AP Calculus BC Unit 7 Notes: Solving Differential Equations (Separable, Exponential, Logistic)

Separation of Variables: General and Particular Solutions

A differential equation is an equation that relates a function to one or more of its derivatives. In AP Calculus, you most often see first-order differential equations, where the derivative looks like $\frac{dy}{dx}$ or $\frac{dP}{dt}$. Solving a differential equation means finding a function (or family of functions) whose derivative behavior matches the equation.

A powerful and commonly testable technique is separation of variables. You use it when the differential equation can be rearranged so that all the $y$-terms are on one side and all the $x$-terms are on the other side. Once separated, you integrate both sides to “undo” the derivatives.

What “separable” means (and why it matters)

A first-order differential equation is separable if it can be written (or rearranged) in the form

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

where $g(x)$ is a function of $x$ only and $h(y)$ is a function of $y$ only. This matters because it lets you treat the derivative as a fraction (with care) and move factors around:

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

and then

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

At that point, each side involves only one variable, so you can integrate both sides.

Conceptually, separation of variables works because you’re rewriting the statement “the rate of change of $y$ depends on both $x$ and $y$” into a form where changes in $y$ can be matched to changes in $x$ through integrals.

General solution vs. particular solution

When you integrate, you introduce an arbitrary constant. That’s because many different functions can have the same derivative “pattern” except for a vertical shift or scaling.

• A general solution is a whole family of solutions that includes an arbitrary constant (often $C$).
• A particular solution is the single solution you get after using an initial condition, like $y(0)=5$, to determine that constant.

In AP questions, you’re often expected to find the general solution first, then apply an initial condition to produce the particular solution.

The standard separation procedure (mechanism)

Suppose you’re given a separable differential equation.

1. Rewrite the equation so that you clearly see which parts depend on $x$ and which depend on $y$.
2. Separate: move all $y$ expressions (including $dy$) to one side and all $x$ expressions (including $dx$) to the other.
3. Integrate both sides.
4. Combine constants into a single constant $C$.
5. Solve for $y$ if possible (sometimes an implicit solution is acceptable or even expected).
6. Apply the initial condition (if given) to find the particular solution.

A common point of confusion: you’ll sometimes see an implicit step where $\frac{dy}{dx}$ is treated like a fraction. This is justified here because the separation comes from algebraic manipulation consistent with the chain rule, but you still need to separate correctly before integrating.

Important notation (so you don’t get tripped up)

In calculus, the derivative can be written in different ways. These are equivalent when $y$ depends on $x$:

On the AP exam, $\frac{dy}{dx}$ and Leibniz notation are common in differential equations because they make separation more visually natural.

Example 1: Solving a separable differential equation (general solution)

Solve

$\frac{dy}{dx} = xy^2$

Step 1: Separate variables. Put all $y$ terms with $dy$ and all $x$ terms with $dx$.

$\frac{1}{y^2}dy = xdx$

Step 2: Integrate both sides.

$\int y^{-2}dy = \int xdx$

$-y^{-1} = \frac{x^2}{2} + C$

Step 3: Solve for $y$.

$-\frac{1}{y} = \frac{x^2}{2} + C$

Multiply by $-1$ and absorb the sign into the constant (renaming constants is allowed):

$\frac{1}{y} = -\frac{x^2}{2} + C$

Invert:

$y = \frac{1}{C - \frac{x^2}{2}}$

This is a general solution because it contains an arbitrary constant $C$.

What can go wrong here: forgetting that $\int y^{-2}dy = -y^{-1}+C$ (the negative sign is easy to miss), or incorrectly integrating $\int xdx$.

Example 2: Particular solution using an initial condition

Solve the initial value problem

$\frac{dy}{dx} = xy$

with $y(0)=3$.

Separate:

$\frac{1}{y}dy = xdx$

Integrate:

$\int \frac{1}{y}dy = \int xdx$

A key detail: $\int \frac{1}{y}dy = \ln|y| + C$, not $\ln(y)$ without absolute value.

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

Exponentiate to solve for $y$:

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

Rewrite $e^C$ as a positive constant $K$:

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

This implies

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

where $A$ can be any nonzero real constant (positive or negative), which accounts for the absolute value.

Apply the initial condition $y(0)=3$:

$3 = Ae^{0}$

$A=3$

So the particular solution is

$y = 3e^{\frac{x^2}{2}}$

What can go wrong here: dropping the absolute value too early and accidentally excluding negative solutions, or mishandling exponent rules when solving for $y$.

