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AP Calculus BC Flashcards: Initial Conditions And Separation Of Variables

Study Initial Conditions And Separation Of Variables in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

← Back to flashcard decks

What this deck covers

This deck focuses on Initial Conditions And Separation Of Variables, 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: Initial Conditions And Separation Of Variables

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QUESTION

What equation results from integrating $\frac{dy}{y} = \frac{1}{x} dx$ ?

Tap or drag to reveal answer

ANSWER

$\ln |y| = \ln |x| + C$ . Standard result from integrating separated variables.

Swipe Right = I Know It! 🎉

Swipe Left = Still Learning

All flashcards

Flashcard 1: What equation results from integrating $\frac{dy}{y} = \frac{1}{x} dx$ ?

Answer: $\ln |y| = \ln |x| + C$ . Standard result from integrating separated variables.

Flashcard 2: Convert $\frac{dy}{dx} = x^2y^3$ into a separable form.

Answer: $\frac{dy}{y^3} = x^2 dx$ . Move $y^3$ to the left side and $x^2$ to the right.

Flashcard 3: Find the particular solution of $\frac{dy}{dx} = 2x$ with $y(0) = 4$ .

Answer: $y = x^2 + 4$ . Integrate to get $y = x^2 + C$ , then use initial condition.

Flashcard 4: What is the particular solution for $\frac{dy}{dx} = -3y$ with $y(2) = 1$ ?

Answer: $y = e^{-3x + 6}$ . From $y = Ce^{-3x}$ , use $y(2) = 1$ to find $C = e^6$ .

Flashcard 5: What is the role of the integration constant $C$ in differential equations?

Answer: Represents the family of solutions. Each value of $C$ gives a different particular solution.

Flashcard 6: What is the general solution of $\frac{dy}{dx} = 1$ ?

Answer: $y = x + C$ . Direct integration of a constant function.

Flashcard 7: Determine $C$ for the particular solution of $\frac{dy}{dx} = 5y$ with $y(0) = 7$ .

Answer: $C = 7$ . From $y = Ce^{5x}$ with $y(0) = 7$ , so $C = 7$ .

Flashcard 8: Solve $\frac{dy}{dx} = x^3y^2$ by separating variables.

Answer: $\frac{dy}{y^2} = x^3 dx$ . Move $y^2$ to left side and $x^3$ to right side.

Flashcard 9: Given $\frac{dy}{dx} = 3x^2$ , what is the particular solution with $y(0) = 2$ ?

Answer: $y = x^3 + 2$ . Integrate to get $y = x^3 + C$ , then use $y(0) = 2$ to find $C = 2$ .

Flashcard 10: State the integration result of $\frac{dy}{y} = \frac{1}{x} dx$ .

Answer: $\text{ln}|y| = \text{ln}|x| + C$ . Direct integration of the separated variables.

Flashcard 11: What is the general solution for $\frac{dy}{dx} = -y$ ?

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

Flashcard 12: What is the role of $C$ in the solution $y = Ce^{kx}$ ?

Answer: Constant for initial conditions. Determines the specific solution from the family of solutions.

Flashcard 13: Given $\frac{dy}{dx} = 4x$ , find $C$ if $y(1) = 6$ .

Answer: $C = 4$ . From $y = 2x^2 + C$ with $y(1) = 6$ , so $C = 4$ .

Flashcard 14: Solve $\frac{dy}{dx} = \frac{y}{x}$ for $y$ using separation of variables.

Answer: $y = Cx$ . Separate to $\frac{dy}{y} = \frac{dx}{x}$ and integrate both sides.

Flashcard 15: What is the particular solution for $\frac{dy}{dx} = 2y$ with $y(0) = 3$ ?

Answer: $y = 3e^{2x}$ . From $y = Ce^{2x}$ with condition $y(0) = 3$ giving $C = 3$ .

Flashcard 16: Determine the particular solution of $\frac{dy}{dx} = y$ with $y(0) = 5$ .

Answer: $y = 5e^x$ . From general solution $y = Ce^x$ , use condition to find $C = 5$ .

Flashcard 17: What is the general solution to $\frac{dy}{dx} = -ky$ ?

