Recipe

Build and solve a linear rate-in/rate-out model

Use when: a word problem describes a quantity $y$ that increases by an incoming rate and decreases by a rate proportional to $y$ itself — a tank, a circuit, an account balance

Goal: the standard-form linear equation for $y$ , then $y(t)$ itself via the integrating-factor method

Taught in Applications of the integrating-factor method

Quick recipe

Identify the rate in and the rate out
Assemble $\dfrac{dy}{dt}=(\text{rate in})-(\text{rate out})$
Rewrite in standard form and read off $P(t)$ , $Q(t)$
Solve using the integrating-factor method, then apply any initial condition

1
Write down the incoming rate as given in the problem, and the outgoing rate — usually a constant times the current amount $y$ itself.

Example

A tank holds $y(t)$ kg of a chemical. It flows in at a constant $8$ kg/min, and flows out at a rate proportional to the amount present, $0.2y$ kg/min.

Why: the "leaving depends on what's currently there" clause is exactly the $P(t)y$ term that will make the resulting equation linear.
2
Write the balance equation directly from Step 1's two rates.

Example

$\dfrac{dy}{dt}=8-0.2y$

Why: this rate-in-minus-rate-out balance is the physical law every one of these models shares, regardless of what $y$ represents.
3
Move the $y$ term to the left side so the equation reads $\dfrac{dy}{dt}+P(t)y=Q(t)$ .

Example

$\dfrac{dy}{dt}+0.2y=8$
so $P(t)=0.2$ (constant) and $Q(t)=8$ .

Why: standard form is what "Solve a first-order linear equation" (Section 2's recipe) needs before it can even start.
4
Run Section 2's recipe on the standard-form equation: build $\mu(t)=e^{\int P(t)\,dt}$ , multiply through, integrate, solve for $y$ , and apply $y(t_0)=y_0$ if given.

Example

$\mu(t)=e^{0.2t}$ ; $\dfrac{d}{dt}\left[e^{0.2t}y\right]=8e^{0.2t}$ ; integrating: $e^{0.2t}y=40e^{0.2t}+C$ , so $y=40+Ce^{-0.2t}$ . If $y(0)=50$ : $50=40+C$ , so $C=10$ , giving $\boxed{y=40+10e^{-0.2t}}$

Why: this is exactly Section 2's recipe, run start to finish, on the standard-form equation Step 3 just produced.

Common mistakes

Writing the rate out as a plain constant instead of proportional to the current amount $y$ , silently losing the $P(t)y$ term that makes the equation linear.
Reading $P(t)$ directly off the rate-out expression before moving it to the left side, missing the sign flip.
Solving for the constant using the wrong condition — an amount asked about later in the problem instead of the actual starting amount at $t_0$ .
Stopping at the standard-form equation from Step 3 as if it were already the answer, instead of carrying out Step 4's solving steps.