AP Calculus AB Unit 7 Study Guide: Differential Equations (Slope Fields, Separable IVPs, Euler’s Method, Exponential & Logistic Models)

What Differential Equations Are and How They Model Change

A differential equation is an equation that relates an unknown function (often written as %%LATEX0%% as a function of %%LATEX1%%) to one or more of its derivatives (like \frac{dy}{dx}). In plain language: instead of telling you the function directly, it tells you how the function changes.

This connects nicely to ideas you’ve already seen in related rates: in both topics, the main object is a rate of change (a derivative). The difference is that in differential equations, you’re often using a rule about rates to reconstruct the original function.

That idea matters because many real-world situations are naturally described by change rules rather than explicit formulas. For example:

• A population might grow at a rate proportional to its current size.
• The amount of medicine in your bloodstream might decrease at a rate proportional to how much is present.
• The temperature of coffee might cool at a rate proportional to the difference between the coffee’s temperature and room temperature.

In each case, you often know (or assume) a rule for the rate of change, and you want to find the function itself.

Differential equation vs. explicit function

In earlier units, you often started with a function %%LATEX3%% and then found %%LATEX4%%. Differential equations reverse the direction:

• You start with something like \frac{dy}{dx}=g(x,y).
• You use calculus (integration, plus algebra) to work back to y.

A key point: a differential equation typically has infinitely many solutions, a whole family of functions, because differentiation “loses information” (constants disappear). To pick one specific solution, you need an initial condition.

Notation you must recognize

Differential equations appear in several equivalent notations. You’re expected to translate among them fluently.

You’ll see both %%LATEX14%% and %%LATEX15%% on AP-style questions.

Solution, general solution, particular solution

A solution to a differential equation is a function %%LATEX16%% that makes the equation true when you substitute %%LATEX17%% and its derivative(s).

A general solution describes a family of solutions using an arbitrary constant %%LATEX18%%. A particular solution is the one function you get after using an initial condition to find %%LATEX19%%.

For example, if solving leads to

y=x^2+C

that’s a general solution. If you also know %%LATEX21%%, then %%LATEX22%% so C=5 and the particular solution is

y=x^2+5

Checking a proposed solution (a skill that shows up often)

To verify that a function solves a differential equation:

1. Compute the derivative(s) required by the differential equation.
2. Substitute the function and its derivative(s) into the differential equation.
3. Confirm both sides match for all x in the domain.

This is conceptually simple but easy to mess up with derivative errors.

Example: verify a solution

Check whether

y=e^{2x}

solves

\frac{dy}{dx}=2y

Compute derivative:

\frac{dy}{dx}=2e^{2x}

Right-hand side:

2y=2e^{2x}

They match, so y=e^{2x} is a solution.

Exam Focus

• Typical question patterns:
  • “Show that the function %%LATEX31%% satisfies the differential equation %%LATEX32%%.”
  • “Write a differential equation that models the situation described (rate proportional to…, limited by carrying capacity, etc.).”
  • “Find the particular solution given an initial condition.”
• Common mistakes:
  • Forgetting that solutions are functions: plugging in a single point instead of substituting the whole function.
  • Mixing up %%LATEX33%% and %%LATEX34%% (for example, substituting %%LATEX35%% where %%LATEX36%% belongs).
  • Differentiation slips when verifying (especially chain rule with exponentials).

Slope Fields (Direction Fields) and Qualitative Solutions

A slope field (also called a direction field) is a picture that shows the slope %%LATEX37%% at many points (x, y) without explicitly solving the differential equation. At each point, you draw a small line segment with slope equal to the value of %%LATEX38%% in

\frac{dy}{dx}=g(x,y)

This matters because many differential equations are hard (or impossible in a typical AP setting) to solve symbolically, but you can still understand behavior: where solutions increase or decrease, where they level off, and how initial conditions change the resulting curve.

How to construct a slope field (plug-and-draw)

To construct a slope field, you plug in the coordinates (the x-value, the y-value, or both, depending on the differential equation) into the right-hand side and sketch a short segment with that slope.

For example, for

\frac{dy}{dx}=x

the slope at any point with %%LATEX41%% is %%LATEX42%% (because the slope depends only on x).

How to read a slope field

When you look at the small segments:

• Segments slanting upward mean \frac{dy}{dx}>0 so solution curves increase.
• Segments slanting downward mean \frac{dy}{dx}

A solution curve is a smooth curve that is always tangent to the tiny segments. You can sketch one solution by “following the flow” of the field.

