Initial Conditions and Separation of Variables - AP Calculus AB
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What is the general solution of $\frac{dy}{dx} = 1$?
What is the general solution of $\frac{dy}{dx} = 1$?
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$y = x + C$. Direct integration of the constant function gives $x + C$.
$y = x + C$. Direct integration of the constant function gives $x + C$.
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What is the role of the constant of integration in solving separable equations?
What is the role of the constant of integration in solving separable equations?
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It accounts for the family of solutions. Different values of $C$ give different solution curves.
It accounts for the family of solutions. Different values of $C$ give different solution curves.
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Find the particular solution for $\frac{dy}{dx} = 6x$ with $y(2) = 5$.
Find the particular solution for $\frac{dy}{dx} = 6x$ with $y(2) = 5$.
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$y = 3x^2 - 7$. Integrate: $y = 3x^2 + C$, apply $y(2) = 5$ to find $C$.
$y = 3x^2 - 7$. Integrate: $y = 3x^2 + C$, apply $y(2) = 5$ to find $C$.
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Determine the particular solution for $\frac{dy}{dx} = x$ with $y(0) = 1$.
Determine the particular solution for $\frac{dy}{dx} = x$ with $y(0) = 1$.
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$y = \frac{x^2}{2} + 1$. Integrate: $y = \frac{x^2}{2} + C$, apply $y(0) = 1$ to find $C = 1$.
$y = \frac{x^2}{2} + 1$. Integrate: $y = \frac{x^2}{2} + C$, apply $y(0) = 1$ to find $C = 1$.
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Given $\frac{dy}{dx} = y \sin(x)$, find the general solution.
Given $\frac{dy}{dx} = y \sin(x)$, find the general solution.
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$y = C e^{-\cos(x)}$. Separate: $\frac{dy}{y} = \sin(x) dx$, then integrate both sides.
$y = C e^{-\cos(x)}$. Separate: $\frac{dy}{y} = \sin(x) dx$, then integrate both sides.
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Solve $\frac{dy}{dx} = \frac{x}{y}$ given $y(1) = 2$. What is the particular solution?
Solve $\frac{dy}{dx} = \frac{x}{y}$ given $y(1) = 2$. What is the particular solution?
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$y^2 = x^2 + 3$. Separate and integrate: $\int y dy = \int x dx$, then apply initial condition.
$y^2 = x^2 + 3$. Separate and integrate: $\int y dy = \int x dx$, then apply initial condition.
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Solve $\frac{dy}{dx} = 2y$ for $y(0) = 3$. What is the particular solution?
Solve $\frac{dy}{dx} = 2y$ for $y(0) = 3$. What is the particular solution?
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$y = 3e^{2x}$. Separate: $\frac{dy}{y} = 2dx$, integrate, apply initial condition.
$y = 3e^{2x}$. Separate: $\frac{dy}{y} = 2dx$, integrate, apply initial condition.
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Identify the integral needed to solve $\frac{dy}{dx} = x^2y$ after separating variables.
Identify the integral needed to solve $\frac{dy}{dx} = x^2y$ after separating variables.
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$\frac{1}{y} dy = x^2 dx$. Rearranging gives $\frac{dy}{y} = x^2 dx$ for integration.
$\frac{1}{y} dy = x^2 dx$. Rearranging gives $\frac{dy}{y} = x^2 dx$ for integration.
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What is the particular solution of $\frac{dy}{dx} = 2x - 1$ if $y(0) = 3$?
What is the particular solution of $\frac{dy}{dx} = 2x - 1$ if $y(0) = 3$?
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$y = x^2 - x + 3$. Integrate: $y = x^2 - x + C$, apply $y(0) = 3$ gives $C = 3$.
$y = x^2 - x + 3$. Integrate: $y = x^2 - x + C$, apply $y(0) = 3$ gives $C = 3$.
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What is the first step in solving a differential equation using separation of variables?
What is the first step in solving a differential equation using separation of variables?
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Separate the variables to opposite sides of the equation. This isolates each variable for independent integration.
Separate the variables to opposite sides of the equation. This isolates each variable for independent integration.
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State the general form of a separable differential equation.
State the general form of a separable differential equation.
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An equation of the form $\frac{dy}{dx} = g(x)h(y)$.. The function can be factored into separate $x$ and $y$ terms.
An equation of the form $\frac{dy}{dx} = g(x)h(y)$.. The function can be factored into separate $x$ and $y$ terms.
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Given $\frac{dy}{dx} = 3x$, find the particular solution if $y(0) = 2$.
Given $\frac{dy}{dx} = 3x$, find the particular solution if $y(0) = 2$.
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$y = \frac{3}{2}x^2 + 2$. Integrate $3x$ and apply $y(0) = 2$ to find $C = 2$.
$y = \frac{3}{2}x^2 + 2$. Integrate $3x$ and apply $y(0) = 2$ to find $C = 2$.
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What is the purpose of integrating both sides after separation of variables?
What is the purpose of integrating both sides after separation of variables?
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To find the antiderivatives, leading to the general solution. Integration reverses differentiation to recover the function.
To find the antiderivatives, leading to the general solution. Integration reverses differentiation to recover the function.
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Find the particular solution of $\frac{dy}{dx} = 2xy$ with $y(0) = 1$.
Find the particular solution of $\frac{dy}{dx} = 2xy$ with $y(0) = 1$.
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$y = e^{x^2}$. Separate: $\frac{dy}{y} = 2x dx$, integrate, apply $y(0) = 1$.