A subtle but important point: equilibrium (constant) solutions

Sometimes you divide by an expression involving $y$ to separate variables. If that expression can be zero, you may accidentally lose solutions.

For example, if

$\frac{dy}{dx} = y(1-y)$

and you divide by $y(1-y)$, you are assuming $y \neq 0$ and $y \neq 1$. But $y=0$ and $y=1$ are actually valid constant (equilibrium) solutions because they make $\frac{dy}{dx}=0$ everywhere.

A good habit: before dividing by something involving $y$, quickly ask, “Could that be zero? If so, does it create a constant solution?”

Exam Focus

• Typical question patterns:
  • “Solve the differential equation” followed by “use $y(a)=b$ to find the particular solution.”
  • “Find $y$ as a function of $x$” and then evaluate $y$ at a new input.
  • “Show that the solution can be written as …” (often expecting separation, integration, and algebraic rearrangement).
• Common mistakes:
  • Forgetting the constant of integration $C$ or failing to use the initial condition to determine it.
  • Incorrect integration, especially $\int \frac{1}{y}dy$ and negative exponents.
  • Dividing by a factor like $y$ and losing an equilibrium solution (not always tested, but it’s a classic trap).

Exponential Models with Differential Equations

An exponential growth/decay model is based on the idea that the rate of change of a quantity is proportional to the amount present. This is one of the most important modeling ideas in calculus because it turns a real-world statement about “rate” into a differential equation you can solve.

The core idea: “proportional to the amount present”

If $y(t)$ is a quantity depending on time $t$, “the rate of change is proportional to the amount present” translates to

$\frac{dy}{dt} = ky$

Here:

• $y(t)$ is the amount at time $t$.
• $k$ is a constant of proportionality.
  • $k>0$ means growth.
  • $k<0$ means decay.
• Units: $k$ has units of “per unit time” (for example, per year).

This differential equation is separable, so you can solve it with the method from the previous section.

Solving the exponential differential equation

Start with

$\frac{dy}{dt} = ky$

Separate variables:

$\frac{1}{y}dy = kdt$

Integrate:

$\int \frac{1}{y}dy = \int kdt$

$\ln|y| = kt + C$

Exponentiate:

$|y| = e^{kt+C}$

Rewrite $e^C$ as a positive constant $A$:

$|y| = Ae^{kt}$

So

$y = Ce^{kt}$

where $C$ can be any nonzero real constant. In most applied settings, $y$ represents a positive amount, so you usually take $C>0$.

Why initial conditions matter in models

In modeling problems, you rarely want the whole family $y=Ce^{kt}$. You want the specific model that matches the initial amount:

If $y(0)=y_0$, then

$y_0 = Ce^{0}$

$C=y_0$

so

$y(t) = y_0e^{kt}$

This gives a clear interpretation: $y_0$ is the starting amount, and $k$ controls how quickly it grows or decays.

Doubling time and half-life (how they connect)

Even when a problem doesn’t explicitly ask for $k$, you may be given a doubling time (growth) or half-life (decay). These come from solving for the time when the amount is multiplied by a factor.

If $y(t)=y_0e^{kt}$:

• Doubling time $T_d$ satisfies $2y_0 = y_0e^{kT_d}$, so

$2 = e^{kT_d}$

$\ln 2 = kT_d$

$T_d = \frac{\ln 2}{k}$

• Half-life $T_{1/2}$ satisfies $\frac{1}{2}y_0 = y_0e^{kT_{1/2}}$, so

$\frac{1}{2} = e^{kT_{1/2}}$

$\ln\left(\frac{1}{2}\right) = kT_{1/2}$

$T_{1/2} = \frac{\ln\left(\frac{1}{2}\right)}{k}$

For decay, $k<0$, which makes $T_{1/2}$ positive as expected.

Example 1: Find an exponential model from data

A population follows $\frac{dP}{dt} = kP$. You are told $P(0)=500$ and $P(3)=800$. Find $P(t)$.

From the general solution,

$P(t) = Ce^{kt}$

Use $P(0)=500$:

$500 = Ce^{0}$

$C=500$

So

$P(t)=500e^{kt}$

Use $P(3)=800$:

$800 = 500e^{3k}$

$\frac{8}{5} = e^{3k}$

Take natural log:

$\ln\left(\frac{8}{5}\right)=3k$

$k=\frac{1}{3}\ln\left(\frac{8}{5}\right)$

Therefore,

$P(t)=500e^{\left(\frac{1}{3}\ln\left(\frac{8}{5}\right)\right)t}$

It’s common (and often cleaner) to rewrite using exponent properties:

$P(t)=500\left(e^{\ln\left(\frac{8}{5}\right)}\right)^{t/3}$

Since $e^{\ln(a)}=a$ for $a>0$,

$P(t)=500\left(\frac{8}{5}\right)^{t/3}$

Both forms are equivalent. AP problems usually accept either, as long as it’s correct.