Answer: $y = Ce^{-kx}$ . Standard exponential decay solution with rate $k$ .

Flashcard 18: What is the general solution for $\frac{dy}{dx} = x^2$ ?

Answer: $y = \frac{x^3}{3} + C$ . Direct integration of the polynomial function.

Flashcard 19: Find the particular solution for $\frac{dy}{dx} = 3x^2$ with $y(0) = 1$ .

Answer: $y = x^3 + 1$ . Integrate $3x^2$ and apply the initial condition at $x = 0$ .

Flashcard 20: What is the solution to $\frac{dy}{dx} = 0$ ?

Answer: $y = C$ . When the derivative is zero, the function is constant.

Flashcard 21: Solve $\frac{dy}{dx} = -2xy$ for $y$ using separation of variables.

Answer: $y = Ce^{-x^2}$ . Separate to $\frac{dy}{y} = -2x \, dx$ and integrate both sides.

Flashcard 22: Given $y = Ce^{x^2}$ , find $C$ if $y(0) = 1$ .

Answer: $C = 1$ . Substitute $x = 0$ and $y = 1$ into the general solution.

Flashcard 23: Find the particular solution to $\frac{dy}{dx} = \frac{2y}{x}$ with $y(1) = 3$ .

Answer: $y = 3x^2$ . Separate to $\frac{dy}{y} = \frac{2dx}{x}$ , integrate to get $y = Cx^2$ .

Flashcard 24: Solve for $y$ given $\frac{dy}{dx} = x$ with initial condition $y(1) = 3$ .

Answer: $y = \frac{x^2}{2} + \frac{5}{2}$ . Integrate $\int x \, dx = \frac{x^2}{2} + C$ and apply condition.

Flashcard 25: What is the purpose of using initial conditions in solving differential equations?

Answer: To find particular solutions. They eliminate the arbitrary constant $C$ from general solutions.

Flashcard 26: Given $\frac{dy}{dx} = 2y$ , find the particular solution with $y(0) = 1$ .

Answer: $y = e^{2x}$ . From $y = Ce^{2x}$ with initial condition $y(0) = 1$ .

Flashcard 27: What is the general solution to $\frac{dy}{dx} = ky$ using separation of variables?

Answer: $y = Ce^{kx}$ . Standard form for exponential growth/decay differential equations.

Flashcard 28: How do you integrate $y' = \frac{dy}{dx} = x^2 y$ using separation of variables?

Answer: $\frac{dy}{y} = x^2 dx$ . Divide both sides by $y$ and multiply by $dx$ .

Flashcard 29: What is the first step in solving a differential equation using separation of variables?

Answer: Separate the variables. This allows variables to be on different sides for integration.

Flashcard 30: What is the solution to $\frac{dy}{dx} = 4y$ using separation of variables?

Answer: $y = Ce^{4x}$ . Standard exponential solution with growth rate $k = 4$ .

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AP Calculus BC Flashcards: Initial Conditions And Separation Of Variables

Study Initial Conditions And Separation Of Variables in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

← Back to flashcard decks

What this deck covers

This deck focuses on Initial Conditions And Separation Of Variables, 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: Initial Conditions And Separation Of Variables

/ 30

0 reviewed

0% Complete

0 reviewing

QUESTION

What equation results from integrating $\frac{dy}{y} = \frac{1}{x} dx$ ?

Tap or drag to reveal answer

ANSWER

$\ln |y| = \ln |x| + C$ . Standard result from integrating separated variables.

Swipe Right = I Know It! 🎉

Swipe Left = Still Learning

All flashcards

Flashcard 1: What equation results from integrating $\frac{dy}{y} = \frac{1}{x} dx$ ?

Answer: $\ln |y| = \ln |x| + C$ . Standard result from integrating separated variables.

Flashcard 2: Convert $\frac{dy}{dx} = x^2y^3$ into a separable form.

Answer: $\frac{dy}{y^3} = x^2 dx$ . Move $y^3$ to the left side and $x^2$ to the right.

Flashcard 3: Find the particular solution of $\frac{dy}{dx} = 2x$ with $y(0) = 4$ .

Answer: $y = x^2 + 4$ . Integrate to get $y = x^2 + C$ , then use initial condition.

