4.5 First-order Linear Equations

First-order Linear Equations

First-order linear equations are differential equations where the dependent variable and its derivative both appear to the first power. They show up constantly in modeling: population growth, radioactive decay, cooling, mixing problems. The key technique for solving them is the integrating factor method, which this section is all about.

Standard Form of Linear Equations

Before you can solve a first-order linear equation, you need to get it into standard form:

$\frac{dy}{dx} + P(x)y = Q(x)$

Here, $P(x)$ and $Q(x)$ are functions of $x$ only (or constants). The requirements are:

• The coefficient of $\frac{dy}{dx}$ must be 1. If it's not, divide the entire equation through until it is.
• The coefficient of $y$ must depend only on $x$, not on $y$.
• Everything on the right side must depend only on $x$.

For example, if you're given $2\frac{dy}{dx} + 6xy = 4x$, divide everything by 2 first to get $\frac{dy}{dx} + 3xy = 2x$. Now it's in standard form with $P(x) = 3x$ and $Q(x) = 2x$.

Integrating Factors for Equation Solving

The integrating factor is the tool that makes first-order linear equations solvable in a systematic way. The idea: multiply both sides of the equation by a carefully chosen function that turns the left side into the derivative of a product, which you can then integrate directly.

Given the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, here's the process:

1. Compute the integrating factor: $\mu(x) = e^{\int P(x)\,dx}$ You don't need a constant of integration here; any single antiderivative of $P(x)$ works.

2. Multiply both sides by $\mu(x)$: $\mu(x)\frac{dy}{dx} + \mu(x)P(x)y = \mu(x)Q(x)$

3. Recognize the left side as a product rule derivative. This is the whole point of the integrating factor. The left side equals: $\frac{d}{dx}\bigl[\mu(x)\,y\bigr] = \mu(x)Q(x)$

4. Integrate both sides with respect to $x$: $\mu(x)\,y = \int \mu(x)\,Q(x)\,dx + C$

5. Solve for $y$: $y = \frac{1}{\mu(x)}\left(\int \mu(x)\,Q(x)\,dx + C\right)$

Why does this work? The integrating factor $\mu(x)$ is specifically constructed so that $\mu'(x) = \mu(x)P(x)$. That's what makes the left side collapse into a single derivative via the product rule. If you ever forget the formula, you can re-derive it from this condition.

Worked Example

Solve $\frac{dy}{dx} + 2y = e^{3x}$.

1. It's already in standard form: $P(x) = 2$, $Q(x) = e^{3x}$.
2. Integrating factor: $\mu(x) = e^{\int 2\,dx} = e^{2x}$.
3. Multiply both sides by $e^{2x}$: $e^{2x}\frac{dy}{dx} + 2e^{2x}y = e^{5x}$
4. The left side is $\frac{d}{dx}[e^{2x}y]$, so: $\frac{d}{dx}[e^{2x}y] = e^{5x}$
5. Integrate: $e^{2x}y = \frac{1}{5}e^{5x} + C$.
6. Solve for $y$: $y = \frac{1}{5}e^{3x} + Ce^{-2x}$.

Real-World Applications of Linear Equations

First-order linear equations model situations where a quantity changes at a rate that depends on the quantity itself and possibly some external input. Common applications include population dynamics, radioactive decay, heat transfer, and mixing problems.

To apply the method to a word problem:

1. Identify the variables (what's changing, and with respect to what).
2. Translate the problem's description into a differential equation.
3. Note any initial conditions (e.g., "at time $t = 0$, the temperature is 90°C").
4. Put the equation in standard form and solve using the integrating factor.
5. Use the initial condition to find $C$, then interpret your answer.

Classic Models

• Exponential growth/decay: $\frac{dP}{dt} = kP$, where $k > 0$ gives growth and $k < 0$ gives decay. This is separable and linear. Solution: $P(t) = P_0 e^{kt}$.
• Radioactive decay: $\frac{dN}{dt} = -\lambda N$, where $\lambda$ is the decay constant. If a substance has a half-life of 5 years, then $\lambda = \frac{\ln 2}{5} \approx 0.1386 \text{ yr}^{-1}$.
• Newton's law of cooling: $\frac{dT}{dt} = -k(T - T_a)$, where $T_a$ is the ambient temperature. Rewriting: $\frac{dT}{dt} + kT = kT_a$. This is linear with $P(t) = k$ and $Q(t) = kT_a$. For instance, a cup of coffee at 90°C in a 20°C room cools according to this model.
• Mixing problems: A tank holds a solution, and liquid flows in and out at given rates. The equation typically looks like $\frac{dA}{dt} = (\text{rate in}) - (\text{rate out})$, where $A(t)$ is the amount of solute. The "rate out" term usually depends on $A$, making the equation linear.

Types of First-Order Differential Equations

First-order linear equations are one type among several. Knowing the landscape helps you pick the right technique:

• Separable equations: Variables can be separated to opposite sides and integrated independently. Example: $\frac{dy}{dx} = xy$ becomes $\frac{dy}{y} = x\,dx$.
• Homogeneous equations: Can be reduced to separable form using the substitution $v = y/x$.
• Exact equations: Have the form $M(x,y)\,dx + N(x,y)\,dy = 0$ where $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Solved by finding a potential function.
• Bernoulli equations: Have the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$ where $n \neq 0, 1$. These are nonlinear, but the substitution $v = y^{1-n}$ transforms them into a linear equation you can solve with an integrating factor.

A given equation might fit more than one category. For example, $\frac{dP}{dt} = kP$ is both separable and linear. Use whichever method you find more straightforward.