AP Calculus AB Flashcards: Verifying Solutions For Differential Equations

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 AB.

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.

Flashcard 1: What is a particular solution?

Answer: A solution with specified initial conditions. General solution evaluated at given conditions.

Flashcard 2: What is a particular solution?

Answer: A solution with specified initial conditions. General solution evaluated at given conditions.

Flashcard 3: Identify the solution form for $y'' + y = 0$.

Answer: General solution: $y = C_1 \text{cos}(x) + C_2 \text{sin}(x)$. Characteristic equation $r^2 + 1 = 0$ gives $r = \pm i$.

Flashcard 4: What is a separable differential equation?

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

Flashcard 5: What is the general solution for $y'' - 4y = 0$?

Answer: $y = C_1 e^{2x} + C_2 e^{-2x}$. Characteristic equation $r^2 - 4 = 0$ gives $r = \pm 2$.

Flashcard 6: Identify the order of the differential equation: $y'' + 3y' - 5y = 0$.

Answer: Order 2. Highest derivative is the second, so order is 2.

Flashcard 7: Verify if $y = \frac{1}{2}e^{2x}$ solves $y' = y$.

Answer: No, $y' = e^{2x} \neq y$. $y' = e^{2x} \neq \frac{1}{2}e^{2x} = y$.

Flashcard 8: What is the general solution for $y' = -y$?

Answer: $y = Ce^{-x}$. Exponential decay with rate constant 1.

Flashcard 9: Verify if $y = C \text{cos}(2x)$ solves $y'' + 4y = 0$.

Answer: Yes, it satisfies $y'' + 4y = 0$. $y'' = -4C\cos(2x)$, so $y'' + 4y = 0$.

Flashcard 10: Verify if $y = x^2$ satisfies $y'' = 2$.

Answer: Yes, $y'' = 2$. Second derivative of $x^2$ is constant 2.

Flashcard 11: What is the solution form for $y'' + 9y = 0$?

Answer: General solution: $y = C_1 \text{cos}(3x) + C_2 \text{sin}(3x)$. Characteristic equation $r^2 + 9 = 0$ gives $r = \pm 3i$.

Flashcard 12: Verify if $y = \frac{1}{2}e^{2x}$ solves $y' = y$.

Answer: No, $y' = e^{2x} \neq y$. $y' = e^{2x} \neq \frac{1}{2}e^{2x} = y$.

Flashcard 13: State the general solution for $y' = k$.

Answer: $y = kt + C$. Integrate the constant $k$ with respect to time.

Flashcard 14: Is $y = x^2 + C$ a solution to $y'' = 2$?

Answer: Yes, $y'' = 2$. Second derivative of $x^2 + C$ is constant 2.

Flashcard 15: What is the general solution for $y'' = 0$?

Answer: $y = C_1x + C_2$. Double integration of zero gives linear function.

Flashcard 16: What is a differential equation?

Answer: An equation involving derivatives of a function. Mathematical relationship involving rates of change.

Flashcard 17: Determine if $y = \text{sin}(x)$ solves $y'' + y = 0$.

Answer: Yes, $y'' + y = 0$. $y'' = -\sin x$, so $y'' + y = 0$.

Flashcard 18: What is the general solution for $y' = 2y$?

Answer: $y = Ce^{2x}$. Exponential growth with rate constant 2.

Flashcard 19: What is the solution for $y' = 0$?

Answer: $y = C$. Derivative of constant is zero.

Flashcard 20: What is a linear differential equation?

Answer: An equation of the form $a_n(x)y^{(n)} + ... + a_1(x)y' + a_0(x)y = g(x)$. Coefficients are functions of $x$, equation is first-degree in $y$.

Flashcard 21: Determine if $y = \frac{1}{x}$ solves $xy' = -y^2$.

Answer: Yes, $y' = -\frac{1}{x^2}$. Check: $x(-\frac{1}{x^2}) = -\frac{1}{x} = -(\frac{1}{x})^2$.

Flashcard 22: Verify $y = x^3$ for $y'' = 6x$.

Answer: Yes, $y'' = 6x$. Second derivative of $x^3$ is $6x$.

Flashcard 23: Which function is a solution to $y' = 3x^2$?