Flashcard 4: What is the particular solution for $\frac{dy}{dx} = -3y$ with $y(2) = 1$ ?

Answer: $y = e^{-3x + 6}$ . From $y = Ce^{-3x}$ , use $y(2) = 1$ to find $C = e^6$ .

Flashcard 5: What is the role of the integration constant $C$ in differential equations?

Answer: Represents the family of solutions. Each value of $C$ gives a different particular solution.

Flashcard 6: What is the general solution of $\frac{dy}{dx} = 1$ ?

Answer: $y = x + C$ . Direct integration of a constant function.

Flashcard 7: Determine $C$ for the particular solution of $\frac{dy}{dx} = 5y$ with $y(0) = 7$ .

Answer: $C = 7$ . From $y = Ce^{5x}$ with $y(0) = 7$ , so $C = 7$ .

Flashcard 8: Solve $\frac{dy}{dx} = x^3y^2$ by separating variables.

Answer: $\frac{dy}{y^2} = x^3 dx$ . Move $y^2$ to left side and $x^3$ to right side.

Flashcard 9: Given $\frac{dy}{dx} = 3x^2$ , what is the particular solution with $y(0) = 2$ ?

Answer: $y = x^3 + 2$ . Integrate to get $y = x^3 + C$ , then use $y(0) = 2$ to find $C = 2$ .

Flashcard 10: State the integration result of $\frac{dy}{y} = \frac{1}{x} dx$ .

Answer: $\text{ln}|y| = \text{ln}|x| + C$ . Direct integration of the separated variables.

Flashcard 11: What is the general solution for $\frac{dy}{dx} = -y$ ?

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

Flashcard 12: What is the role of $C$ in the solution $y = Ce^{kx}$ ?

Answer: Constant for initial conditions. Determines the specific solution from the family of solutions.

Flashcard 13: Given $\frac{dy}{dx} = 4x$ , find $C$ if $y(1) = 6$ .

Answer: $C = 4$ . From $y = 2x^2 + C$ with $y(1) = 6$ , so $C = 4$ .

Flashcard 14: Solve $\frac{dy}{dx} = \frac{y}{x}$ for $y$ using separation of variables.

Answer: $y = Cx$ . Separate to $\frac{dy}{y} = \frac{dx}{x}$ and integrate both sides.

Flashcard 15: What is the particular solution for $\frac{dy}{dx} = 2y$ with $y(0) = 3$ ?

Answer: $y = 3e^{2x}$ . From $y = Ce^{2x}$ with condition $y(0) = 3$ giving $C = 3$ .

Flashcard 16: Determine the particular solution of $\frac{dy}{dx} = y$ with $y(0) = 5$ .

Answer: $y = 5e^x$ . From general solution $y = Ce^x$ , use condition to find $C = 5$ .

Flashcard 17: What is the general solution to $\frac{dy}{dx} = -ky$ ?

Answer: $y = Ce^{-kx}$ . Standard exponential decay solution with rate $k$ .

Flashcard 18: What is the general solution for $\frac{dy}{dx} = x^2$ ?

Answer: $y = \frac{x^3}{3} + C$ . Direct integration of the polynomial function.

Flashcard 19: Find the particular solution for $\frac{dy}{dx} = 3x^2$ with $y(0) = 1$ .

Answer: $y = x^3 + 1$ . Integrate $3x^2$ and apply the initial condition at $x = 0$ .

Flashcard 20: What is the solution to $\frac{dy}{dx} = 0$ ?

Answer: $y = C$ . When the derivative is zero, the function is constant.

Flashcard 21: Solve $\frac{dy}{dx} = -2xy$ for $y$ using separation of variables.

Answer: $y = Ce^{-x^2}$ . Separate to $\frac{dy}{y} = -2x \, dx$ and integrate both sides.

Flashcard 22: Given $y = Ce^{x^2}$ , find $C$ if $y(0) = 1$ .

Answer: $C = 1$ . Substitute $x = 0$ and $y = 1$ into the general solution.

Flashcard 23: Find the particular solution to $\frac{dy}{dx} = \frac{2y}{x}$ with $y(1) = 3$ .

