Initial Conditions and Separation of Variables - AP Calculus BC
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What equation results from integrating $\frac{dy}{y} = \frac{1}{x} dx$?
What equation results from integrating $\frac{dy}{y} = \frac{1}{x} dx$?
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$\ln |y| = \ln |x| + C$. Standard result from integrating separated variables.
$\ln |y| = \ln |x| + C$. Standard result from integrating separated variables.
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Convert $\frac{dy}{dx} = x^2y^3$ into a separable form.
Convert $\frac{dy}{dx} = x^2y^3$ into a separable form.
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$\frac{dy}{y^3} = x^2 dx$. Move $y^3$ to the left side and $x^2$ to the right.
$\frac{dy}{y^3} = x^2 dx$. Move $y^3$ to the left side and $x^2$ to the right.
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Find the particular solution of $\frac{dy}{dx} = 2x$ with $y(0) = 4$.
Find the particular solution of $\frac{dy}{dx} = 2x$ with $y(0) = 4$.
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$y = x^2 + 4$. Integrate to get $y = x^2 + C$, then use initial condition.
$y = x^2 + 4$. Integrate to get $y = x^2 + C$, then use initial condition.
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What is the particular solution for $\frac{dy}{dx} = -3y$ with $y(2) = 1$?
What is the particular solution for $\frac{dy}{dx} = -3y$ with $y(2) = 1$?
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$y = e^{-3x + 6}$. From $y = Ce^{-3x}$, use $y(2) = 1$ to find $C = e^6$.
$y = e^{-3x + 6}$. From $y = Ce^{-3x}$, use $y(2) = 1$ to find $C = e^6$.
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What is the role of the integration constant $C$ in differential equations?
What is the role of the integration constant $C$ in differential equations?
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Represents the family of solutions. Each value of $C$ gives a different particular solution.
Represents the family of solutions. Each value of $C$ gives a different particular solution.
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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 a constant function.
$y = x + C$. Direct integration of a constant function.
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Determine $C$ for the particular solution of $\frac{dy}{dx} = 5y$ with $y(0) = 7$.
Determine $C$ for the particular solution of $\frac{dy}{dx} = 5y$ with $y(0) = 7$.
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$C = 7$. From $y = Ce^{5x}$ with $y(0) = 7$, so $C = 7$.
$C = 7$. From $y = Ce^{5x}$ with $y(0) = 7$, so $C = 7$.
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Solve $\frac{dy}{dx} = x^3y^2$ by separating variables.
Solve $\frac{dy}{dx} = x^3y^2$ by separating variables.
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$\frac{dy}{y^2} = x^3 dx$. Move $y^2$ to left side and $x^3$ to right side.
$\frac{dy}{y^2} = x^3 dx$. Move $y^2$ to left side and $x^3$ to right side.
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Given $\frac{dy}{dx} = 3x^2$, what is the particular solution with $y(0) = 2$?
Given $\frac{dy}{dx} = 3x^2$, what is the particular solution with $y(0) = 2$?
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$y = x^3 + 2$. Integrate to get $y = x^3 + C$, then use $y(0) = 2$ to find $C = 2$.
$y = x^3 + 2$. Integrate to get $y = x^3 + C$, then use $y(0) = 2$ to find $C = 2$.
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State the integration result of $\frac{dy}{y} = \frac{1}{x} dx$.
State the integration result of $\frac{dy}{y} = \frac{1}{x} dx$.
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$\text{ln}|y| = \text{ln}|x| + C$. Direct integration of the separated variables.
$\text{ln}|y| = \text{ln}|x| + C$. Direct integration of the separated variables.
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What is the general solution for $\frac{dy}{dx} = -y$?
What is the general solution for $\frac{dy}{dx} = -y$?
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$y = Ce^{-x}$. Standard exponential decay with rate $1$.
$y = Ce^{-x}$. Standard exponential decay with rate $1$.
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What is the role of $C$ in the solution $y = Ce^{kx}$?
What is the role of $C$ in the solution $y = Ce^{kx}$?
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Constant for initial conditions. Determines the specific solution from the family of solutions.
Constant for initial conditions. Determines the specific solution from the family of solutions.
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Given $\frac{dy}{dx} = 4x$, find $C$ if $y(1) = 6$.
Given $\frac{dy}{dx} = 4x$, find $C$ if $y(1) = 6$.
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$C = 4$. From $y = 2x^2 + C$ with $y(1) = 6$, so $C = 4$.
$C = 4$. From $y = 2x^2 + C$ with $y(1) = 6$, so $C = 4$.
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Solve $\frac{dy}{dx} = \frac{y}{x}$ for $y$ using separation of variables.
