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AP Calculus BC Flashcards: Verifying Solutions For Differential Equations

Study Verifying Solutions For 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 Verifying Solutions For 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: Verifying Solutions For Differential Equations

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QUESTION

What is the integrating factor for dydx+Py=Q\frac{dy}{dx} + Py = Qdxdy​+Py=Q?

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ANSWER

e∫P dxe^{\textstyle\int P \, dx}e∫Pdx. Factor to make coefficient of dydx\frac{dy}{dx}dxdy​ equal to 111.

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Flashcard 1: What is the integrating factor for dydx+Py=Q\frac{dy}{dx} + Py = Qdxdy​+Py=Q?

Answer: e∫P dxe^{\textstyle\int P \, dx}e∫Pdx. Factor to make coefficient of dydx\frac{dy}{dx}dxdy​ equal to 111.

Flashcard 2: Is y=2x2+3y = 2x^2 + 3y=2x2+3 a solution to y′=4xy' = 4xy′=4x?

Answer: Yes, it is a solution. Derivative: dydx=4x\frac{dy}{dx} = 4xdxdy​=4x matches the equation.

Flashcard 3: Identify the particular solution for y′=2yy' = 2yy′=2y given y(0)=5y(0) = 5y(0)=5.

Answer: y=5e2xy = 5e^{2x}y=5e2x. Using initial condition y(0)=5y(0) = 5y(0)=5 with y=Ce2xy = Ce^{2x}y=Ce2x.

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

Answer: y=C1cos(x)+C2sin(x)y = C_1 \text{cos}(x) + C_2 \text{sin}(x)y=C1​cos(x)+C2​sin(x). Solution using characteristic equation r2+1=0r^2 + 1 = 0r2+1=0.

Flashcard 5: What is the general form of a first-order differential equation?

Answer: dydx=f(x,y)\frac{dy}{dx} = f(x, y)dxdy​=f(x,y). Standard form where fff depends on both xxx and yyy.

Flashcard 6: What type of differential equation is y′+3y=6y' + 3y = 6y′+3y=6?

Answer: First-order linear. Has form y′+Py=Qy' + Py = Qy′+Py=Q with constant coefficient.

Flashcard 7: What is the characteristic equation for y′′−4y=0y'' - 4y = 0y′′−4y=0?

Answer: r2−4=0r^2 - 4 = 0r2−4=0. Replace y′′y''y′′ with r2r^2r2 and set equal to zero.

Flashcard 8: Verify if y=1+ln(x)y = 1 + \text{ln}(x)y=1+ln(x) satisfies xy′=1xy' = 1xy′=1.

Answer: Yes, it satisfies the equation. y′=1xy' = \frac{1}{x}y′=x1​, so x⋅1x=1x \cdot \frac{1}{x} = 1x⋅x1​=1 ✓

Flashcard 9: Verify if y=11−xy = \frac{1}{1-x}y=1−x1​ is a solution to y′=y2y' = y^2y′=y2.

Answer: Yes, it is a solution. y′=1(1−x)2y' = \frac{1}{(1-x)^2}y′=(1−x)21​ and y2=1(1−x)2y^2 = \frac{1}{(1-x)^2}y2=(1−x)21​ ✓

Flashcard 10: Verify if y=1xy = \frac{1}{x}y=x1​ is a solution to xy′+2y=0xy' + 2y = 0xy′+2y=0.

Answer: No, it is not a solution. y′=−1x2y' = -\frac{1}{x^2}y′=−x21​, so x(−1x2)+2(1x)=1x≠0x(-\frac{1}{x^2}) + 2(\frac{1}{x}) = \frac{1}{x} \neq 0x(−x21​)+2(x1​)=x1​=0

Flashcard 11: What is the form of a second-order linear homogeneous differential equation?

Answer: ay′′+by′+cy=0a y'' + b y' + c y = 0ay′′+by′+cy=0. Standard form with constant coefficients and zero right side.

Flashcard 12: What is the form of a non-homogeneous differential equation?

Answer: ay′′+by′+cy=g(x)a y'' + b y' + c y = g(x)ay′′+by′+cy=g(x). Same as homogeneous but with non-zero right side.

Flashcard 13: Determine if y=exxy = \frac{e^x}{x}y=xex​ is a solution to x2y′−xy=exx^2y' - xy = e^xx2y′−xy=ex.

Answer: Yes, it is a solution. Computing y′y'y′ using quotient rule verifies the equation.

Flashcard 14: What is the general solution to dydx=y\frac{dy}{dx} = ydxdy​=y?

Answer: y=Cexy = Ce^xy=Cex, where CCC is a constant. Solution to the simplest exponential differential equation.

