First-Order Equations

If $\tfrac{dy}{dx} = g(x)h(y)$, move all $y$’s left and $x$’s right: $\tfrac{dy}{h(y)} = g(x)\,dx$, integrate both sides, and solve for $y$. The prototype $y' = ky$ gives exponential growth/decay $y = y_0e^{kx}$, and half-life problems are the case $k = -\tfrac{\ln 2}{t_{1/2}}$.

For $y' + p(x)y = q(x)$, multiply through by $\mu = e^{\int p\,dx}$: the left side collapses to $(\mu y)'$ by design, and one integration finishes it. Example: $y' + y = e^x$ with $\mu = e^x$ gives $(e^xy)' = e^{2x}$, hence $y = \tfrac{e^x}{2} + Ce^{-x}$.

With $\mu = e^{\int p\,dx}$, note $\mu' = p\mu$ (chain rule). Then $(\mu y)' = \mu y' + \mu'y = \mu\left(y' + py\right)$ — the product rule REBUILDS the left side of $y' + py = q$ exactly. So the equation becomes $(\mu y)' = \mu q$; integrate once and divide by $\mu$. The factor is engineered so the product rule does the collapsing. $\blacksquare$

If $\tfrac{dy}{dx} = g(x)h(y)$ with $h(y) \ne 0$, divide: $\tfrac{y'(x)}{h(y(x))} = g(x)$. Integrate both sides in $x$; on the left, substitute $u = y(x)$, $du = y'(x)dx$ (the chain rule in reverse): $\int\tfrac{du}{h(u)} = \int g(x)\,dx$.

$y' = ky$ separates to $\int\tfrac{dy}{y} = \int k\,dx$, so $\ln|y| = kx + C$ and $y = y_0e^{kx}$ — and this is the ONLY solution: if $z$ also solves it, then $(z e^{-kx})' = z'e^{-kx} - kze^{-kx} = 0$, so $ze^{-kx}$ is constant. Setting $y(t_{1/2}) = \tfrac{y_0}{2}$ gives $e^{kt_{1/2}} = \tfrac12$, i.e. $k = -\tfrac{\ln 2}{t_{1/2}}$. $\blacksquare$