Answer: $y = 3x^2$ . Separate to $\frac{dy}{y} = \frac{2dx}{x}$ , integrate to get $y = Cx^2$ .

Flashcard 24: Solve for $y$ given $\frac{dy}{dx} = x$ with initial condition $y(1) = 3$ .

Answer: $y = \frac{x^2}{2} + \frac{5}{2}$ . Integrate $\int x \, dx = \frac{x^2}{2} + C$ and apply condition.

Flashcard 25: What is the purpose of using initial conditions in solving differential equations?

Answer: To find particular solutions. They eliminate the arbitrary constant $C$ from general solutions.

Flashcard 26: Given $\frac{dy}{dx} = 2y$ , find the particular solution with $y(0) = 1$ .

Answer: $y = e^{2x}$ . From $y = Ce^{2x}$ with initial condition $y(0) = 1$ .

Flashcard 27: What is the general solution to $\frac{dy}{dx} = ky$ using separation of variables?

Answer: $y = Ce^{kx}$ . Standard form for exponential growth/decay differential equations.

Flashcard 28: How do you integrate $y' = \frac{dy}{dx} = x^2 y$ using separation of variables?

Answer: $\frac{dy}{y} = x^2 dx$ . Divide both sides by $y$ and multiply by $dx$ .

Flashcard 29: What is the first step in solving a differential equation using separation of variables?

Answer: Separate the variables. This allows variables to be on different sides for integration.

Flashcard 30: What is the solution to $\frac{dy}{dx} = 4y$ using separation of variables?

Answer: $y = Ce^{4x}$ . Standard exponential solution with growth rate $k = 4$ .

Opening subject page...

AP Calculus BC Flashcards: Initial Conditions And Separation Of Variables

What this deck covers

How to use these flashcards

AP Calculus BC Flashcards: Initial Conditions And Separation Of Variables

All flashcards

Flashcard 1: What equation results from integrating dyy=1xdx\frac{dy}{y} = \frac{1}{x} dxydy​=x1​dx?

Flashcard 2: Convert dydx=x2y3\frac{dy}{dx} = x^2y^3dxdy​=x2y3 into a separable form.

Flashcard 3: Find the particular solution of dydx=2x\frac{dy}{dx} = 2xdxdy​=2x with y(0)=4y(0) = 4y(0)=4.

Flashcard 4: What is the particular solution for dydx=−3y\frac{dy}{dx} = -3ydxdy​=−3y with y(2)=1y(2) = 1y(2)=1?

Flashcard 5: What is the role of the integration constant CCC in differential equations?

Flashcard 6: What is the general solution of dydx=1\frac{dy}{dx} = 1dxdy​=1?

Flashcard 7: Determine CCC for the particular solution of dydx=5y\frac{dy}{dx} = 5ydxdy​=5y with y(0)=7y(0) = 7y(0)=7.

Flashcard 8: Solve dydx=x3y2\frac{dy}{dx} = x^3y^2dxdy​=x3y2 by separating variables.

Flashcard 9: Given dydx=3x2\frac{dy}{dx} = 3x^2dxdy​=3x2, what is the particular solution with y(0)=2y(0) = 2y(0)=2?

Flashcard 10: State the integration result of dyy=1xdx\frac{dy}{y} = \frac{1}{x} dxydy​=x1​dx.

Flashcard 11: What is the general solution for dydx=−y\frac{dy}{dx} = -ydxdy​=−y?

Flashcard 12: What is the role of CCC in the solution y=Cekxy = Ce^{kx}y=Cekx?

Flashcard 13: Given dydx=4x\frac{dy}{dx} = 4xdxdy​=4x, find CCC if y(1)=6y(1) = 6y(1)=6.

Flashcard 14: Solve dydx=yx\frac{dy}{dx} = \frac{y}{x}dxdy​=xy​ for yyy using separation of variables.

Flashcard 15: What is the particular solution for dydx=2y\frac{dy}{dx} = 2ydxdy​=2y with y(0)=3y(0) = 3y(0)=3?

Flashcard 16: Determine the particular solution of dydx=y\frac{dy}{dx} = ydxdy​=y with y(0)=5y(0) = 5y(0)=5.

