Difficulty: Intermediate | Prerequisites: Basic integration techniques, understanding of derivatives, familiarity with slope fields.
Separable differential equations are one of the first families of DEs you can solve exactly, and they appear on every AB Calculus exam. The core move is algebraic: rearrange so that each variable lives on its own side of the equation, then integrate both sides. Once you have the general solution, an initial condition pins down the constant and gives you a particular solution.
Before you can solve the DE, you often need to approximate its behaviour at a nearby point. That is where tangent-line approximation (linearisation) comes in, and the second derivative tells you whether your straight-line guess overshoots or undershoots the true curve. These three skills, separating variables, linearising, and checking concavity, form a single toolkit that the AP exam tests as one multi-part free-response question.
A separable DE can be split so each variable is on one side, then integrated. To approximate a solution without solving, use the tangent line at a known point. Check the second derivative to find out whether your approximation is too high or too low.
Separable differential equation
A first-order DE that can be written in the form f(y) dy = g(t) dt, so that all y-terms (including dy) are on one side and all t-terms (including dt) are on the other. Think of it as sorting laundry: every piece goes to one side or the other before you do anything else.
Tangent-line approximation (linearisation)
Using the tangent line at a known point (t₀, y₀) to estimate the value of y at a nearby point. The formula is y ≈ y(t₀) + y′(t₀)(t − t₀). In simple terms, you are pretending the curve is a straight line over a short interval.
Particular solution
The one specific solution to a DE that satisfies a given initial condition. The general solution contains an arbitrary constant C; plugging in the initial condition fixes C and gives the particular solution.
Initial condition
A known pair (t₀, y₀) that the solution must pass through. It pins the general family of curves down to a single curve.
Overestimate vs. underestimate
When a tangent-line approximation gives a value above the true curve, it is an overestimate; below, an underestimate. Concavity (the sign of the second derivative) determines which: if the curve is concave up, the tangent line lies below the curve, so the approximation underestimates; if concave down, the tangent line lies above, so it overestimates.
Start with a first-order DE in the form dH/dt = (1/12)(H − 10).
Separate: move all H-terms to the left, all t-terms to the right.
1/(H − 10) dH = (1/12) dt
Integrate both sides.
ln|H − 10| = t/12 + C
Solve for H by exponentiating.
H − 10 = Ae^(t/12), where A = ±e^C
H = Ae^(t/12) + 10
Apply the initial condition to find A.
If H(0) = 30, then 30 = A·1 + 10, so A = 20.
Particular solution: H(t) = 20e^(t/12) + 10.
You need two things: the function value and the derivative at the starting point.
At t = 0 with H(0) = 30:
dH/dt = (1/12)(30 − 10) = 20/12 = 5/3
Build the tangent line: y = (5/3)t + 30.
Evaluate at the target: H(1.5) ≈ (5/3)(1.5) + 30 = 2.5 + 30 = 32.5.
This gives an approximate height of 32.5 feet after 1.5 months.
Compute d²H/dt² by differentiating dH/dt with respect to t.
Since dH/dt = (1/12)(H − 10), apply the chain rule:
d²H/dt² = (1/12)(dH/dt) = (1/12) · (1/12)(H − 10) = (1/144)(H − 10)
On the interval [0, 1.5], H > 10 (the water level starts at 30 and is rising).
Therefore d²H/dt² > 0, which means the curve is concave up.
A tangent line to a concave-up curve lies below the curve.
Conclusion: the approximation of 32.5 feet is an underestimate.
With H(t) = 20e^(t/12) + 10, find the height after 2 years (t = 24 months):
H(24) = 20e^(24/12) + 10 = 20e² + 10
H(24) ≈ 20(7.389) + 10 ≈ 147.78 + 10 ≈ 157.78 feet.
\frac{dH}{dt} = \frac{1}{12}(H - 10)\text{Tangent-line approximation: } y \approx y(t_0) + y'(t_0)(t - t_0)\text{General solution: } H(t) = Ae^{t/12} + 10\text{Second derivative (chain rule on DE): } \frac{d^2H}{dt^2} = \frac{1}{12} \cdot \frac{dH}{dt} = \frac{1}{144}(H - 10)This exact type of DE models any quantity whose rate of change is proportional to how far it sits from an equilibrium value: water rising behind a dam, a hot object cooling toward room temperature (Newton's Law of Cooling), or a chemical concentration approaching a saturation level. The tangent-line technique is the basis of Euler's method, which engineers use to simulate systems that have no closed-form solution.