The AP exam might require you to sketch a solution curve given a slope field. The key is to “flow” with the slopes: your curve should look like it is made of tangent pieces aligned with the little segments. Because this is done by hand, it doesn’t have to be exact, but it should not cross the direction segments abruptly.

Isoclines (where slopes are equal)

An isocline is a curve in the xy-plane along which the slope field has the same slope value. If

\frac{dy}{dx}=g(x,y)=k

then the set of points satisfying %%LATEX48%% is an isocline for slope %%LATEX49%%. Isoclines help you organize slope fields: find where slope is %%LATEX50%%, where slope is %%LATEX51%%, etc.

Equilibrium (constant) solutions

An equilibrium solution (also called a constant solution) is a solution where %%LATEX52%% stays constant as %%LATEX53%% changes, meaning

\frac{dy}{dx}=0

If the differential equation is autonomous (depends only on y), like

\frac{dy}{dx}=f(y)

then equilibria occur at values of %%LATEX57%% where %%LATEX58%%. On a slope field, these show up as horizontal segments across an entire horizontal line.

Equilibria matter because they often represent long-term behavior: populations stabilizing, temperatures approaching room temperature, and so on.

Stability (qualitative)

If you have an autonomous differential equation %%LATEX59%% and an equilibrium at %%LATEX60%%:

• The equilibrium is stable if solution curves near %%LATEX61%% move toward %%LATEX62%% over time.
• It is unstable if nearby solutions move away.

You can often tell from the sign of %%LATEX63%% above and below %%LATEX64%%:

• If %%LATEX65%% just below %%LATEX66%% (solutions increase) and %%LATEX67%% just above %%LATEX68%% (solutions decrease), then solutions are pushed toward a: stable.
• Reverse signs imply unstable.

Example: stability from signs

Suppose

\frac{dy}{dx}=y(3-y)

Equilibria: solve %%LATEX71%% giving %%LATEX72%% and y=3.

Test intervals:

• If %%LATEX74%%, then %%LATEX75%% and %%LATEX76%% so %%LATEX77%%: solutions increase.
• If %%LATEX78%%, then %%LATEX79%% so \frac{dy}{dx}

So solutions move toward %%LATEX81%% from both sides: %%LATEX82%% is stable. Near %%LATEX83%%, if %%LATEX84%% then %%LATEX85%% so solutions decrease (move away from %%LATEX86%%), and if %%LATEX87%% they increase (also away from %%LATEX88%%), so y=0 is unstable.

Exam Focus

• Typical question patterns:
  • “Sketch the slope field for \frac{dy}{dx}=g(x,y) and sketch the solution curve through a given initial point.”
  • “Use the slope field to estimate %%LATEX91%% given %%LATEX92%%.”
  • “Identify equilibrium solutions and classify stability from a direction field or from the sign of f(y).”
• Common mistakes:
  • Drawing solution curves that cross direction segments instead of staying tangent.
  • Treating the slope as depending only on %%LATEX94%% or only on %%LATEX95%% when it depends on both.
  • Confusing an isocline (same slope) with a solution curve (a specific integral curve).
  • Forgetting that a slope field is built by plugging points into the differential equation (for example, for %%LATEX96%%, all points with the same %%LATEX97%% share the same slope).

Initial Value Problems (IVPs) and Particular Solutions

An initial value problem (IVP) is a differential equation together with an initial condition that specifies the value of the solution at a particular input.

A typical IVP looks like:

\frac{dy}{dx}=g(x,y)

y(x_0)=y_0

The initial condition is what selects one unique member from the family of solutions.

Why IVPs matter

In applications, you almost always know some starting measurement: the initial population, initial temperature, initial amount of a drug, initial position, etc. The differential equation gives a general rule of change, but the initial condition pins down the actual situation.

Existence and uniqueness (conceptual)

In AP Calculus AB, you’re generally expected to know the idea that under reasonable conditions (specifically when %%LATEX100%% and its partial derivative with respect to %%LATEX101%% are continuous near the initial point), an IVP has a unique solution through that point. Practically:

• A well-behaved slope field should show exactly one solution curve passing through the initial point.

If a slope field shows multiple curves that could pass through the same point (or a cusp/vertical tangent situation where the differential equation isn’t defined nicely), uniqueness can fail.

Using initial conditions after solving

A very common workflow:

1. Solve the differential equation to get a general solution with a constant C.
2. Plug the initial condition into the general solution to find C.
3. Write the particular solution.

