Lesson 4-2: First-Order Differential Equations

Master separable and linear first-order differential equations with integrating factors, logistic growth models, and real-world applications.

Learning Objectives

Separable Equations

\frac{dy}{dx} = f(x) g(y)

\frac{dy}{g(y)} = f(x) dx

Separate variables and integrate

Linear Equations

y' + P(x) y = Q(x)

\mu(x) = e^{\int P(x) dx}

Use integrating factor

1. Separate variables:

\frac{dy}{g(y)} = f(x) dx

2. Integrate both sides:

\int \frac{dy}{g(y)} = \int f(x) dx + C

3. Solve for y and apply initial conditions

1. Find integrating factor:

\mu(x) = e^{\int P(x) dx}

2. Multiply equation by μ(x):

\mu(x) y' + \mu(x) P(x) y = \mu(x) Q(x)

3. Left side becomes derivative of μ(x)y:

\frac{d}{dx}[\mu(x) y] = \mu(x) Q(x)

Solve $\frac{dy}{dx} = 2xy$ with $y(0) = 1$ .

Solution:

\frac{dy}{y} = 2x dx

\ln|y| = x^2 + C

y = Ce^{x^2}

y = e^{x^2}

Solve $y' + 2y = e^{-x}$ with $y(0) = 1$ .

Solution:

\mu(x) = e^{\int 2 dx} = e^{2x}

e^{2x} y' + 2e^{2x} y = e^{x}

\frac{d}{dx}[e^{2x} y] = e^{x}

y = e^{-2x} + Ce^{-2x}

\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)

Models population growth with carrying capacity K and growth rate r.

RC\frac{dV}{dt} + V = V_0

Describes voltage across capacitor in RC circuit with source voltage V₀.

Solve the separable equation: $\frac{dy}{dx} = \frac{x^2}{y}$ with $y(1) = 2$ .

y = \sqrt{\frac{2x^3}{3} + \frac{10}{3}}

Solve the linear equation: $y' + 3y = 6x$ with $y(0) = 0$ .

y = 2x - \frac{2}{3} + \frac{2}{3}e^{-3x}

Separable equations can be solved by separating variables and integrating.
Linear equations require the integrating factor method for solution.
Initial conditions are essential for finding unique solutions.
Differential equations model many real-world phenomena in science and engineering.

Continue to Lesson 4-3 for parametric and polar systems.