Students often forget to separate variables completely before integrating. Every y-piece (including dy) must be on one side and every t-piece (including dt) on the other. If you integrate with mixed variables, the antiderivatives will be wrong.
A common error is concluding that "concave up means the approximation is too high." It is the opposite. A tangent to a concave-up curve sits below the curve, so the linear estimate underestimates.
When exponentiating ln|H − 10| = t/12 + C, students sometimes write H − 10 = e^(t/12) + e^C. This is incorrect; the exponent of a sum is a product: H − 10 = e^C · e^(t/12).
Students sometimes plug the initial condition into the general solution before solving for the constant. The constant C from integration is not the same as the initial value. You must first solve for C, then write the particular solution.
⚠️ Separable DEs with tangent-line approximation and concavity analysis appear as a multi-part free-response question on nearly every AP Calculus AB exam.
⚠️ You will be asked to justify whether an approximation is an overestimate or underestimate. The expected justification: state the sign of the second derivative on the interval, name the concavity, and conclude. A bare answer with no reasoning earns zero credit.
⚠️ When solving the DE, show every step of the separation and integration. Graders award points for the setup (correct separation), the antiderivatives, applying the initial condition, and the final explicit solution.
True or False: To separate dH/dt = (1/12)(H − 10), you move dH to the left and dt to the right.
Fill in the blank: The tangent-line approximation uses the formula y ≈ ____.
True or False: If d²H/dt² > 0 on an interval, a tangent-line approximation on that interval is an overestimate.
Fill in the blank: After integrating 1/(H − 10) dH = (1/12) dt, the left side becomes ____.
True or False: The initial condition is applied before integrating both sides of the separated equation.
Answers: 1. True. 2. y(t₀) + y′(t₀)(t − t₀). 3. False (it is an underestimate). 4. ln|H − 10|. 5. False (integrate first, then apply the initial condition).
Q: Given dH/dt = (1/12)(H − 10) with H(0) = 30, find the tangent-line approximation for H(1.5).
A: At t = 0, dH/dt = (1/12)(30 − 10) = 5/3. The tangent line is y = (5/3)t + 30. So H(1.5) ≈ (5/3)(1.5) + 30 = 32.5 feet.
Q: Explain why the tangent-line approximation from the previous question is an underestimate.
A: d²H/dt² = (1/144)(H − 10). Since H > 10 on [0, 1.5], the second derivative is positive, so the graph is concave up. A tangent line to a concave-up graph lies below the curve, making the approximation an underestimate.
Q: Solve dH/dt = (1/12)(H − 10) with H(0) = 30 and find H(24).
A: Separate and integrate: ln|H − 10| = t/12 + C. At t = 0, ln|20| = C. So ln(H − 10) = t/12 + ln 20, giving H − 10 = 20e^(t/12), hence H(t) = 20e^(t/12) + 10. At t = 24: H(24) = 20e² + 10 ≈ 157.78 feet.
Q: For a general separable DE of the form dy/dx = f(x)g(y), outline the solution method in order.
A: (1) Rewrite as (1/g(y)) dy = f(x) dx. (2) Integrate both sides. (3) Solve for y if possible. (4) Apply the initial condition to determine the constant of integration.
This material connects directly to slope fields, where you sketch the direction of solutions without solving the DE. Knowing how to separate and solve gives you the curves that the slope field hints at.
Tangent-line approximation is the one-step version of Euler's method, which extends the idea to many small steps and appears in BC Calculus.
The concavity argument (second derivative test for over/underestimate) reuses the same reasoning you apply when determining whether a critical point is a local max or min in optimisation problems.
Separable differential equations, separation of variables, tangent line approximation, linearisation, linearization, local linear approximation, Euler's method, concavity, second derivative test, overestimate, underestimate, initial value problem, IVP, particular solution, general solution, Newton's Law of Cooling, exponential growth, AB Calculus free response, AP Calc DE problem, slope fields, dH/dt, rate of change proportional to difference from equilibrium.