Example: particular solution from an IVP

Solve the IVP:

\frac{dy}{dx}=6x

y(2)=3

Step 1: integrate both sides with respect to x:

y=\int 6x\,dx

y=3x^2+C

Step 2: use the initial condition y(2)=3:

3=3(2^2)+C

3=12+C

C=-9

So the particular solution is:

y=3x^2-9

A subtle but important habit: when you integrate, always include +C (or you will lose the ability to satisfy the initial condition).

Exam Focus

• Typical question patterns:
  • “Find the particular solution to the differential equation that satisfies y(a)=b.”
  • “Given a slope field and a point, sketch the solution and estimate a value.”
  • “Determine whether the solution is increasing/decreasing at a point using the differential equation.”
• Common mistakes:
  • Forgetting the constant of integration, then being unable to match the initial condition.
  • Plugging the initial condition into the differential equation instead of into the solved function.
  • Mixing up the meaning of y(a)=b (it means the point (a, b) lies on the solution curve).

Euler’s Method: Approximating Solutions Numerically

Often you cannot (or are not asked to) solve a differential equation explicitly. Euler’s method is a numerical technique that approximates the solution to an IVP by using tangent line steps.

Suppose you have:

\frac{dy}{dx}=g(x,y)

and an initial condition y(x_0)=y_0.

Euler’s method uses the idea: near %%LATEX119%%, the solution curve looks like its tangent line. The slope at (x0, y0) is %%LATEX120%%, so if you take a small step of size %%LATEX121%% in %%LATEX122%%, the change in y is approximately slope times run:

\Delta y\approx g(x_0,y_0)h

So the next approximate point is:

x_1=x_0+h

y_1=y_0+hg(x_0,y_0)

Then you repeat: compute the slope at (x1, y1), step again, and so on.

The Euler update formulas

If you label the approximate points (x_n,y_n), then:

x_{n+1}=x_n+h

y_{n+1}=y_n+hg(x_n,y_n)

Here:

• %%LATEX130%% is the step size (for example, %%LATEX131%% or 0.5).
• Smaller h usually gives better accuracy but requires more steps.

Why Euler’s method makes sense (conceptually)

Euler’s method is repeated local linearization. Every step uses a tangent line approximation. It’s the same idea as using the linearization formula from derivatives:

y(x+h)\approx y(x)+y'(x)h

but with %%LATEX135%% replaced by %%LATEX136%% because the slope depends on both variables.

Error behavior (what you should know)

Euler’s method can accumulate error because:

• Each step uses an approximation.
• If the approximation is slightly off, the next slope calculation is based on a slightly wrong point.

You should still expect these truths:

• Decreasing h tends to improve the approximation.
• Over many steps, errors can grow.

AP questions often test whether you can execute the algorithm accurately and interpret what the approximation means.

Example: Euler’s method computation

Approximate y(0.3) for the IVP:

\frac{dy}{dx}=x+y

y(0)=1

Use step size h=0.1.

Start: %%LATEX142%%, %%LATEX143%%.

Step to x_1=0.1:

Slope at (0, 1):

g(0,1)=0+1=1

Update:

y_1=1+0.1(1)=1.1

Step to x_2=0.2:

Slope at (0.1, 1.1):

g(0.1,1.1)=0.1+1.1=1.2

Update:

y_2=1.1+0.1(1.2)=1.22

Step to x_3=0.3:

Slope at (0.2, 1.22):

g(0.2,1.22)=0.2+1.22=1.42

Update:

y_3=1.22+0.1(1.42)=1.362

So Euler’s method gives

y(0.3)\approx 1.362

A common interpretation point: this number is not exact; it is an approximation produced by three tangent-line steps.

Exam Focus

• Typical question patterns:
  • “Use Euler’s method with step size %%LATEX154%% to approximate %%LATEX155%% at a given x value.”
  • “Set up a table of values for Euler’s method and fill in the missing entries.”
  • “Compare two Euler approximations with different step sizes and decide which should be more accurate.”
• Common mistakes:
  • Using the wrong point when computing slope (mixing %%LATEX157%% with %%LATEX158%%).
  • Forgetting to multiply by step size h.
  • Rounding too early and compounding error; keep several decimals until the end unless instructed otherwise.

Separable Differential Equations: Solving by Separating Variables

A separable differential equation is one where you can algebraically rearrange the equation so that all %%LATEX160%%-terms are on one side and all %%LATEX161%%-terms are on the other. This is the main symbolic solving technique emphasized in AP Calculus AB for differential equations.

