1. First-Order Linear Equations: Standard Form

First-order linear differential equations are among the most important equations in applied mathematics. They appear in countless applications from physics to biology to economics.

Definition 2.1: First-Order Linear ODE

A first-order linear ordinary differential equation has the form:

\frac{dx}{dt} + p(t)x = q(t)

where $p(t)$ and $q(t)$ are given functions of $t$ . This is the standard form. The equation is homogeneous if $q(t) = 0$ and nonhomogeneous otherwise.

Remark:

Any equation $a(t)\frac{dx}{dt} + b(t)x = c(t)$ with $a(t) \neq 0$ can be converted to standard form by dividing by $a(t)$ .

Theorem 2.1: Existence and Uniqueness for Linear First-Order ODEs

Let $p(t)$ and $q(t)$ be continuous on an open interval $I = (a, b)$ . Then for any $t_0 \in I$ and any $x_0$ , the IVP:

\frac{dx}{dt} + p(t)x = q(t), \quad x(t_0) = x_0

has a unique solution that exists on the entire interval $I$ .

Proof of Theorem 2.1:

Existence: Constructed via integrating factor method (Section 2).

Uniqueness: If $x_1$ and $x_2$ are both solutions, let $y = x_1 - x_2$ . Then $y' + p(t)y = 0$ with $y(t_0) = 0$ . The homogeneous solution is $y = Ce^{-\int p\,dt}$ . Using $y(t_0) = 0$ gives $C = 0$ , so $y = 0$ . ∎

2. The Integrating Factor Method

The integrating factor method is the primary technique for solving first-order linear ODEs. The idea is to multiply by a function that makes the left side a perfect derivative.

Definition 2.2: Integrating Factor

For $\frac{dx}{dt} + p(t)x = q(t)$ , the integrating factor is:

\mu(t) = e^{\int p(t)\,dt}

Theorem 2.2: Integrating Factor Solution Formula

The general solution is:

x(t) = \frac{1}{\mu(t)}\left[\int \mu(t)q(t)\,dt + C\right]

where $\mu(t) = e^{\int p(t)\,dt}$ .

Proof of Theorem 2.2:

Multiply by $\mu(t)$ : $\mu x' + \mu p x = \mu q$ . Since $\mu' = p\mu$ , the left side is $(\mu x)'$ . Integrating: $\mu x = \int \mu q\,dt + C$ . ∎

Algorithm 2.1: Integrating Factor Method

Input: $x' + p(t)x = q(t)$ , possibly with IVP

1. Compute $\mu(t) = e^{\int p(t)\,dt}$

2. Multiply both sides by $\mu$

3. Recognize left side as $(\mu x)'$

4. Integrate: $\mu x = \int \mu q\,dt + C$

5. Solve for $x(t)$

Example 2.2: Integrating Factor with IVP

Solve: $x' + 2x = e^{-t}$ , $x(0) = 1$

Solution: $\mu = e^{2t}$ . Then $(e^{2t}x)' = e^t$ , so $e^{2t}x = e^t + C$ .

Thus $x = e^{-t} + Ce^{-2t}$ . Using $x(0) = 1$ : $C = 0$ . Answer: $x = e^{-t}$

Example 2.3: Variable Coefficient

Solve: $tx' + 2x = t^3$ , $t > 0$

Standard form: $x' + \frac{2}{t}x = t^2$ . Integrating factor: $\mu = t^2$ .

$(t^2 x)' = t^4$ , so $t^2 x = \frac{t^5}{5} + C$ . Answer: $x = \frac{t^3}{5} + \frac{C}{t^2}$

3. Variation of Constants

Variation of constants provides an alternative perspective based on the solution structure.

Theorem 2.3: Structure of Solutions

The general solution of $x' + px = q$ is $x = x_h + x_p$ , where $x_h = Ce^{-\int p\,dt}$ is the homogeneous solution and $x_p$ is any particular solution.

