Differential Equations
Unit 7 of AP Calculus AB, worth 6–12% of the exam. 13 questions below, each with the working. Every answer was checked by a second pass before it was published.
Modelling with differential equations, verifying solutions, slope fields, separation of variables, exponential models.
How this unit is tested
Start by learning to translate a verbal statement about a rate of change into an equation involving $\frac{dy}{dx}$ or $\frac{dy}{dt}$ — phrases like 'proportional to' become $\frac{dy}{dt}=ky$ or $\frac{dy}{dt}=k(y-L)$. Once you can write a differential equation, practice verifying whether a given function is a solution: differentiate the candidate function, substitute both the function and its derivative into the original equation, and confirm both sides are identically equal for all $x$ (not just at one point).
For slope fields, remember that a differential equation of the form $\frac{dy}{dx}=f(x,y)$ gives you a slope at every point $(x,y)$ in the plane. To read a slope field, plug specific $(x,y)$ coordinates into $f(x,y)$ to get the slope of the short segment drawn there; to sketch a solution curve, start at the initial point and follow the direction of the segments. Look for symmetry, horizontal segments (where $f(x,y)=0$), and vertical or steep segments to match a picture to its equation.
Separation of variables is the core algebraic technique of the unit: get all $y$ terms (including $dy$) on one side and all $x$ terms (including $dx$) on the other, integrate both sides separately, and add a single constant of integration. If an initial condition is given, substitute it immediately after integrating to solve for the constant — this turns the general solution into the particular solution. Many AP problems accept an implicit solution (an equation relating $x$ and $y$) as a final answer; only solve explicitly for $y$ when the problem asks for it or when it's needed to answer a numerical question.
Finally, recognize that $\frac{dy}{dt}=ky$ always has solution $y=Ce^{kt}$ — memorize this shortcut so you don't have to re-derive it via separation every time, but know how to derive it because the AP exam sometimes asks you to. Practice extracting $k$ from data (doubling time, half-life, two data points) and then using the exponential formula to answer growth/decay questions.
What you have to know
Differential equation and solution
A differential equation is an equation involving a derivative, such as $\frac{dy}{dx}=f(x,y)$. A function $y=g(x)$ is a solution if substituting $g(x)$ for $y$ and $g'(x)$ for $\frac{dy}{dx}$ makes the equation true for all $x$ in the domain.
General vs. particular solution
The general solution of a differential equation contains an arbitrary constant $C$ and represents a family of curves. A particular solution satisfies one additional initial condition, which is used to solve for the specific value of $C$.
Slope field
A slope field for $\frac{dy}{dx}=f(x,y)$ consists of short line segments drawn at grid points $(x,y)$, each with slope equal to $f(x,y)$ evaluated at that point; a solution curve is tangent to the field at every point it passes through.
Separation of variables
If $\frac{dy}{dx}=g(x)h(y)$, divide both sides by $h(y)$ to write $\frac{1}{h(y)}dy=g(x)dx$, then integrate both sides and add a constant of integration to obtain the general solution.
Exponential growth/decay model
If $\frac{dy}{dt}=ky$ for a constant $k$, then $y=Ce^{kt}$ for some constant $C$; $k>0$ gives growth and $k<0$ gives decay. The constant $C$ equals the value of $y$ when $t=0$.
13 practice questions
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Verify that $y=3e^{2x}$ is a solution of the differential equation $\frac{dy}{dx}-2y=0$.
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Answer. Yes, it is a solution.
Differentiate: $\frac{dy}{dx}=6e^{2x}$. Substitute: $6e^{2x}-2(3e^{2x})=6e^{2x}-6e^{2x}=0$, which matches the right side of the equation for all x. -
Which differential equation is satisfied by every function of the form $y=Ce^{-5x}$?
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Answer. $\frac{dy}{dx}=-5y$
Differentiating gives $\frac{dy}{dx}=-5Ce^{-5x}=-5y$, so the family satisfies $\frac{dy}{dx}=-5y$ for any constant C. -
A rumor spreads through a school of 800 students. The rate at which the rumor spreads is proportional to the product of the number of students who have heard it, y, and the number who have not, (800-y). Write a differential equation modeling dy/dt.
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Answer. $\frac{dy}{dt}=ky(800-y)$
The phrase 'proportional to the product of' two quantities becomes k times the product of those two quantities, giving the stated equation with proportionality constant k. -
For the differential equation $\frac{dy}{dx}=x-y$, find the slope indicated by the slope field at the point (1,3).
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Answer. -2
Substitute x=1 and y=3 into $x-y$: $1-3=-2$. This is the slope of the short segment drawn at that point in the slope field. -
For the differential equation $\frac{dy}{dx}=xy$, at which points does the slope field show horizontal segments?