The general form you’re aiming for is:

\frac{dy}{dx}=f(x)g(y)

Then you “separate”:

\frac{1}{g(y)}\,dy=f(x)\,dx

and integrate both sides.

Why separation works

At a conceptual level, separating variables is leveraging the fact that integration “undoes” differentiation. If you can express the derivative as a product of an %%LATEX164%%-only part and a %%LATEX165%%-only part, then you can integrate each side with respect to its own variable.

A frequent point of confusion is the notation. When you write

\frac{dy}{dx}=f(x)g(y)

you can treat it like a fraction for the purpose of rearranging:

\frac{1}{g(y)}\frac{dy}{dx}=f(x)

and then multiply both sides by dx:

\frac{1}{g(y)}dy=f(x)dx

This is a standard technique in AP Calculus, even though in more advanced math courses you learn a more rigorous justification.

The general solving process (with key habits)

A reliable checklist is:

1. Separate variables so that the left side involves only %%LATEX170%% and %%LATEX171%%, and the right side involves only %%LATEX172%% and %%LATEX173%%.
2. Integrate both sides.
3. Add a constant of integration (usually just on one side).
4. Solve for y if possible (sometimes leaving an implicit solution is acceptable).
5. Apply initial conditions to find the constant and get a particular solution.

A helpful memory trick for these problems is SIPPY:

• S: Separate (put %%LATEX175%% and %%LATEX176%% on separate sides)
• I: Integrate (remove the derivative)
• P: Plus C (include the constant)
• P: Plug in your initial condition
• Y: Y equals (solve to isolate y, if possible)

Two habits prevent many errors:

• Make sure you integrate with respect to the correct variable.
• Do not drop absolute values when integrating \frac{1}{y}.

Example 1: basic separation

Solve:

\frac{dy}{dx}=xy

Step 1: separate:

\frac{1}{y}dy=x\,dx

Step 2: integrate:

\int \frac{1}{y}dy=\int x\,dx

\ln|y|=\frac{x^2}{2}+C

Step 3: solve for y by exponentiating:

|y|=e^{\frac{x^2}{2}+C}

Rewrite %%LATEX185%% as a new constant %%LATEX186%%:

|y|=Ke^{\frac{x^2}{2}}

This means %%LATEX188%% could be positive or negative; you can absorb the sign into a nonzero constant %%LATEX189%%:

y=Ae^{\frac{x^2}{2}}

That is the general solution.

Example 2: separation with an initial condition

Solve the IVP:

\frac{dy}{dx}=(1+x^2)y

y(0)=4

Separate:

\frac{1}{y}dy=(1+x^2)dx

Integrate:

\ln|y|=x+\frac{x^3}{3}+C

Exponentiate:

y=Ae^{x+\frac{x^3}{3}}

Use y(0)=4:

4=Ae^{0}

So A=4 and the particular solution is:

y=4e^{x+\frac{x^3}{3}}

Example 3: separable IVP from the SIPPY checklist

Solve:

\frac{dy}{dx}=\frac{4x}{y}

with

y(0)=5

Separate by multiplying both sides by y\,dx:

y\,dy=4x\,dx

Integrate both sides:

\int y\,dy=\int 4x\,dx

\frac{y^2}{2}=2x^2+C

Plus C is essential here.

Plug in the initial condition y(0)=5:

\frac{25}{2}=0+C

So

C=\frac{25}{2}

Now Y equals: first rewrite the equation, then solve for y.

Multiply both sides by 2:

y^2=4x^2+25

Take the square root:

y=\pm \sqrt{4x^2+25}

Since y(0)=5 is positive, we choose the positive branch:

y=\sqrt{4x^2+25}

Implicit solutions are sometimes the natural stopping point

Not every separable differential equation solves neatly for y. For example, you might end with something like

y+\ln|y|=x^2+C

That is still a valid solution description. On AP-style tasks, if the question asks you to “solve” without specifying explicit form, an implicit solution is often acceptable.

Exam Focus

• Typical question patterns:
  • “Solve the differential equation by separation of variables.”
  • “Find the particular solution that satisfies y(x_0)=y_0.”
  • “Show that a given function is a solution by substituting into the differential equation.”
• Common mistakes:
  • Separating incorrectly (for example, leaving an %%LATEX217%% on the %%LATEX218%% side).
  • Forgetting absolute value in %%LATEX219%% after integrating %%LATEX220%%.
  • Losing constant solutions (equilibria) by dividing by an expression that could be zero (for example, dividing by %%LATEX221%% without noting %%LATEX222%% may be a solution).
  • Forgetting the “Plus C” step, which prevents you from fitting an initial condition.

