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AP Calculus BC Flashcards: Modeling Situations With Differential Equations

Study Modeling Situations With Differential Equations in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Modeling Situations With Differential Equations, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus BC Flashcards: Modeling Situations With Differential Equations

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QUESTION

Define a linear differential equation.

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ANSWER

An equation where the dependent variable and derivatives appear linearly. Dependent variable appears to first power only.

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All flashcards

Flashcard 1: Define a linear differential equation.

Answer: An equation where the dependent variable and derivatives appear linearly. Dependent variable appears to first power only.

Flashcard 2: Write the differential equation for exponential growth.

Answer: dydt=ky\frac{dy}{dt} = kydtdy​=ky. Rate proportional to current amount.

Flashcard 3: What is the integrating factor for dy/dx+2y=exdy/dx + 2y = e^xdy/dx+2y=ex?

Answer: e2xe^{2x}e2x. Exponential of integral of coefficient: e∫2dxe^{\int 2 dx}e∫2dx.

Flashcard 4: What is a particular solution?

Answer: A solution satisfying both the differential equation and initial conditions. Specific solution meeting initial conditions.

Flashcard 5: Identify the dependent variable in dz/dt=z2+t2dz/dt = z^2 + t^2dz/dt=z2+t2.

Answer: The dependent variable is zzz. Function being differentiated (zzz depends on ttt).

Flashcard 6: Identify the dependent variable in dy/dx=x2+y2dy/dx = x^2 + y^2dy/dx=x2+y2.

Answer: The dependent variable is yyy. Variable being differentiated (yyy depends on xxx).

Flashcard 7: What is the integrating factor for y′+p(x)y=g(x)y' + p(x)y = g(x)y′+p(x)y=g(x)?

Answer: e∫p(x)dxe^{\int p(x) \text{d}x}e∫p(x)dx. Multiplier that makes equation exact.

Flashcard 8: State a real-world example of a first-order differential equation.

Answer: Newton's Law of Cooling. Temperature change proportional to difference.

Flashcard 9: What is the general solution to y′′+4y=0y'' + 4y = 0y′′+4y=0?

Answer: y=C1cos(2x)+C2sin(2x)y = C_1 \text{cos}(2x) + C_2 \text{sin}(2x)y=C1​cos(2x)+C2​sin(2x). Characteristic equation r2+4=0r^2 + 4 = 0r2+4=0 gives r=±2ir = ±2ir=±2i.

Flashcard 10: What method is used to solve y′=kyy' = kyy′=ky?

Answer: Separation of variables. Rearrange to separate yyy and xxx terms.

Flashcard 11: What is the general solution to y′′+9y=0y'' + 9y = 0y′′+9y=0?

Answer: y=C1cos(3x)+C2sin(3x)y = C_1 \text{cos}(3x) + C_2 \text{sin}(3x)y=C1​cos(3x)+C2​sin(3x). Characteristic equation r2+9=0r^2 + 9 = 0r2+9=0 gives r=±3ir = ±3ir=±3i.

Flashcard 12: What does it mean for a function to be a solution to a differential equation?

Answer: Substituting it into the equation yields a true statement. The function satisfies the equation identically.

Flashcard 13: What is the characteristic equation of y′′+2y′+5y=0y'' + 2y' + 5y = 0y′′+2y′+5y=0?

Answer: r2+2r+5=0r^2 + 2r + 5 = 0r2+2r+5=0. Substitute y′′=r2y'' = r^2y′′=r2, y′=ry' = ry′=r into equation.

Flashcard 14: Define a non-homogeneous differential equation.

Answer: An equation with non-zero terms not involving the dependent variable. Contains terms independent of dependent variable.

Flashcard 15: Identify the homogeneous part of y′′+5y′+6y=exy'' + 5y' + 6y = e^xy′′+5y′+6y=ex.

Answer: y′′+5y′+6yy'' + 5y' + 6yy′′+5y′+6y. Left side when right side equals zero.

Flashcard 16: Solve y′=1xy' = \frac{1}{x}y′=x1​ for yyy.

Answer: y=ln∣x∣+Cy = \text{ln}|x| + Cy=ln∣x∣+C. Antiderivative of 1x\frac{1}{x}x1​ is ln⁡∣x∣\ln|x|ln∣x∣.

Flashcard 17: What function solves y′−2y=0y' - 2y = 0y′−2y=0?

Answer: y=Ce2xy = Ce^{2x}y=Ce2x. Exponential solution with coefficient k=2k = 2k=2.

Flashcard 18: What is the solution to y′′−y=0y'' - y = 0y′′−y=0?

Answer: y=C1ex+C2e−xy = C_1e^x + C_2e^{-x}y=C1​ex+C2​e−x. Characteristic equation r2−1=0r^2 - 1 = 0r2−1=0 gives r=±1r = ±1r=±1.

Flashcard 19: Identify the linear part of y′′+4y′+4y=5xy'' + 4y' + 4y = 5xy′′+4y′+4y=5x.

Answer: y′′+4y′+4yy'' + 4y' + 4yy′′+4y′+4y. Terms involving yyy and its derivatives.

Flashcard 20: What is the solution for dy/dx=ydy/dx = ydy/dx=y using separation of variables?

Answer: y=Cexy = Ce^{x}y=Cex. Separate variables: dyy=dx\frac{dy}{y} = dxydy​=dx, integrate.

Flashcard 21: What is the characteristic equation of y′′+4y′+4y=0y'' + 4y' + 4y = 0y′′+4y′+4y=0?

Answer: r2+4r+4=0r^2 + 4r + 4 = 0r2+4r+4=0. Replace y′′y''y′′ with r2r^2r2, y′y'y′ with rrr.

Flashcard 22: State the general solution form for y′=kyy' = kyy′=ky.

Answer: y=Cekty = Ce^{kt}y=Cekt, where CCC is a constant. Exponential growth/decay model solution.

Flashcard 23: Define a separable differential equation.

Answer: An equation where variables can be separated on opposite sides. Variables can be moved to opposite sides.

Flashcard 24: What is a homogeneous differential equation?

Answer: An equation where 000 is a solution when all terms set to 000. All terms involve the dependent variable.

Flashcard 25: Identify the type of differential equation: y′′+y=tan⁡(x)y'' + y = \tan(x)y′′+y=tan(x).

Answer: Second-order non-homogeneous differential equation. Second-order with non-zero right side.

Flashcard 26: Solve for yyy in the differential equation dydx=0\frac{dy}{dx} = 0dxdy​=0.

Answer: y=Cy = Cy=C, where CCC is a constant. Zero derivative means constant function.

Flashcard 27: What is the order of d2y/dx2+3y′+4y=x2d^2y/dx^2 + 3y' + 4y = x^2d2y/dx2+3y′+4y=x2?

Answer: Order is 2. Highest derivative is second order.

Flashcard 28: Identify the independent variable in dy/dx=x2+y2dy/dx = x^2 + y^2dy/dx=x2+y2.

Answer: The independent variable is xxx. Variable with respect to which we differentiate.

Flashcard 29: What type of differential equation is y′+p(x)y=g(x)y' + p(x)y = g(x)y′+p(x)y=g(x)?

Answer: First-order linear differential equation. Standard form with yyy and y′y'y′ terms only.

Flashcard 30: Find the derivative of y=Cekty = Ce^{kt}y=Cekt.

Answer: y′=kCekty' = kCe^{kt}y′=kCekt. Apply chain rule: ddt(Cekt)=kCekt\frac{d}{dt}(Ce^{kt}) = kCe^{kt}dtd​(Cekt)=kCekt.