Linear Equations

A linear first-order equation is simply one where $y$ and $y'$ each show up on their own, to the first power: never squared, multiplied together or tucked inside a function. "Linear" because the left side is a combination like $ay' + by$, the same shape as $ax + by$ in the equation of a line (here the coefficients may depend on $t$), and "first order" because the highest derivative is $y'$. These model anything with a steady input and a loss proportional to how much is there. The formulas look heavy, but they're just three moves: $\mu(t) = e^{\int p(t)\,dt}$ (a helper function, the integrating factor, that we multiply the equation by), $\int\mu(t)g(t)\,dt + C$ (integrate what's left) and $\frac{1}{\mu(t)}$ (divide the helper back out). Multiply, integrate, divide; bear with it.

In the last unit, equations like $y' = f(x)$ were easy because the left side was already a single derivative, so you could just integrate. An equation like $y' + p(t)y = g(t)$ doesn't look like that: the $p(t)y$ term mixes in $y$ itself. So what can we do now? Try to make the left side into one derivative.

Why it works. The product rule says $(\mu y)' = \mu y' + \mu' y$. That has the same shape as our left side times $\mu$, which is $\mu y' + \mu p y$. The two match exactly if $\mu' = \mu p$, and that equation is easy to solve: it gives $\mu = e^{\int p\,dt}$ (no constant needed, since any one $\mu$ that works is enough). After multiplying by $\mu$, the equation reads $(\mu y)' = \mu g$, and the right side has no $y$ in it, so integrating finishes the job.

A common mistake. In $ty' + 2y = t^2 - t + 1$, it's tempting to read off $p = 2$, because that's the number sitting next to $y$. It isn't. The whole trick assumed the coefficient of $y'$ is $1$, so you have to divide by $t$ first, and then $p = \frac2t$. Skip that step and the left side won't collapse into $(\mu y)'$.

Steps:

1. Put the equation in standard form $y' + p(t)y = g(t)$; divide by the coefficient of $y'$ if necessary.
2. Compute $\mu = e^{\int p\,dt}$.
3. Rewrite as $(\mu y)' = \mu g$ and integrate both sides. This is where the $+C$ goes, so don't forget it.
4. Solve for $y$, then apply the initial condition.

Linear first-order equations model anything with a steady input and proportional loss: a drug infusion, a charging capacitor (RC circuit), a lake with a pollutant flowing in and out.