Differential Equations as Models: Setting Up Equations From Words

A big part of Unit 7 is translating a description of a changing quantity into a differential equation. In modeling, the derivative represents a rate with units that tell you what the derivative means.

If %%LATEX223%% is a quantity (population, mass, temperature) and %%LATEX224%% is time, then

\frac{dy}{dx}

means “rate of change of %%LATEX226%% per unit time.” If %%LATEX227%% is measured in people and %%LATEX228%% in days, then %%LATEX229%% has units people per day.

Proportionality models

The phrase “proportional to” is a huge clue. If a rate is proportional to a quantity, you typically write:

\frac{dy}{dx}=ky

where k is the constant of proportionality.

If it’s proportional to the difference between a quantity and some baseline A, you often write:

\frac{dy}{dx}=k(y-A)

The sign of %%LATEX234%% determines whether the quantity moves toward or away from %%LATEX235%%.

Interpreting parameter meanings

In a model like

\frac{dy}{dx}=ky

• %%LATEX237%% means growth (the larger %%LATEX238%% is, the faster it increases).
• %%LATEX239%% means decay (the larger %%LATEX240%% is, the faster it decreases).

In models like logistic growth, parameters have concrete interpretations (carrying capacity, growth rate), and AP problems may ask what those parameters mean in context.

Example: build a differential equation

“A bacteria culture grows at a rate proportional to the number of bacteria present.”

Let %%LATEX241%% be the number of bacteria at time %%LATEX242%%. “Rate proportional to the amount present” translates to:

\frac{dB}{dt}=kB

If it additionally says “it doubles every 3 hours,” that would be used later to solve for k after solving the differential equation.

Example: rate depends on both time and amount

If the statement gives a rate explicitly as a function of both variables, you write it directly. For instance:

“The rate of change of %%LATEX245%% is given by %%LATEX246%%.”

That already is the differential equation. Your job might then be slope field analysis or numerical approximation rather than symbolic solution.

Exam Focus

• Typical question patterns:
  • “Write a differential equation that models the situation described.”
  • “Identify what the constant k represents and determine its sign.”
  • “Use units to interpret \frac{dy}{dx} and to check whether a model makes sense.”
• Common mistakes:
  • Confusing proportional to %%LATEX249%% with proportional to %%LATEX250%% (reading too quickly and assigning the wrong variables).
  • Choosing the wrong sign for decay or cooling situations.
  • Ignoring units, which often reveal an incorrect setup.

Exponential Growth and Decay as Differential Equations

Exponential growth and decay are among the most important applications of differential equations because they arise from one simple assumption:

The rate of change of a quantity is proportional to the amount of that quantity present.

Mathematically, if %%LATEX251%% is the quantity and %%LATEX252%% is time:

\frac{dy}{dt}=ky

Solving the exponential differential equation

This is separable:

\frac{1}{y}dy=k\,dt

Integrate:

\ln|y|=kt+C

Exponentiate:

y=Ae^{kt}

Here %%LATEX257%% is a nonzero constant. If you are modeling a quantity like population, you typically take %%LATEX258%%.

If an initial condition is given, such as y(0)=y_0, then:

y_0=Ae^{0}

So A=y_0 and the solution becomes:

y=y_0e^{kt}

Connecting to familiar exponential forms

You may also be familiar with discrete forms like %%LATEX263%%. The differential equation model naturally leads to the base-%%LATEX264%% form %%LATEX265%%. They describe the same kind of behavior, with %%LATEX266%% acting as a continuous growth rate.

Doubling time and half-life

If a quantity grows exponentially, the doubling time is constant (it does not depend on the current amount). If a quantity decays exponentially, the half-life is constant.

Using y=y_0e^{kt}:

Doubling means 2y_0=y_0e^{kT}, so

2=e^{kT}

\ln 2=kT

T=\frac{\ln 2}{k}

Half-life means \frac{1}{2}y_0=y_0e^{kT}, so

\frac{1}{2}=e^{kT}

\ln\left(\frac{1}{2}\right)=kT

T=\frac{\ln\left(\frac{1}{2}\right)}{k}

Since decay has k