Flashcard 15: What is the characteristic equation for y′′+3y′+2y=0y'' + 3y' + 2y = 0y′′+3y′+2y=0?

Answer: r2+3r+2=0r^2 + 3r + 2 = 0r2+3r+2=0. Replace y′′y''y′′ with r2r^2r2, y′y'y′ with rrr, and yyy with 111.

Flashcard 16: What is the general solution to y′′−5y′+6y=0y'' - 5y' + 6y = 0y′′−5y′+6y=0?

Answer: y=C1e2x+C2e3xy = C_1 e^{2x} + C_2 e^{3x}y=C1​e2x+C2​e3x. Using characteristic roots r=2,3r = 2, 3r=2,3 from factoring.

Flashcard 17: Determine if y=x3y = x^3y=x3 is a solution for y′′=6xy'' = 6xy′′=6x.

Answer: Yes, it is a solution. y′=3x2y' = 3x^2y′=3x2, y′′=6xy'' = 6xy′′=6x, so equation holds.

Flashcard 18: What is the order of y′′+2xy′+y=0y'' + 2xy' + y = 0y′′+2xy′+y=0?

Answer: Second order. Highest derivative is the second derivative.

Flashcard 19: Determine if y=x+1y = x + 1y=x+1 is a solution for y′=1y' = 1y′=1.

Answer: Yes, it is a solution. Derivative of x+1x + 1x+1 is constant 111.

Flashcard 20: Verify if y=x3+xy = x^3 + xy=x3+x is a solution to y′=3x2+1y' = 3x^2 + 1y′=3x2+1.

Answer: Yes, it is a solution. Derivative matches: ddx(x3+x)=3x2+1\frac{d}{dx}(x^3 + x) = 3x^2 + 1dxd​(x3+x)=3x2+1.

Flashcard 21: Verify if y=sin(x)y = \text{sin}(x)y=sin(x) satisfies y′′+y=0y'' + y = 0y′′+y=0.

Answer: Yes, it satisfies the equation. y′=cos⁡(x)y' = \cos(x)y′=cos(x), y′′=−sin⁡(x)y'' = -\sin(x)y′′=−sin(x), so y′′+y=0y'' + y = 0y′′+y=0 ✓

Flashcard 22: What is the order of the differential equation y′′′+2y′=0y''' + 2y' = 0y′′′+2y′=0?

Answer: Third order. Highest derivative is the third derivative.

Flashcard 23: Does y=3x+2y = 3x + 2y=3x+2 solve y′=3y' = 3y′=3?

Answer: Yes, it solves the equation. Derivative of 3x+23x + 23x+2 is constant 333.

Flashcard 24: Verify if y=e−xy = e^{-x}y=e−x is a solution to y′=−yy' = -yy′=−y.

Answer: Yes, it is a solution. Derivative: ddx(e−x)=−e−x=−y\frac{d}{dx}(e^{-x}) = -e^{-x} = -ydxd​(e−x)=−e−x=−y ✓

Flashcard 25: What is the characteristic equation for y′′+9y=0y'' + 9y = 0y′′+9y=0?

Answer: r2+9=0r^2 + 9 = 0r2+9=0. Replace y′′y''y′′ with r2r^2r2 and yyy with 111.

Flashcard 26: What is the order of y(4)+2y′′+y=0y^{(4)} + 2y'' + y = 0y(4)+2y′′+y=0?

Answer: Fourth order. Highest derivative is the fourth derivative.

Flashcard 27: Verify if y=x22y = \frac{x^2}{2}y=2x2​ satisfies y′′=1y'' = 1y′′=1.

Answer: Yes, it satisfies the equation. y′=xy' = xy′=x, y′′=1y'' = 1y′′=1, so equation is satisfied.

Flashcard 28: What is the general solution of dydx=ky\frac{dy}{dx} = kydxdy​=ky?

Answer: y=Cekxy = Ce^{kx}y=Cekx, where CCC is a constant. Solution to exponential growth/decay differential equation.

Flashcard 29: Identify the particular solution for y′=5yy' = 5yy′=5y given y(0)=2y(0) = 2y(0)=2.

Answer: y=2e5xy = 2e^{5x}y=2e5x. Using initial condition y(0)=2y(0) = 2y(0)=2 with y=Ce5xy = Ce^{5x}y=Ce5x.

Flashcard 30: Verify if y=e2xy = e^{2x}y=e2x is a solution to dydx=2y\frac{dy}{dx} = 2ydxdy​=2y.

Answer: Yes, it is a solution. Taking derivative: dydx=2e2x=2y\frac{dy}{dx} = 2e^{2x} = 2ydxdy​=2e2x=2y ✓