Algorithm 2.2: Variation of Constants

1. Solve homogeneous: $x_h = e^{-\int p\,dt}$

2. Assume $x = C(t) \cdot x_h$

3. Substitute and solve for $C(t)$

Example 2.4: Variation of Constants

Solve: $x' + x = e^{2t}$

Homogeneous: $x_h = e^{-t}$ . Assume $x = C(t)e^{-t}$ .

Substituting: $C'e^{-t} = e^{2t}$ , so $C' = e^{3t}$ , $C = \frac{e^{3t}}{3} + K$ .

Answer: $x = \frac{e^{2t}}{3} + Ke^{-t}$

4. Bernoulli Equations

Definition 2.3: Bernoulli Equation

A Bernoulli equation has the form:

\frac{dx}{dt} + p(t)x = q(t)x^\alpha

where $\alpha \neq 0, 1$ .

Theorem 2.4: Linearization of Bernoulli Equations

The substitution $y = x^{1-\alpha}$ transforms the Bernoulli equation into:

\frac{dy}{dt} + (1-\alpha)p(t)y = (1-\alpha)q(t)

Proof of Theorem 2.4:

Let $y = x^{1-\alpha}$ . Then $y' = (1-\alpha)x^{-\alpha}x'$ . From $x' = -px + qx^\alpha$ :

y' = (1-\alpha)x^{-\alpha}(-px + qx^\alpha) = -(1-\alpha)py + (1-\alpha)q

Rearranging gives the linear equation. ∎

Example 2.5: Bernoulli Equation

Solve: $x' + x = x^3$

Here $\alpha = 3$ . Substitute $y = x^{-2}$ : $y' = -2x^{-3}x'$ .

Transform: $y' - 2y = -2$ . Solve: $y = 1 + Ce^{2t}$ .

Answer: $x = \pm(1 + Ce^{2t})^{-1/2}$ and $x = 0$ .

5. Riccati Equations

Definition 2.4: Riccati Equation

A Riccati equation has the form:

\frac{dx}{dt} = P(t)x^2 + Q(t)x + R(t)

Theorem 2.5: Reduction of Riccati Equations

If $x_1(t)$ is a known particular solution, the substitution $x = x_1 + \frac{1}{y}$ transforms the Riccati equation into a linear equation:

\frac{dy}{dt} + (2Px_1 + Q)y = -P

Example 2.6: Riccati Equation

Solve: $x' = x^2 - 2tx + t^2 + 1$ , given $x_1 = t$ .

Verify: $x_1' = 1$ , $t^2 - 2t^2 + t^2 + 1 = 1$ ✓

Substitute $x = t + 1/y$ . With $P=1$ , $Q=-2t$ : $y' + 0 \cdot y = -1$ .

Solve: $y = -t + C$ . Answer: $x = t + \frac{1}{C - t}$

6. Applications

6.1 Mixing Problems

Example 2.7: Salt Mixing Problem

Problem: A tank has 200 L of brine with 50 kg salt. Brine (0.5 kg/L) flows in at 6 L/min; solution flows out at 6 L/min. Find $S(t)$ .

Model: $\frac{dS}{dt} = 3 - \frac{3S}{100}$ , $S(0) = 50$ .

Integrating factor: $e^{3t/100}$ . Solution: $S = 100 + Ce^{-3t/100}$ .

Using IC: $C = -50$ . Answer: $S(t) = 100 - 50e^{-0.03t}$

6.2 Newton's Law of Cooling

Example 2.8: Cooling Problem

Problem: Coffee at 90°C in a 20°C room cools to 70°C in 5 min. Find $T(t)$ .

Model: $\frac{dT}{dt} = -k(T - 20)$ , $T(0) = 90$ .

Solution: $T = 20 + Ce^{-kt}$ . Using $T(0) = 90$ : $C = 70$ .

Using $T(5) = 70$ : $70 = 20 + 70e^{-5k}$ , so $e^{-5k} = 5/7$ , $k = \frac{\ln(7/5)}{5}$ .

Answer: $T(t) = 20 + 70e^{-kt}$ where $k \approx 0.0673$

Note:

In both applications, the solution approaches an equilibrium: salt concentration matches input (0.5 kg/L → 100 kg total); temperature approaches room temperature (20°C).

First-Order Linear Equations