0.11060$, $y(1)\approx 0.31606$ and $y(2)\approx 0.49084$. The derivative $y^{\prime }=t{e}^{-t^2}$ verifies $y^{\prime }+2ty=t$, including at zero, where both coefficient and input vanish. No division by $t$ is necessary. The general solution is $y=1/2+C{e}^{-t^2}$: $1/2$ is a constant particular solution, and $C{e}^{-t^2}$ solves the associated homogeneous linear differential equation. Although this special forcing term admits a constant trajectory, this is a nonautonomous differential equation: it depends explicitly on time. Merely setting its right-hand side to zero at one instant would not establish a constant solution of a differential equation; at $t=0$ every state has zero slope.

The direction field shows the selected trajectory leaving its initially horizontal tangent line. For $t\ge 0$, the difference between any two solutions of a differential equation is multiplied by $e^{-t^2}$, rather than $e^{-2t}$ in the constant-rate example. This dependence on elapsed history is clearer if the exact formula is rewritten asFor $t\ge t_0$, each contribution from the input at $s$ is weighted by the accumulated coefficient between $s$ and $t$. The homogeneous solution remembers the initial condition; the particular solution given by the second term accumulates the forcing term with zero initial value. This decomposition differs from choosing a constant particular solution, but gives the same total. Initial-data memory tends to zero when ${\int }_{t_0}^{t}a\to +\infty$; a negative accumulated coefficient instead amplifies it. The reciprocal relation between integrating factor and homogeneous solution is developed in MIT's notes on solutions to linear first-order equations.

Why does the trick work? The fundamental theorem of calculus and chain rule give ${\mu }^{\prime }=a\mu$. Consequently the product rule givesIntegrating this identity and dividing by the positive exponential function $\mu$ produces the stated formula. Any nonzero constant multiple of $\mu$ works equally well; normalization is a convenience. This is the beauty of the method: a sum involving an unknown function becomes one exact derivative, leaving only an integral of known functions. See Oregon State's product-rule derivation. The related integrating factor for a differential one-form extends the idea of making an expression exact beyond this linear setting.

On an interval where $A\ne 0$, the displayed factor givesThis follows from $(wA{)}^{\prime }=wB$. It gives a useful self-adjoint form but does not solve a general second-order ordinary differential equation, since the $wCy$ term remains. If $C=0$, it does reduce to one integration for $y^{\prime }$. Another reducible class has $C=B^{\prime }-A^{\prime \prime }$, because the original left side is then $[A{y}^{\prime }+(B-A^{\prime })y{]}^{\prime }$. These are sufficient special cases, not a classification of all solvable equations.

For $y^{\prime }+a(x)y=H(x)$ with continuous $a$, the integrating factor $\mu =\exp (\int a)$ gives the displayed solution. It is continuous at the step and solves the differential equation classically on each side; the derivative jumps by one. The value assigned to the Heaviside step function at the single point zero does not affect the integral. A discontinuity in $y$ would instead produce an unintended delta term in its distributional derivative.

For the differential one-form $fdy+gdx$, a nonzero integrating factor $\mu$ makes $\mu fdy+\mu gdx$ an exact differential form locally. If it equals $dF$, then $F_y=\mu f$, $F_x=\mu g$, and equality of mixed partial derivatives gives the displayed compatibility condition. Conversely this condition is local sufficiency for smooth coefficients on a small rectangle; global exactness additionally depends on the domain. Zeros of the multiplying factor can change the solution set, so equivalence is asserted only where the factor is nonzero.

For a differential one-form $Pdx+Qdy$, seek a nonzero integrating factor $\mu (x)$. Expanding $(\mu P{)}_y=(\mu Q{)}_x$ gives the displayed equation where $Q\ne 0$. Such a factor exists locally when $(P_y-Q_x)/Q$ depends only on $x$, and is $\mu =\exp \int [(P_y-Q_x)/Q]dx$. Likewise, a factor depending only on $y$ obeys ${\mu }^{\prime }/\mu =(Q_x-P_y)/P$ where $P\ne 0$. The resulting product is an exact differential form on a sufficiently small regular domain; singular lines of the factor need separate treatment.

For a differential one-form $Mdx+Ndy$, a monomial integrating factor exists on a chosen domain away from the axes when constants $\alpha ,\beta$ makeThis follows by expanding the exactness condition ${\partial }_y(\mu M)={\partial }_x(\mu N)$. When $M,N$ are sums of monomials, matching their coefficients often reduces the search for $\alpha ,\beta$ to a small linear system. Integrating $dF=\mu Mdx+\mu Ndy$ then gives the implicit solution $F=C$. Curves excluded by the multiplying factor, especially solutions on coordinate axes, must be tested in the original equation.

For a first-order linear equation whose homogeneous solution grows at infinity, boundedness fixes the integration constant by rewriting the particular integral as a tail integral from the current point to infinity.