Example 2: Interpreting the constant $k$

Suppose a substance decays according to $\frac{dA}{dt}=kA$. If $k=-0.2$ (per day), then

$A(t)=A_0e^{-0.2t}$

The negative sign tells you the amount decreases, and the magnitude $0.2$ controls how quickly. A common mistake is to write $A(t)=A_0e^{0.2t}$ just because you associate exponentials with growth—always check the sign of $k$.

Exam Focus

• Typical question patterns:
  • Given $\frac{dy}{dt}=ky$ with one or two data points, find $y(t)$ or find $k$.
  • Given a half-life or doubling time, write an exponential function and use it to predict a future value.
  • Interpret the meaning of $k$ in context (growth rate vs. decay rate, units).
• Common mistakes:
  • Forgetting that the solution is $y=Ce^{kt}$ (not $y=Ce^{t}$ unless $k=1$).
  • Using base-10 logs instead of natural logs when solving for $k$.
  • Losing the sign: decay requires $k<0$, and that affects algebra and interpretation.

Logistic Models with Differential Equations

Exponential models assume “more present means proportionally faster growth,” forever. That’s often unrealistic—populations run out of resources, infections slow as fewer susceptible people remain, and growth levels off. The logistic model modifies exponential growth by building in a limiting value called a carrying capacity.

The key idea: growth slows as you approach a maximum

A standard logistic differential equation is

$\frac{dP}{dt} = kP\left(1-\frac{P}{L}\right)$

where:

• $P(t)$ is the population (or quantity).
• $k>0$ is a growth constant.
• $L>0$ is the carrying capacity, the long-term maximum sustainable value.

The factor $\left(1-\frac{P}{L}\right)$ is the “slowdown” term:

• If $P$ is small compared to $L$, then $\frac{P}{L}$ is near 0, so $1-\frac{P}{L}$ is near 1, and the equation behaves almost like exponential growth $\frac{dP}{dt} \approx kP$.
• If $P$ is close to $L$, then $1-\frac{P}{L}$ is near 0, so growth slows dramatically.
• If $P > L$, then $1-\frac{P}{L}$ becomes negative, making $\frac{dP}{dt}<0$, so the model pushes the population back down toward $L$.

This built-in feedback is why logistic growth tends to level off.

Equilibrium solutions and what they mean

Set $\frac{dP}{dt}=0$:

$kP\left(1-\frac{P}{L}\right)=0$

This happens when:

$P=0$

or

$1-\frac{P}{L}=0$

which gives

$P=L$

So the logistic model has two equilibrium (constant) solutions: $P(t)=0$ and $P(t)=L$.

In many contexts, $P=0$ is an unstable equilibrium (tiny positive populations grow away from 0), while $P=L$ is stable (solutions approach it). You don’t usually need formal stability analysis in AP Calculus, but you should be able to interpret that $L$ is the long-term limiting value.

Solving the logistic differential equation (separation of variables)

Because it’s separable, you can solve it analytically.

Start with

$\frac{dP}{dt} = kP\left(1-\frac{P}{L}\right)$

Rewrite the right-hand side in a form that’s easier to separate:

$\frac{dP}{dt} = kP\left(\frac{L-P}{L}\right)$

$\frac{dP}{dt} = \frac{k}{L}P(L-P)$

Separate variables:

$\frac{1}{P(L-P)}dP = \frac{k}{L}dt$

To integrate the left side, you use partial fractions:

$\frac{1}{P(L-P)} = \frac{1}{L}\left(\frac{1}{P} + \frac{1}{L-P}\right)$

So

$\int \frac{1}{P(L-P)}dP = \int \frac{1}{L}\left(\frac{1}{P} + \frac{1}{L-P}\right)dP$

Integrate term-by-term (and be careful: $\int \frac{1}{L-P}dP = -\ln|L-P| + C$ because the derivative of $L-P$ is $-1$):

$\frac{1}{L}\left(\ln|P| - \ln|L-P|\right) = \frac{k}{L}t + C$

Multiply both sides by $L$:

$\ln|P| - \ln|L-P| = kt + C$

Combine logs:

$\ln\left|\frac{P}{L-P}\right| = kt + C$

Exponentiate:

$\left|\frac{P}{L-P}\right| = e^{kt+C}$

Let $A=e^C>0$ and absorb sign into a new constant (in typical population contexts $0<P<L$ so the fraction is positive anyway):

$\frac{P}{L-P} = Ae^{kt}$

Solve for $P$:

$P = Ae^{kt}(L-P)$

$P = ALe^{kt} - Ae^{kt}P$

$P + Ae^{kt}P = ALe^{kt}$

$P(1+Ae^{kt}) = ALe^{kt}$

$P = \frac{ALe^{kt}}{1+Ae^{kt}}$

A common equivalent form divides numerator and denominator by $Ae^{kt}$:

$P = \frac{L}{1+Be^{-kt}}$

where $B=\frac{1}{A}$. This form is often easiest for applying an initial condition.

Applying an initial condition in the logistic model

If $P(0)=P_0$, use

$P(0) = \frac{L}{1+Be^{0}}$

$P_0 = \frac{L}{1+B}$

Solve for $B$:

$1+B = \frac{L}{P_0}$

$B = \frac{L}{P_0}-1$

So the particular solution can be written as

$P(t)=\frac{L}{1+\left(\frac{L}{P_0}-1\right)e^{-kt}}$

This is a great form to remember because it directly shows how $P_0$ sets the curve.

Shape facts you’re expected to understand (not just the formula)

The logistic model has a characteristic S-shape when $0<P_0<L$:

• Early on, growth is almost exponential.
• Growth speeds up until a point, then slows.
• The population approaches $L$ asymptotically.

A particularly important, testable idea: the growth rate $\frac{dP}{dt}$ is maximized when $P=\frac{L}{2}$. You can see this without heavy calculus by noticing $P(L-P)$ is a parabola in $P$ with its maximum at the midpoint.

That midpoint is also the inflection point of the logistic solution curve (where concavity changes), which sometimes appears in interpretation questions.

Example 1: Logistic model with an initial condition

A population follows

$\frac{dP}{dt} = 0.4P\left(1-\frac{P}{1000}\right)$

and $P(0)=100$. Write $P(t)$.

Here $k=0.4$ and $L=1000$, so

$P(t)=\frac{L}{1+Be^{-kt}}$

$P(t)=\frac{1000}{1+Be^{-0.4t}}$

Use $P(0)=100$:

$100 = \frac{1000}{1+B}$

$1+B = 10$

$B=9$

So

$P(t)=\frac{1000}{1+9e^{-0.4t}}$

Common pitfall: mixing up where the negative sign goes. The standard approach with $k>0$ leads to $e^{-kt}$ in the denominator form; if you instead keep $e^{kt}$, your constant will change form, but the final model should still behave correctly (increasing toward $L$ when $P_0<L$).

Example 2: When does the population reach a certain size?

Using

$P(t)=\frac{1000}{1+9e^{-0.4t}}$

find when $P(t)=500$.

Set up the equation:

$500 = \frac{1000}{1+9e^{-0.4t}}$

Invert both sides:

$\frac{1}{500} = \frac{1+9e^{-0.4t}}{1000}$

Multiply by 1000:

$2 = 1+9e^{-0.4t}$

$1 = 9e^{-0.4t}$

$\frac{1}{9} = e^{-0.4t}$

Take natural log:

$\ln\left(\frac{1}{9}\right) = -0.4t$

$t = \frac{\ln(9)}{0.4}$

This is a typical AP-style “solve for time” step after building the model.

What goes wrong most often in logistic algebra

Logistic solutions are algebra-heavy, so many errors are not calculus errors—they’re equation-solving errors.

Common issues include:

• Losing parentheses when solving for $P$ (for example, distributing incorrectly when you have $P(1+Ae^{kt})$).
• Sign mistakes in the logarithm step, especially with $\ln|L-P|$.
• Forgetting that equilibrium solutions $P=0$ and $P=L$ exist even though separation divides by $P(L-P)$.

Exam Focus

• Typical question patterns:
  • Given a logistic differential equation, solve for $P(t)$ using an initial condition.
  • Interpret parameters: identify the carrying capacity $L$ and explain what it means in context.
  • Solve for the time when $P$ hits a threshold, or find when the growth rate is greatest (often at $P=\frac{L}{2}$).
• Common mistakes:
  • Treating logistic growth as exponential and predicting it grows without bound.
  • Dropping equilibrium solutions by dividing by $P$ or $L-P$ without noting $P=0$ or $P=L$.
  • Incorrect partial fraction setup or missing the negative sign in $\int \frac{1}{L-P}dP$.