$y = e^{x^2}$. Separate: $\frac{dy}{y} = 2x dx$, integrate, apply $y(0) = 1$.
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How do you determine the constant of integration using an initial condition?
How do you determine the constant of integration using an initial condition?
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Substitute the initial condition into the general solution. This replaces $C$ with a specific numerical value.
Substitute the initial condition into the general solution. This replaces $C$ with a specific numerical value.
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Find $C$ for the solution $y = Ce^{3x}$ given $y(0) = 5$.
Find $C$ for the solution $y = Ce^{3x}$ given $y(0) = 5$.
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$C = 5$. At $x = 0$: $5 = Ce^0 = C \cdot 1$, so $C = 5$.
$C = 5$. At $x = 0$: $5 = Ce^0 = C \cdot 1$, so $C = 5$.
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What does 'separable' mean in the context of differential equations?
What does 'separable' mean in the context of differential equations?
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Variables can be separated on opposite sides. The equation can be written as $f(x)dx = g(y)dy$.
Variables can be separated on opposite sides. The equation can be written as $f(x)dx = g(y)dy$.
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Determine the particular solution for $\frac{dy}{dx} = x^3$ with $y(1) = 4$.
Determine the particular solution for $\frac{dy}{dx} = x^3$ with $y(1) = 4$.
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$y = \frac{x^4}{4} + \frac{15}{4}$. Integrate: $y = \frac{x^4}{4} + C$, apply $y(1) = 4$ to find $C$.
$y = \frac{x^4}{4} + \frac{15}{4}$. Integrate: $y = \frac{x^4}{4} + C$, apply $y(1) = 4$ to find $C$.
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What does it mean if a solution is 'particular'?
What does it mean if a solution is 'particular'?
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It satisfies the differential equation and initial condition. It's the unique solution from the family that meets given conditions.
It satisfies the differential equation and initial condition. It's the unique solution from the family that meets given conditions.
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Solve $\frac{dy}{dx} = y$ for $y(0) = 2$. What is the particular solution?
Solve $\frac{dy}{dx} = y$ for $y(0) = 2$. What is the particular solution?
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$y = 2e^x$. Separate: $\frac{dy}{y} = dx$, integrate, apply initial condition.
$y = 2e^x$. Separate: $\frac{dy}{y} = dx$, integrate, apply initial condition.
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What is the initial condition in a differential equation problem?
What is the initial condition in a differential equation problem?
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A value that specifies the solution at a particular point, e.g., $y(x_0) = y_0$. This constraint determines the unique particular solution.
A value that specifies the solution at a particular point, e.g., $y(x_0) = y_0$. This constraint determines the unique particular solution.
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What is the general solution for $\frac{dy}{dx} = 0$?
What is the general solution for $\frac{dy}{dx} = 0$?
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$y = C$. Zero derivative means the function is constant.
$y = C$. Zero derivative means the function is constant.
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Find $C$ for $y = \frac{x^3}{3} + C$ given $y(1) = 2$.
Find $C$ for $y = \frac{x^3}{3} + C$ given $y(1) = 2$.
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$C = \frac{5}{3}$. At $x = 1$: $2 = \frac{1}{3} + C$, so $C = \frac{5}{3}$.
$C = \frac{5}{3}$. At $x = 1$: $2 = \frac{1}{3} + C$, so $C = \frac{5}{3}$.
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Find the particular solution of $\frac{dy}{dx} = 4y$ with $y(1) = 3$.
Find the particular solution of $\frac{dy}{dx} = 4y$ with $y(1) = 3$.
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$y = 3e^{4(x-1)}$. General solution $y = Ce^{4x}$, apply $y(1) = 3$ to find $C$.
$y = 3e^{4(x-1)}$. General solution $y = Ce^{4x}$, apply $y(1) = 3$ to find $C$.
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What is the general solution of $\frac{dy}{dx} = y$?
What is the general solution of $\frac{dy}{dx} = y$?
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$y = Ce^x$. This is the exponential growth equation with rate $1$.
$y = Ce^x$. This is the exponential growth equation with rate $1$.
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Explain why initial conditions are necessary for finding particular solutions.
Explain why initial conditions are necessary for finding particular solutions.
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To determine the specific value of the integration constant. Without them, we only get the general solution family.
To determine the specific value of the integration constant. Without them, we only get the general solution family.
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Describe the integration process in separation of variables.
Describe the integration process in separation of variables.
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Integrate both sides after separating variables. This finds antiderivatives of both separated expressions.
Integrate both sides after separating variables. This finds antiderivatives of both separated expressions.
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What is the role of the integration constant when finding the general solution?
What is the role of the integration constant when finding the general solution?
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Represents an arbitrary constant for family of solutions. It parameterizes all possible solutions before applying conditions.
Represents an arbitrary constant for family of solutions. It parameterizes all possible solutions before applying conditions.
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What is the general solution of $\frac{dy}{dx} = 4$?
What is the general solution of $\frac{dy}{dx} = 4$?
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$y = 4x + C$. Direct integration of constant $4$ gives $4x + C$.
$y = 4x + C$. Direct integration of constant $4$ gives $4x + C$.
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What technique is used to solve $\frac{dy}{dx} = \frac{x}{y}$?
What technique is used to solve $\frac{dy}{dx} = \frac{x}{y}$?
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Separation of variables. The equation has form $\frac{dy}{dx} = f(x)g(y)$ where variables separate.
Separation of variables. The equation has form $\frac{dy}{dx} = f(x)g(y)$ where variables separate.
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