Differential Equations: AP Calc AB Unit 7

Why this unit matters

Differential equations describe systems where change depends on current state, including growth, cooling, and mixing behavior.

What you will learn

Interpret slope fields to compare families of possible solutions.
Solve separable first-order differential equations.
Use initial conditions to determine a specific solution curve.

Understand the core ideas

A differential equation links a function to one or more of its derivatives, so it describes how a quantity changes rather than giving the quantity directly. In AP Calculus AB, first-order equations are emphasized through slope fields, separable equations, and basic growth or decay models. A slope field gives local line segments that show possible solution behavior before you solve algebraically. This is useful for checking whether a symbolic solution makes sense.

For separable equations, isolate y terms with dy and x terms with dx, integrate both sides, then solve for the constant using the initial condition. This workflow turns a family of curves into one specific solution. Exponential growth and decay equations of the form dy/dt = ky have solutions $y = Ce^($ kt), with k > 0 for growth and k < 0 for decay. In AB, accuracy depends on clean algebra and careful constant handling, especially when converting ln|y| expressions into exponential form.

Key terms

Differential equation: An equation involving an unknown function and its derivatives.
Slope field: A plot of short segments that shows the local slope dy/dx at many points in the plane.
Separable equation: A differential equation that can be rearranged into g(y)dy = h(x)dx for direct integration.
Initial condition: A given function value at a specific input used to determine the constant in a solution family.

Solve a separable initial value problem

Solve dy/dx = 2xy with y(0) = 3.

Separate variables: (1/y)dy = 2x dx.
Integrate both sides: integral (1/y)dy = integral 2x dx, so ln $|y| = x^2 + C$ .
Exponentiate: y = C* $e^(x^2)$ , where C can be any nonzero constant.
Apply y(0) = 3: 3 = C* $e^0$ , so C = 3 and $y = 3e^(x^2)$ .

Result: The solution satisfying the initial condition is

y = 3e^(x^2)

A common misconception

Claim: After integrating ln $|y| = x^2 + C$ , the absolute value bars can be dropped without changing anything.

Correction: You must handle sign through the constant when exponentiating. Writing $y = Ce^(x^2)$ captures both positive and negative branches.

Lessons in this unit

From verbal model to differential equationTranslate rate statements into equations involving derivatives.
Slope fields and qualitative behaviorUse local slope patterns to predict long-run tendencies.
Separable equationsSeparate variables, integrate both sides, and apply constants correctly.
Exponential growth and decay solutionsUse dy/dt = ky with initial value data to model changing quantities.

Study task

Solve dy/dx = 2xy with y(0) = 3 and explain how the initial condition changes a family of solutions into one function.

Unit checkpoint

Solve dy/dx = 4x with y(1) = 5.

Integrate to get $y = 2x^2 + C$ . Use y(1) = 5, so 2 + C = 5 and C = 3. The solution is $y = 2x^2 + 3$ .

Differential Equations