Flashcard 17: What is the general solution to dydx=−ky\frac{dy}{dx} = -kydxdy​=−ky?

Flashcard 18: What is the general solution for dydx=x2\frac{dy}{dx} = x^2dxdy​=x2?

Flashcard 19: Find the particular solution for dydx=3x2\frac{dy}{dx} = 3x^2dxdy​=3x2 with y(0)=1y(0) = 1y(0)=1.

Flashcard 20: What is the solution to dydx=0\frac{dy}{dx} = 0dxdy​=0?

Flashcard 21: Solve dydx=−2xy\frac{dy}{dx} = -2xydxdy​=−2xy for yyy using separation of variables.

Flashcard 22: Given y=Cex2y = Ce^{x^2}y=Cex2, find CCC if y(0)=1y(0) = 1y(0)=1.

Flashcard 23: Find the particular solution to dydx=2yx\frac{dy}{dx} = \frac{2y}{x}dxdy​=x2y​ with y(1)=3y(1) = 3y(1)=3.

Flashcard 24: Solve for yyy given dydx=x\frac{dy}{dx} = xdxdy​=x with initial condition y(1)=3y(1) = 3y(1)=3.

Flashcard 25: What is the purpose of using initial conditions in solving differential equations?

Flashcard 26: Given dydx=2y\frac{dy}{dx} = 2ydxdy​=2y, find the particular solution with y(0)=1y(0) = 1y(0)=1.

Flashcard 27: What is the general solution to dydx=ky\frac{dy}{dx} = kydxdy​=ky using separation of variables?

Flashcard 28: How do you integrate y′=dydx=x2yy' = \frac{dy}{dx} = x^2 yy′=dxdy​=x2y using separation of variables?

Flashcard 29: What is the first step in solving a differential equation using separation of variables?

Flashcard 30: What is the solution to dydx=4y\frac{dy}{dx} = 4ydxdy​=4y using separation of variables?

Opening subject page...

AP Calculus BC Flashcards: Initial Conditions And Separation Of Variables

What this deck covers

How to use these flashcards

AP Calculus BC Flashcards: Initial Conditions And Separation Of Variables

All flashcards

Flashcard 1: What equation results from integrating dyy=1xdx\frac{dy}{y} = \frac{1}{x} dxydy​=x1​dx?

Flashcard 2: Convert dydx=x2y3\frac{dy}{dx} = x^2y^3dxdy​=x2y3 into a separable form.

Flashcard 3: Find the particular solution of dydx=2x\frac{dy}{dx} = 2xdxdy​=2x with y(0)=4y(0) = 4y(0)=4.

Flashcard 4: What is the particular solution for dydx=−3y\frac{dy}{dx} = -3ydxdy​=−3y with y(2)=1y(2) = 1y(2)=1?

Flashcard 5: What is the role of the integration constant CCC in differential equations?

Flashcard 6: What is the general solution of dydx=1\frac{dy}{dx} = 1dxdy​=1?

Flashcard 7: Determine CCC for the particular solution of dydx=5y\frac{dy}{dx} = 5ydxdy​=5y with y(0)=7y(0) = 7y(0)=7.

Flashcard 8: Solve dydx=x3y2\frac{dy}{dx} = x^3y^2dxdy​=x3y2 by separating variables.

Flashcard 9: Given dydx=3x2\frac{dy}{dx} = 3x^2dxdy​=3x2, what is the particular solution with y(0)=2y(0) = 2y(0)=2?

Flashcard 10: State the integration result of dyy=1xdx\frac{dy}{y} = \frac{1}{x} dxydy​=x1​dx.

Flashcard 11: What is the general solution for dydx=−y\frac{dy}{dx} = -ydxdy​=−y?

Flashcard 12: What is the role of CCC in the solution y=Cekxy = Ce^{kx}y=Cekx?

Flashcard 13: Given dydx=4x\frac{dy}{dx} = 4xdxdy​=4x, find CCC if y(1)=6y(1) = 6y(1)=6.

Flashcard 14: Solve dydx=yx\frac{dy}{dx} = \frac{y}{x}dxdy​=xy​ for yyy using separation of variables.

