First-Order Equations

Why moving the differentials is legitimate: if $y(x)$ is a solution, then $\dfrac{y'(x)}{h(y(x))} = g(x)$ holds for every $x$, and integrating both sides in $x$ turns the left side into $\int \frac{dy}{h(y)}$ by the substitution $u = y(x)$ — the chain rule run backward. So every solution satisfies $H(y) = G(x) + C$ with $H' = \tfrac1h$ and $G' = g$, and the constant $C$ labels which curve you are on. The division needs $h(y) \ne 0$: each root $y^*$ of $h$ is a constant solution $y \equiv y^*$ that the division discards, and you add it back by hand.

Why $\mu$ has that formula: we want the left side $\mu y' + \mu p\,y$ to be the derivative of the product $\mu y$, which is $\mu y' + \mu' y$. The two agree exactly when $\mu' = p\mu$ — a separable equation for $\mu$ whose solution is $\mu = e^{\int p\,dx}$. Any antiderivative will do: a different constant of integration only rescales $\mu$, and the rescaling cancels when you divide at the end.

The check is decisive because it is the definition: a function is a solution exactly when substituting it makes the two sides agree for every $x$ and the initial value matches. It catches the typical slips. The tempting wrong answer $y = e^{x^2/2} + 1$ (constant added instead of multiplied) has $y' = xe^{x^2/2}$ but $xy = xe^{x^2/2} + x$; the two sides differ by $x$, so it is not a solution.

Nothing in $y' = y^2$ looks dangerous, yet its solutions reach infinity at a finite $x$. The slope is the square of the height, so the higher the curve, the faster it climbs, and climbing from $1$ to $\infty$ costs only $\int_1^\infty \frac{dy}{y^2} = 1$ unit of $x$. A linear equation $y' + p(x)y = q(x)$ with continuous $p$ and $q$ cannot do this — its solutions exist wherever $p$ and $q$ are continuous — so blow-up is a nonlinear effect, and the way to find it is to solve and look at where the formula fails.

Step 3 — use the data now, while the constant is still visible: $-1 = 0 + C$, so $C = -1$ and $-\tfrac1y = x - 1$, i.e. $y = \dfrac{1}{1 - x}$.

Step 4 — the interval: the formula has a vertical asymptote at $x = 1$. A solution is a continuous function through $(0, 1)$, so it lives on $(-\infty, 1)$ only; the branch of the same formula for $x > 1$ is a different curve.

Step 5 — verify: $y' = \dfrac{1}{(1-x)^2} = y^2$ ✓ and $y(0) = 1$ ✓.

Step 3 — the data: $\arctan 0 = 0$, so $C = 0$.

Step 4 — solve for $y$, which is where the interval appears: $\arctan y$ only takes values in $\left(-\tfrac\pi2, \tfrac\pi2\right)$, so $\arctan y = x^2$ is possible only while $x^2 < \tfrac\pi2$, and there $y = \tan(x^2)$.

Step 5 — the solution exists for $|x| < \sqrt{\pi/2} \approx 1.2533$ and tends to $+\infty$ at both ends.

Step 6 — verify: $y' = 2x\sec^2(x^2) = 2x\left(1 + \tan^2(x^2)\right) = 2x(1 + y^2)$ ✓.

The recipe $\mu = e^{\int p}$ came from matching $\mu y' + \mu p y$ against $(\mu y)' = \mu y' + \mu' y$, and that match has a bare $\mu y'$ in it, so it only works when $y'$ has coefficient $1$. The shape $y = y_p + \tfrac{C}{\mu}$ also has a reason: the difference of two solutions solves $z' + pz = 0$, whose solutions are exactly $Ce^{-\int p}$. When $p > 0$ that homogeneous part decays, so solutions forget their starting values and merge onto one curve; when $p < 0$ they spread apart.

Step 3 — multiply the standard form by $x^2$: $x^2y' + 2xy = 4x^3$, and the left side is $(x^2y)'$ — confirm this with the product rule before going on, since it is the whole point of $\mu$.

