To verify y = g(x) is a solution:

1. Find y', y'', etc.
2. Substitute into differential equation
3. Verify equality holds

Graphical representation showing $\frac{dy}{dx}$ at various points (x, y).

• Each small line segment has slope = dy/dx at that point
• Solution curves follow the slope field pattern

For differential equation of form $\frac{dy}{dx} = g(x)h(y)$:

1. Separate: $\frac{dy}{h(y)} = g(x)dx$
2. Integrate both sides: $\int \frac{1}{h(y)}dy = \int g(x)dx$
3. Solve for y (if possible)

General solution contains +C (arbitrary constant). Particular solution: Use initial condition y(x0) = y0 to find C.

For $\frac{dy}{dt} = ky$: $y = y_0 e^{kt}$

• k > 0: Exponential growth
• k < 0: Exponential decay
• Half-life: $t_{1/2} = \frac{\ln 2}{|k|}$

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

where:

• P = population at time t
• k = growth rate constant
• L = carrying capacity (maximum sustainable population)

Maximum population that environment can sustain.

• Growth rate maximum when P = L/2
• Population approaches L as t -> ∞