Answer: Any antiderivative of $3x^2$, e.g., $y = x^3 + C$. Integrate $3x^2$ to get $x^3 + C$.

Flashcard 24: Verify if $y = e^{3x}$ solves $y' = 3y$.

Answer: Yes, $y' = 3e^{3x}$. Derivative matches 3 times the function.

Flashcard 25: Determine if $y = \text{cos}(x)$ satisfies $y'' + y = 0$.

Answer: Yes, $y'' + y = 0$. $y'' = -\cos x$, so $y'' + y = 0$.

Flashcard 26: What is the general solution for $y'' = 0$?

Answer: $y = C_1x + C_2$. Double integration of zero gives linear function.

Flashcard 27: State the general solution for $y' = k$.

Answer: $y = kt + C$. Integrate the constant $k$ with respect to time.

Flashcard 28: Identify the solution form for $y'' + y = 0$.

Answer: General solution: $y = C_1 \text{cos}(x) + C_2 \text{sin}(x)$. Characteristic equation $r^2 + 1 = 0$ gives $r = \pm i$.

Flashcard 29: What is a solution to a differential equation?

Answer: A function satisfying the differential equation. When substituted, makes the equation true.

Flashcard 30: Verify if $y = Ce^{-2x}$ solves $y' + 2y = 0$.

Answer: Yes, it satisfies $y' + 2y = 0$. $y' = -2Ce^{-2x} = -2y$, so $y' + 2y = 0$.

Flashcard 31: Determine if $y = \text{sin}(x)$ solves $y'' + y = 0$.

Answer: Yes, $y'' + y = 0$. $y'' = -\sin x$, so $y'' + y = 0$.

Flashcard 32: Verify $y = x^3$ for $y'' = 6x$.

Answer: Yes, $y'' = 6x$. Second derivative of $x^3$ is $6x$.

Flashcard 33: Is $y = Ce^{3x}$ a solution for $y' = 3y$?

Answer: Yes, $y' = 3y$. Derivative of $Ce^{3x}$ is $3Ce^{3x} = 3y$.

Flashcard 34: What is a solution to a differential equation?

Answer: A function satisfying the differential equation. When substituted, makes the equation true.

Flashcard 35: Verify $y = 2x - 1$ for $y' = 2$.

Answer: Yes, $y' = 2$. Derivative of linear function $2x - 1$ is 2.

Flashcard 36: Find the derivative of $y = \frac{1}{x}$.

Answer: $y' = -\frac{1}{x^2}$. Power rule: $\frac{d}{dx}[x^{-1}] = -x^{-2}$.

Flashcard 37: What is the solution form for $y'' + 9y = 0$?

Answer: General solution: $y = C_1 \text{cos}(3x) + C_2 \text{sin}(3x)$. Characteristic equation $r^2 + 9 = 0$ gives $r = \pm 3i$.

Flashcard 38: Verify if $y = e^x$ satisfies $y' = y$.

Answer: Yes, $y' = e^x = y$. Derivative of $e^x$ equals itself.

Flashcard 39: Is $y = Ce^{3x}$ a solution for $y' = 3y$?

Answer: Yes, $y' = 3y$. Derivative of $Ce^{3x}$ is $3Ce^{3x} = 3y$.

Flashcard 40: What is the solution for $y' = 0$?

Answer: $y = C$. Derivative of constant is zero.

Flashcard 41: What is the first step in verifying a solution?

Answer: Substitute the function into the differential equation. Direct substitution method for verification.

Flashcard 42: Verify if $y = Ce^{-2x}$ solves $y' + 2y = 0$.

Answer: Yes, it satisfies $y' + 2y = 0$. $y' = -2Ce^{-2x} = -2y$, so $y' + 2y = 0$.

Flashcard 43: Verify if $y = e^{-2x}$ satisfies $y' = -2y$.

Answer: Yes, $y' = -2e^{-2x}$. Derivative equals -2 times the function.

Flashcard 44: Which function is a solution to $y' = 3x^2$?

Answer: Any antiderivative of $3x^2$, e.g., $y = x^3 + C$. Integrate $3x^2$ to get $x^3 + C$.

Flashcard 45: What is the next step after substitution in solution verification?

Answer: Check if the equation holds true. Verify both sides are equal after substitution.