Step 4 — integrate: $x^2y = x^4 + C$, so $y = x^2 + \dfrac{C}{x^2}$.

Step 5 — the data: $2 = 1 + C$, so $C = 1$ and $y = x^2 + \dfrac{1}{x^2}$.

Step 6 — verify in the original form: $xy' = x\left(2x - \dfrac{2}{x^3}\right) = 2x^2 - \dfrac{2}{x^2}$ and $2y = 2x^2 + \dfrac{2}{x^2}$, which add to $4x^2$ ✓.

Step 3 — multiply through: $(e^{-x^2}y)' = xe^{-x^2}$.

Step 4 — integrate with the substitution $w = -x^2$, $dw = -2x\,dx$: $\int xe^{-x^2}dx = -\tfrac12e^{-x^2}$, so $e^{-x^2}y = -\tfrac12e^{-x^2} + C$ and $y = -\tfrac12 + Ce^{x^2}$.

Step 5 — the data: $0 = -\tfrac12 + C$, so $y = \tfrac12\left(e^{x^2} - 1\right)$.

Step 6 — verify: $y' = xe^{x^2}$, and $2xy + x = x\left(e^{x^2} - 1\right) + x = xe^{x^2}$ ✓. Here $p = -2x < 0$ for $x > 0$, so the homogeneous part $Ce^{x^2}$ grows: nearby solutions separate instead of merging.

Step 3 — so $e^xy = \tfrac12e^x(\sin x - \cos x) + C$ and $y = \tfrac12(\sin x - \cos x) + Ce^{-x}$.

Step 4 — the data: $0 = -\tfrac12 + C$, so $C = \tfrac12$ and $y = \tfrac12(\sin x - \cos x) + \tfrac12e^{-x}$.

Step 5 — verify: $y' = \tfrac12(\cos x + \sin x) - \tfrac12e^{-x}$, and adding $y$ leaves $\sin x$ ✓.

Step 6 — read it: the term $Ce^{-x}$ (the transient) dies whatever $C$ is, so every solution settles onto the same steady oscillation $\tfrac12(\sin x - \cos x) = \tfrac{\sqrt2}{2}\sin\!\left(x - \tfrac\pi4\right)$, of amplitude $\tfrac{\sqrt2}{2} \approx 0.707$, lagging the forcing by $\tfrac\pi4$.

The informal "move the $dy$" is shorthand for exactly this substitution — nothing illegal happens. $\blacksquare$

($y' = x - y$ hides its linearity until rearranged: $y' + y = x$.) Some equations are both ($y' = ky$), many are neither — those go to slope fields, Euler, or substitutions. Two habits matter more than any method. First, track the constant from the moment you integrate: it lives inside the exponential as $y = Ce^{kx}$, not tacked on the end as $e^{kx} + C$ — a wrong constant placement is the single most common first-order error. Second, plug the final answer back into the original equation; ten seconds of differentiation catches nearly everything.

Step 3 — exponentiate carefully: $|y| = e^Ce^{x^2/2}$, and absorbing the sign into a new constant $A = \pm e^C$: $y = Ae^{x^2/2}$ — the constant is multiplicative.

Step 4 — apply the initial condition: $2 = Ae^0 = A$, so $y = 2e^{x^2/2}$.

Step 5 — verify: $y' = 2xe^{x^2/2}\cdot 1 = x\cdot y$ ✓ and $y(0) = 2$ ✓. Watch what Step 3 prevented: writing $y = e^{x^2/2} + C$ and "solving" $C = 1$ gives a function that fails the original equation instantly.

Step 3 — integrate: $e^{2x}y = 3e^{2x} + C$.

Step 4 — solve: $y = 3 + Ce^{-2x}$; the condition $y(0) = 0$ gives $C = -3$, so $y = 3 - 3e^{-2x}$.

Step 5 — verify and interpret: $y' = 6e^{-2x}$ and $y' + 2y = 6e^{-2x} + 6 - 6e^{-2x} = 6$ ✓; the solution climbs from $0$ toward the equilibrium $3$ (where $y' = 0$ forces $2y = 6$) — the phase-line picture and the formula telling one story.