Flashcard 15: What is the particular solution for dydx=2y\frac{dy}{dx} = 2ydxdy​=2y with y(0)=3y(0) = 3y(0)=3?

Flashcard 16: Determine the particular solution of dydx=y\frac{dy}{dx} = ydxdy​=y with y(0)=5y(0) = 5y(0)=5.

Flashcard 17: What is the general solution to dydx=−ky\frac{dy}{dx} = -kydxdy​=−ky?

Flashcard 18: What is the general solution for dydx=x2\frac{dy}{dx} = x^2dxdy​=x2?

Flashcard 19: Find the particular solution for dydx=3x2\frac{dy}{dx} = 3x^2dxdy​=3x2 with y(0)=1y(0) = 1y(0)=1.

Flashcard 20: What is the solution to dydx=0\frac{dy}{dx} = 0dxdy​=0?

Flashcard 21: Solve dydx=−2xy\frac{dy}{dx} = -2xydxdy​=−2xy for yyy using separation of variables.

Flashcard 22: Given y=Cex2y = Ce^{x^2}y=Cex2, find CCC if y(0)=1y(0) = 1y(0)=1.

Flashcard 23: Find the particular solution to dydx=2yx\frac{dy}{dx} = \frac{2y}{x}dxdy​=x2y​ with y(1)=3y(1) = 3y(1)=3.

Flashcard 24: Solve for yyy given dydx=x\frac{dy}{dx} = xdxdy​=x with initial condition y(1)=3y(1) = 3y(1)=3.

Flashcard 25: What is the purpose of using initial conditions in solving differential equations?

Flashcard 26: Given dydx=2y\frac{dy}{dx} = 2ydxdy​=2y, find the particular solution with y(0)=1y(0) = 1y(0)=1.

Flashcard 27: What is the general solution to dydx=ky\frac{dy}{dx} = kydxdy​=ky using separation of variables?

Flashcard 28: How do you integrate y′=dydx=x2yy' = \frac{dy}{dx} = x^2 yy′=dxdy​=x2y using separation of variables?

Flashcard 29: What is the first step in solving a differential equation using separation of variables?

Flashcard 30: What is the solution to dydx=4y\frac{dy}{dx} = 4ydxdy​=4y using separation of variables?

Flashcard 1: What equation results from integrating $\frac{dy}{y} = \frac{1}{x} dx$ ?

Flashcard 2: Convert $\frac{dy}{dx} = x^2y^3$ into a separable form.

Flashcard 3: Find the particular solution of $\frac{dy}{dx} = 2x$ with $y(0) = 4$ .

Flashcard 4: What is the particular solution for $\frac{dy}{dx} = -3y$ with $y(2) = 1$ ?

Flashcard 5: What is the role of the integration constant $C$ in differential equations?

Flashcard 6: What is the general solution of $\frac{dy}{dx} = 1$ ?

Flashcard 7: Determine $C$ for the particular solution of $\frac{dy}{dx} = 5y$ with $y(0) = 7$ .

Flashcard 8: Solve $\frac{dy}{dx} = x^3y^2$ by separating variables.

Flashcard 9: Given $\frac{dy}{dx} = 3x^2$ , what is the particular solution with $y(0) = 2$ ?

Flashcard 10: State the integration result of $\frac{dy}{y} = \frac{1}{x} dx$ .

Flashcard 11: What is the general solution for $\frac{dy}{dx} = -y$ ?

Flashcard 12: What is the role of $C$ in the solution $y = Ce^{kx}$ ?

Flashcard 13: Given $\frac{dy}{dx} = 4x$ , find $C$ if $y(1) = 6$ .

Flashcard 14: Solve $\frac{dy}{dx} = \frac{y}{x}$ for $y$ using separation of variables.

Flashcard 15: What is the particular solution for $\frac{dy}{dx} = 2y$ with $y(0) = 3$ ?

Flashcard 16: Determine the particular solution of $\frac{dy}{dx} = y$ with $y(0) = 5$ .

Flashcard 17: What is the general solution to $\frac{dy}{dx} = -ky$ ?

Flashcard 18: What is the general solution for $\frac{dy}{dx} = x^2$ ?

