= Integrating factor
{title2=$\mu(t)=\exp\!\left(\int_{t_0}^t a(s)\,ds\right)$}
{wiki}
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 problem>
$$
y'+a(t)y=b(t),\qquad y(t_0)=y_0.
$$
Define $A(t)=\int_{t_0}^t a(s)\,ds$ and $\mu(t)=e^{A(t)}$. The exact <solution of a differential equation> throughout that <interval> is
$$
y(t)=e^{-A(t)}\left[y_0+\int_{t_0}^t e^{A(s)}b(s)\,ds\right].
$$
The <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 $y'$, first divide by that coefficient on an <interval> where it is nonzero; a zero of that coefficient needs separate attention. See https://openstax.org/books/calculus-volume-2/pages/4-5-first-order-linear-equations[OpenStax's first-order linear equations chapter] for this normalization and the <integrating factor> procedure.
For a concrete balance, solve $y'+2y=3$ with $y(0)=0$. Here $\mu=e^{2t}$, so multiplying the entire <ordinary differential equation> gives $(e^{2t}y)'=3e^{2t}$. Accumulating from zero to $t$ gives
$$
e^{2t}y(t)=\int_0^t3e^{2s}\,ds=\frac32(e^{2t}-1),\qquad y(t)=\frac32(1-e^{-2t}).
$$
The computed values are $y(0.5)\approx0.94818$, $y(1)\approx1.29700$ and $y(2)\approx1.47253$. Its <derivative> is $3e^{-2t}$, which verifies $y'+2y=3$ 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> $y_h=-\frac32e^{-2t}$ 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>.
\Image[/integrating-factor-balance.png]
{title=A steady input becomes a constant particular solution plus a decaying transient}
The first panel adds the <particular solution> and <homogeneous solution>; the second shows the same balance in rates, $3=y'+2y$. This links the algebra to a physical accounting rule. <Newton's law of cooling> has this form with $a=k$ and $b=kT_a$, 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 $y'+2ty=t$, $y(0)=0$, the <integrating factor> is $e^{t^2}$, and
$$
(e^{t^2}y)'=te^{t^2},\qquad e^{t^2}y(t)=\frac12(e^{t^2}-1),\qquad y(t)=\frac12(1-e^{-t^2}).
$$
Thus $y(0.5)\approx0.11060$, $y(1)\approx0.31606$ and $y(2)\approx0.49084$. The <derivative> $y'=te^{-t^2}$ verifies $y'+2ty=t$, including at zero, where both coefficient and input vanish. No division by $t$ is necessary. The <general solution> is $y=1/2+Ce^{-t^2}$: $1/2$ is a constant <particular solution>, and $Ce^{-t^2}$ 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.
\Image[/integrating-factor-variable-coefficient.png]
{title=Variable-rate trajectories share a constant particular solution and forget initial data by a factor of exp minus t squared}
The <direction field> shows the selected trajectory leaving its initially horizontal <tangent line>. For $t\geq0$, 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 as
$$
y(t)=y_0e^{-\int_{t_0}^t a(u)\,du}+\int_{t_0}^t e^{-\int_s^t a(u)\,du}b(s)\,ds.
$$
For $t\geq t_0$, 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 https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/539a2280949de0de6ec2d420067af6b2_MIT18_03SCF11_s5_1text.pdf[MIT's notes on solutions to linear first-order equations].
Why does the trick work? The <fundamental theorem of calculus> and <chain rule> give $\mu'=a\mu$. Consequently the <product rule> gives
$$
(\mu y)'=\mu y'+\mu'y=\mu(y'+ay)=\mu b.
$$
Integrating 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 https://sites.science.oregonstate.edu/math/home/programs/undergrad/CalculusQuestStudyGuides/ode/first/linear/linear.html[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.
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