Differential Equations

Separable Differential Equations

A DE is separable if it can be written as: $\frac{dy}{dx} = f(x)g(y)$

Solve by separating: $\frac{dy}{g(y)} = f(x)\,dx$, then integrate both sides.

Slope Fields

A slope field shows the slope $dy/dx$ at many points. The solution curve follows the slopes.

Euler's Method (BC-specific)

Numerical approximation starting from $(x_0, y_0)$: $y_{n+1} = y_n + f(x_n, y_n) \cdot \Delta x$

Smaller step size → better approximation.

Exponential Growth and Decay

$\frac{dy}{dt} = ky \quad \Rightarrow \quad y = y_0 e^{kt}$

• $k > 0$: exponential growth.
• $k < 0$: exponential decay.

Logistic Growth (BC-specific)

$\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)$

Solution: $P(t) = \frac{L}{1 + Ae^{-kt}}$ where $A = (L - P_0)/P_0$.

• $L$ is the carrying capacity.
• Fastest growth at $P = L/2$.
• As $t \to \infty$, $P \to L$.

Problem: Solve $\frac{dy}{dx} = xy$ with $y(0) = 3$.

Solution:

Separate: $\frac{dy}{y} = x\,dx$

$\ln|y| = \frac{x^2}{2} + C$

$y = Ae^{x^2/2}$

$y(0) = 3 \Rightarrow A = 3$

$y = 3e^{x^2/2}$

• Separable DEs: separate variables, integrate both sides, apply initial conditions.
• Euler's method: step-by-step numerical approximation (BC).
• Logistic growth: $dP/dt = kP(1 - P/L)$; carrying capacity is $L$; fastest growth at $P = L/2$.
• Exponential growth/decay: $y = y_0 e^{kt}$.