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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.