First-Order ODEs

Separable Equations

$$\frac{dy}{dx} = f(x)\, g(y) \quad \Longrightarrow \quad \int \frac{dy}{g(y)} = \int f(x)\, dx$$

Linear Equations and the Integrating Factor

$$\frac{dy}{dt} + p(t)y = g(t), \qquad \mu(t) = e^{\int p(t)\, dt}$$

Multiplying through by $\mu$ makes the left side an exact derivative,

$$\frac{d}{dt}\big(\mu(t) y(t)\big) = \mu(t) g(t) \quad \Longrightarrow \quad y(t) = \frac{1}{\mu(t)}\left(\int \mu(t) g(t)\, dt + c\right)$$

Exact Equations

$$M(x,y) + N(x,y) \frac{dy}{dx} = 0$$

is exact if there is $\psi(x,y)$ with $\psi_x = M$ and $\psi_y = N$. Then $\frac{d}{dx}\psi(x, y(x)) = 0$, so the solutions are the level sets $\psi(x,y) = c$. On a simply connected region where $M, N$ have continuous partials, the equation is exact iff $M_y = N_x$.

Existence and Uniqueness

Linear equations If $p$ and $g$ are continuous on an open interval $\alpha < t < \beta$ containing $t_0$, then

$$y' + p(t)y = g(t), \qquad y(t_0) = y_0$$

has a unique solution on the whole interval.

Picard-Lindelöf If $f$ and $\partial f / \partial y$ are continuous on a rectangle $\alpha < t < \beta$, $\gamma < y < \delta$ containing $(t_0, y_0)$, then

$$y' = f(t, y), \qquad y(t_0) = y_0$$

has a unique solution on some interval $t_0 - \varepsilon < t < t_0 + \varepsilon$.

Peano If $f$ is only continuous on such a rectangle, a solution exists on some interval $(t_0 - \varepsilon, t_0 + \varepsilon)$, but it need not be unique.

Autonomous Equations

$$\frac{dy}{dt} = F(y)$$

• Critical points are the $y^*$ with $F(y^*) = 0$, and each gives an equilibrium solution $y(t) \equiv y^*$.
• Phase line Mark the critical points and the sign of $F$ between them. $y^*$ is asymptotically stable if $F$ changes from $+$ to $-$ across it, unstable if from $-$ to $+$, and semistable if $F$ does not change sign.
• Bifurcation diagram For $y' = F(y, \alpha)$, plot the critical points $y^*$ against the parameter $\alpha$, marking stability. Each vertical slice is the phase line for that $\alpha$.