= Heaviside forcing in a first-order integrating-factor equation
{c}
{title2=$y(x)=\mu(x)^{-1}\bigl[\mu(0)y(0)+\int_0^x\mu(t)H(t)\,dt\bigr]$}
For $y'+a(x)y=H(x)$ 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.
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