Math Core

A differential equation tells you how a quantity changes instead of telling you the quantity itself. Before this unit adds Euler's method and logistic growth, this lesson reviews the AB toolkit you will lean on constantly: reading slope fields, checking solutions, and solving separable equations with an initial condition.

Differential equations and their solutions

A differential equation is an equation that involves a derivative, such as $\dfrac{dy}{dx} = x - y$ or $\dfrac{dP}{dt} = 0.3P$ . A solution is a function that makes the equation true for every $x$ in an interval.

To check a proposed solution, differentiate it and substitute. For example, is $y = x - 1$ a solution of $\dfrac{dy}{dx} = x - y$ ? The left side is $\dfrac{dy}{dx} = 1$ . The right side is $x - (x - 1) = 1$ . The two sides match for every $x$ , so yes.

A differential equation usually has a whole family of solutions, called the general solution, that differ by a constant $C$ . An initial condition such as $y(0) = 3$ picks out one member of the family: the particular solution.

Slope fields

You can see the solutions of $\dfrac{dy}{dx} = f(x, y)$ without solving anything. At each point $(x, y)$ , the equation gives the slope a solution curve must have if it passes through that point. Draw a short segment with that slope at each point of a grid, and you get a slope field.

Definition

Slope field

A slope field for $\dfrac{dy}{dx} = f(x, y)$ is a grid of short line segments. The segment at $(x, y)$ has slope $f(x, y)$ . Every solution curve is tangent to the segments it passes through.

Here is the slope field for $\dfrac{dy}{dx} = x - y$ , together with the particular solution $y = x - 1$ (dashed). Notice how the dashed line runs along the segments instead of cutting across them.

Slope field for dy/dx = x − y with the solution y = x − 1 dashed.Open in grapher →

To read or match a slope field, look for patterns rather than computing every segment:

Where are the segments horizontal? Those points satisfy $f(x, y) = 0$ . For $\dfrac{dy}{dx} = x - y$ that is the line $y = x$ .
Do the slopes depend on $x$ only, $y$ only, or both? If $\dfrac{dy}{dx}$ depends only on $x$ , every vertical column of segments is identical. If it depends only on $y$ (an autonomous equation), every horizontal row is identical.
Where are slopes positive or negative? Check a few convenient points, such as points on the axes.

Separation of variables

When the right side factors into a function of $x$ times a function of $y$ , the equation is separable and you can solve it exactly.

Solving a separable equation

For $\dfrac{dy}{dx} = g(x)\,h(y)$ :

Separate: move every $y$ to the side with $dy$ and every $x$ to the side with $dx$ : $\dfrac{1}{h(y)}\,dy = g(x)\,dx$ .
Integrate both sides, and add one constant $C$ .
Use the initial condition to find $C$ right away.
Solve for $y$ if possible, and choose the sign or branch that matches the initial condition.

On the AP exam, a separable-equation free-response part is scored step by step. Separating the variables correctly earns the first point, and you can't earn later points without it, so write the separated form clearly.

Worked example: A square-root solution

Find the particular solution of $\dfrac{dy}{dx} = \dfrac{2x}{y}$ with $y(0) = 3$ .

Solution. Separate: $y\,dy = 2x\,dx$ . Integrate:

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

Substitute $x = 0$ , $y = 3$ : $\dfrac{9}{2} = C$ . So $y^2 = 2x^2 + 9$ . Taking square roots gives $y = \pm\sqrt{2x^2 + 9}$ . Because $y(0) = 3$ is positive, choose the positive root:

y = \sqrt{2x^2 + 9}.

Worked example: An exponential of a trig function

Solve $\dfrac{dy}{dx} = y\cos x$ with $y(0) = 2$ .

Solution. Separate: $\dfrac{1}{y}\,dy = \cos x\,dx$ . Integrate: $\ln\lvert y \rvert = \sin x + C$ . At $x = 0$ : $\ln 2 = 0 + C$ , so $C = \ln 2$ . Then

\lvert y \rvert = e^{\sin x + \ln 2} = 2e^{\sin x}.

Since $y(0) = 2 \gt 0$ , drop the absolute value: $y = 2e^{\sin x}$ . Check: $\dfrac{dy}{dx} = 2e^{\sin x}\cos x = y\cos x$ .

Common mistake

Add the constant when you integrate, not after you solve for $y$ . Writing $\ln\lvert y \rvert = \sin x$ and then tacking on " $+ C$ " at the end as $y = e^{\sin x} + C$ gives the wrong family. The constant becomes a multiplier after you exponentiate, because $e^{\sin x + C} = e^C e^{\sin x}$ .

Exponential growth and decay

The most important separable equation is $\dfrac{dy}{dt} = ky$ : the rate of change is proportional to the amount present. Separating gives $\ln\lvert y \rvert = kt + C$ , so

y = y_0 e^{kt},

where $y_0 = y(0)$ . If $k \gt 0$ you get exponential growth; if $k \lt 0$ , exponential decay.

Worked example: Finding k from a half-life

A radioactive substance decays at a rate proportional to the amount present. Its half-life is 6 years. What fraction of the original sample remains after 15 years?

Solution. The amount is $A = A_0 e^{kt}$ . After 6 years, half remains: $e^{6k} = \dfrac{1}{2}$ , so $k = \dfrac{\ln(1/2)}{6} = -\dfrac{\ln 2}{6}$ . After 15 years,

\frac{A(15)}{A_0} = e^{15k} = e^{-15\ln 2/6} = 2^{-2.5} \approx 0.177.

About $17.7\%$ of the sample remains.

Tip

Always check a particular solution twice: plug in the initial condition, and differentiate to make sure the equation holds. Both checks take under a minute and catch most sign errors.

Practice

Practice 1

Which differential equation could have the slope field shown?

A slope field on the grid −2 ≤ x ≤ 2, −2 ≤ y ≤ 2.Open in grapher →

Practice 2

For $\dfrac{dy}{dx} = x^2 - 2y$ , find the slope of the slope-field segment at the point $(3, 1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Which function is a solution of $\dfrac{dy}{dx} = x + y$ ?

Practice 4

Find the particular solution $y = f(x)$ of $\dfrac{dy}{dx} = 3x^2 y$ with $f(0) = 5$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 5

Find the particular solution of $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = -2$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 6

Find the particular solution of $\dfrac{dy}{dx} = e^{x - y}$ with $y(0) = \ln 2$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 7

A bacteria population grows at a rate proportional to its size. It starts at 200 and doubles every 5 hours. To the nearest whole number, what is the population after 12 hours?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.