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Answer. Along the x-axis (y=0) and along the y-axis (x=0)
Horizontal segments occur where $\frac{dy}{dx}=0$. Since $xy=0$ whenever x=0 or y=0, the slope field is horizontal along both axes. -
Find the general solution to $\frac{dy}{dx}=\frac{y}{x}$.
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Answer. $y=Ax$ for a constant A
Separate variables: $\frac{dy}{y}=\frac{dx}{x}$. Integrate: $\ln|y|=\ln|x|+C$. Exponentiate: $|y|=e^{C}|x|$, so $y=Ax$ where A absorbs the sign and the constant. -
Solve the initial value problem $\frac{dy}{dx}=-2xy$, $y(0)=5$.
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Answer. $y=5e^{-x^2}$
Separate: $\frac{dy}{y}=-2x\,dx$. Integrate: $\ln|y|=-x^2+C$, so $y=Ae^{-x^2}$. Using $y(0)=5$ gives $A=5$, so $y=5e^{-x^2}$. -
A bacteria culture grows at a rate proportional to its size. It starts at 200 bacteria and grows to 600 after 3 hours. Find the growth constant k.
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Answer. $k=\frac{\ln 3}{3}\approx 0.366$
The model is $y=200e^{kt}$. At t=3, $600=200e^{3k}$, so $e^{3k}=3$ and $k=\frac{\ln 3}{3}$. -
A radioactive substance has a half-life of 10 days. Find its decay constant k in the model $y=y_0e^{kt}$ (t in days).
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Answer. $k=\frac{\ln(0.5)}{10}\approx -0.0693$
Half-life means $y=0.5y_0$ when $t=10$: $0.5=e^{10k}$, so $k=\frac{\ln(0.5)}{10}$, a negative number since the quantity decays. -
A cup of coffee obeys Newton's Law of Cooling, $\frac{dT}{dt}=k(T-70)$, where 70°F is room temperature. If $T(0)=200$ and $T(5)=150$, find $k$.
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Answer. $k=\frac{1}{5}\ln\left(\frac{8}{13}\right)$
Separating gives $T-70=Ae^{kt}$ with $A=200-70=130$. At t=5: $150-70=80=130e^{5k}$, so $e^{5k}=\frac{8}{13}$ and $k=\frac{1}{5}\ln\left(\frac{8}{13}\right)$, which is negative since the coffee cools. -
For the differential equation $\frac{dy}{dx}=y-x$, what happens to the slope field segments along the line y=x?
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Answer. They are horizontal (slope 0)
Substituting y=x into $y-x$ gives 0 for every point on that line, so the slope field shows flat, horizontal segments all along y=x. -
An investment grows continuously according to $\frac{dA}{dt}=0.05A$ with $A(0)=1000$. Find the value of the investment after 10 years.
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Answer. $A(10)=1000e^{0.5}\approx 1648.72$
The solution is $A(t)=1000e^{0.05t}$. Substituting t=10 gives $A(10)=1000e^{0.5}$, which evaluates to approximately 1648.72. -
Find the general solution (an implicit equation relating x and y is acceptable) to $\frac{dy}{dx}=\frac{2x}{1+y^2}$.
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Answer. $y+\frac{y^3}{3}=x^2+C$
Separate variables: $(1+y^2)\,dy=2x\,dx$. Integrate both sides: $y+\frac{y^3}{3}=x^2+C$. This implicit relation is a valid final answer since solving explicitly for y is not required.
What people get wrong
- Forgetting to add the constant of integration when separating variables, or forgetting to solve for it using the initial condition — always write '+C' immediately after integrating, then substitute the given point before doing anything else.
- Dropping the absolute value when integrating $\int \frac{1}{y}dy = \ln|y|+C$, then mishandling the sign when exponentiating — carry the absolute value through and let the constant $A=\pm e^{C}$ absorb the sign, determined later by the initial condition.
- Treating $\frac{dy}{dx}=ky$ as if it produces a linear function $y=kx+C$ instead of the exponential solution $y=Ce^{kt}$ — memorize that this specific equation always gives exponential, not linear, growth.
- Verifying a solution by only checking one point instead of confirming the equation holds identically for all x — differentiate symbolically and simplify both sides completely.
- Choosing the wrong sign when solving explicitly for y from an implicit relation like $y^2=x^2+5$ — always check the sign against the given initial condition, not by default assuming the positive root.
- Making an algebra error when separating variables, such as moving a term to the wrong side or forgetting to distribute — rewrite the equation carefully as (function of y) dy = (function of x) dx before integrating.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Calculus AB. Free, and no account needed to start.