Flashcard 46: Is $y = x^2 + C$ a solution to $y'' = 2$?

Answer: Yes, $y'' = 2$. Second derivative of $x^2 + C$ is constant 2.

Flashcard 47: Verify if $y = C \text{cos}(2x)$ solves $y'' + 4y = 0$.

Answer: Yes, it satisfies $y'' + 4y = 0$. $y'' = -4C\cos(2x)$, so $y'' + 4y = 0$.

Flashcard 48: Verify $y = 2x - 1$ for $y' = 2$.

Answer: Yes, $y' = 2$. Derivative of linear function $2x - 1$ is 2.

Flashcard 49: Verify if $y = x^2$ satisfies $y'' = 2$.

Answer: Yes, $y'' = 2$. Second derivative of $x^2$ is constant 2.

Flashcard 50: Find the derivative of $y = \frac{1}{x}$.

Answer: $y' = -\frac{1}{x^2}$. Power rule: $\frac{d}{dx}[x^{-1}] = -x^{-2}$.

Flashcard 51: Verify if $y = Cx^{-1}$ is a solution for $xy' + y = 0$.

Answer: Yes, it satisfies $xy' + y = 0$. $y' = -Cx^{-2}$, and $xy' + y = 0$ checks out.

Flashcard 52: What is the next step after substitution in solution verification?

Answer: Check if the equation holds true. Verify both sides are equal after substitution.

Flashcard 53: What is the general solution for $y'' - 4y = 0$?

Answer: $y = C_1 e^{2x} + C_2 e^{-2x}$. Characteristic equation $r^2 - 4 = 0$ gives $r = \pm 2$.

Flashcard 54: Verify if $y = Cx^{-1}$ is a solution for $xy' + y = 0$.

Answer: Yes, it satisfies $xy' + y = 0$. $y' = -Cx^{-2}$, and $xy' + y = 0$ checks out.

Flashcard 55: What is a separable differential equation?

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

Flashcard 56: What is a linear differential equation?

Answer: An equation of the form $a_n(x)y^{(n)} + ... + a_1(x)y' + a_0(x)y = g(x)$. Coefficients are functions of $x$, equation is first-degree in $y$.

Flashcard 57: Determine if $y = e^{-x}$ solves $y' = -y$.

Answer: Yes, $y' = -e^{-x}$. Derivative equals negative of the function.

Flashcard 58: What is the first step in verifying a solution?

Answer: Substitute the function into the differential equation. Direct substitution method for verification.

Flashcard 59: Verify if $y = e^x$ satisfies $y' = y$.

Answer: Yes, $y' = e^x = y$. Derivative of $e^x$ equals itself.

Flashcard 60: Identify the order of the differential equation: $y'' + 3y' - 5y = 0$.

Answer: Order 2. Highest derivative is the second, so order is 2.

Flashcard 61: Verify if $y = e^{-2x}$ satisfies $y' = -2y$.

Answer: Yes, $y' = -2e^{-2x}$. Derivative equals -2 times the function.

Flashcard 62: What is a differential equation?

Answer: An equation involving derivatives of a function. Mathematical relationship involving rates of change.

Flashcard 63: Verify if $y = e^{3x}$ solves $y' = 3y$.

Answer: Yes, $y' = 3e^{3x}$. Derivative matches 3 times the function.

Flashcard 64: Determine if $y = \frac{1}{x}$ solves $xy' = -y^2$.

Answer: Yes, $y' = -\frac{1}{x^2}$. Check: $x(-\frac{1}{x^2}) = -\frac{1}{x} = -(\frac{1}{x})^2$.

Flashcard 65: Determine if $y = e^{-x}$ solves $y' = -y$.

Answer: Yes, $y' = -e^{-x}$. Derivative equals negative of the function.

Flashcard 66: What is the general solution for $y' = 2y$?

Answer: $y = Ce^{2x}$. Exponential growth with rate constant 2.

Flashcard 67: Determine if $y = \text{cos}(x)$ satisfies $y'' + y = 0$.

Answer: Yes, $y'' + y = 0$. $y'' = -\cos x$, so $y'' + y = 0$.

Flashcard 68: What is the general solution for $y' = -y$?

Answer: $y = Ce^{-x}$. Exponential decay with rate constant 1.