Flashcard 19: Find the particular solution for $\frac{dy}{dx} = 3x^2$ with $y(0) = 1$ .

Flashcard 20: What is the solution to $\frac{dy}{dx} = 0$ ?

Flashcard 21: Solve $\frac{dy}{dx} = -2xy$ for $y$ using separation of variables.

Flashcard 22: Given $y = Ce^{x^2}$ , find $C$ if $y(0) = 1$ .

Flashcard 23: Find the particular solution to $\frac{dy}{dx} = \frac{2y}{x}$ with $y(1) = 3$ .

Flashcard 24: Solve for $y$ given $\frac{dy}{dx} = x$ with initial condition $y(1) = 3$ .

Flashcard 26: Given $\frac{dy}{dx} = 2y$ , find the particular solution with $y(0) = 1$ .

Flashcard 27: What is the general solution to $\frac{dy}{dx} = ky$ using separation of variables?

Flashcard 28: How do you integrate $y' = \frac{dy}{dx} = x^2 y$ using separation of variables?

Flashcard 30: What is the solution to $\frac{dy}{dx} = 4y$ using separation of variables?

Flashcard 1: What equation results from integrating $\frac{dy}{y} = \frac{1}{x} dx$ ?

Flashcard 2: Convert $\frac{dy}{dx} = x^2y^3$ into a separable form.

Flashcard 3: Find the particular solution of $\frac{dy}{dx} = 2x$ with $y(0) = 4$ .

Flashcard 4: What is the particular solution for $\frac{dy}{dx} = -3y$ with $y(2) = 1$ ?

Flashcard 5: What is the role of the integration constant $C$ in differential equations?

Flashcard 6: What is the general solution of $\frac{dy}{dx} = 1$ ?

Flashcard 7: Determine $C$ for the particular solution of $\frac{dy}{dx} = 5y$ with $y(0) = 7$ .

Flashcard 8: Solve $\frac{dy}{dx} = x^3y^2$ by separating variables.

Flashcard 9: Given $\frac{dy}{dx} = 3x^2$ , what is the particular solution with $y(0) = 2$ ?

Flashcard 10: State the integration result of $\frac{dy}{y} = \frac{1}{x} dx$ .

Flashcard 11: What is the general solution for $\frac{dy}{dx} = -y$ ?

Flashcard 12: What is the role of $C$ in the solution $y = Ce^{kx}$ ?

Flashcard 13: Given $\frac{dy}{dx} = 4x$ , find $C$ if $y(1) = 6$ .

Flashcard 14: Solve $\frac{dy}{dx} = \frac{y}{x}$ for $y$ using separation of variables.

Flashcard 15: What is the particular solution for $\frac{dy}{dx} = 2y$ with $y(0) = 3$ ?

Flashcard 16: Determine the particular solution of $\frac{dy}{dx} = y$ with $y(0) = 5$ .

Flashcard 17: What is the general solution to $\frac{dy}{dx} = -ky$ ?

Flashcard 18: What is the general solution for $\frac{dy}{dx} = x^2$ ?

Flashcard 19: Find the particular solution for $\frac{dy}{dx} = 3x^2$ with $y(0) = 1$ .

Flashcard 20: What is the solution to $\frac{dy}{dx} = 0$ ?

Flashcard 21: Solve $\frac{dy}{dx} = -2xy$ for $y$ using separation of variables.

Flashcard 22: Given $y = Ce^{x^2}$ , find $C$ if $y(0) = 1$ .

Flashcard 23: Find the particular solution to $\frac{dy}{dx} = \frac{2y}{x}$ with $y(1) = 3$ .

Flashcard 24: Solve for $y$ given $\frac{dy}{dx} = x$ with initial condition $y(1) = 3$ .

Flashcard 26: Given $\frac{dy}{dx} = 2y$ , find the particular solution with $y(0) = 1$ .

Flashcard 27: What is the general solution to $\frac{dy}{dx} = ky$ using separation of variables?

Flashcard 28: How do you integrate $y' = \frac{dy}{dx} = x^2 y$ using separation of variables?

Flashcard 30: What is the solution to $\frac{dy}{dx} = 4y$ using separation of variables?