Solve $\frac{dy}{dx} = \frac{y}{x}$ for $y$ using separation of variables.
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$y = Cx$. Separate to $\frac{dy}{y} = \frac{dx}{x}$ and integrate both sides.
$y = Cx$. Separate to $\frac{dy}{y} = \frac{dx}{x}$ and integrate both sides.
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What is the particular solution for $\frac{dy}{dx} = 2y$ with $y(0) = 3$?
What is the particular solution for $\frac{dy}{dx} = 2y$ with $y(0) = 3$?
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$y = 3e^{2x}$. From $y = Ce^{2x}$ with condition $y(0) = 3$ giving $C = 3$.
$y = 3e^{2x}$. From $y = Ce^{2x}$ with condition $y(0) = 3$ giving $C = 3$.
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Determine the particular solution of $\frac{dy}{dx} = y$ with $y(0) = 5$.
Determine the particular solution of $\frac{dy}{dx} = y$ with $y(0) = 5$.
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$y = 5e^x$. From general solution $y = Ce^x$, use condition to find $C = 5$.
$y = 5e^x$. From general solution $y = Ce^x$, use condition to find $C = 5$.
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What is the general solution to $\frac{dy}{dx} = -ky$?
What is the general solution to $\frac{dy}{dx} = -ky$?
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$y = Ce^{-kx}$. Standard exponential decay solution with rate $k$.
$y = Ce^{-kx}$. Standard exponential decay solution with rate $k$.
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What is the general solution for $\frac{dy}{dx} = x^2$?
What is the general solution for $\frac{dy}{dx} = x^2$?
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$y = \frac{x^3}{3} + C$. Direct integration of the polynomial function.
$y = \frac{x^3}{3} + C$. Direct integration of the polynomial function.
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Find the particular solution for $\frac{dy}{dx} = 3x^2$ with $y(0) = 1$.
Find the particular solution for $\frac{dy}{dx} = 3x^2$ with $y(0) = 1$.
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$y = x^3 + 1$. Integrate $3x^2$ and apply the initial condition at $x = 0$.
$y = x^3 + 1$. Integrate $3x^2$ and apply the initial condition at $x = 0$.
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What is the solution to $\frac{dy}{dx} = 0$?
What is the solution to $\frac{dy}{dx} = 0$?
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$y = C$. When the derivative is zero, the function is constant.
$y = C$. When the derivative is zero, the function is constant.
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Solve $\frac{dy}{dx} = -2xy$ for $y$ using separation of variables.
Solve $\frac{dy}{dx} = -2xy$ for $y$ using separation of variables.
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$y = Ce^{-x^2}$. Separate to $\frac{dy}{y} = -2x , dx$ and integrate both sides.
$y = Ce^{-x^2}$. Separate to $\frac{dy}{y} = -2x , dx$ and integrate both sides.
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Given $y = Ce^{x^2}$, find $C$ if $y(0) = 1$.
Given $y = Ce^{x^2}$, find $C$ if $y(0) = 1$.
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$C = 1$. Substitute $x = 0$ and $y = 1$ into the general solution.
$C = 1$. Substitute $x = 0$ and $y = 1$ into the general solution.
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Find the particular solution to $\frac{dy}{dx} = \frac{2y}{x}$ with $y(1) = 3$.
Find the particular solution to $\frac{dy}{dx} = \frac{2y}{x}$ with $y(1) = 3$.
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$y = 3x^2$. Separate to $\frac{dy}{y} = \frac{2dx}{x}$, integrate to get $y = Cx^2$.
$y = 3x^2$. Separate to $\frac{dy}{y} = \frac{2dx}{x}$, integrate to get $y = Cx^2$.
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Solve for $y$ given $\frac{dy}{dx} = x$ with initial condition $y(1) = 3$.
Solve for $y$ given $\frac{dy}{dx} = x$ with initial condition $y(1) = 3$.
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$y = \frac{x^2}{2} + \frac{5}{2}$. Integrate $\int x , dx = \frac{x^2}{2} + C$ and apply condition.
$y = \frac{x^2}{2} + \frac{5}{2}$. Integrate $\int x , dx = \frac{x^2}{2} + C$ and apply condition.
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What is the purpose of using initial conditions in solving differential equations?
What is the purpose of using initial conditions in solving differential equations?
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To find particular solutions. They eliminate the arbitrary constant $C$ from general solutions.
To find particular solutions. They eliminate the arbitrary constant $C$ from general solutions.
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Given $\frac{dy}{dx} = 2y$, find the particular solution with $y(0) = 1$.
Given $\frac{dy}{dx} = 2y$, find the particular solution with $y(0) = 1$.
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$y = e^{2x}$. From $y = Ce^{2x}$ with initial condition $y(0) = 1$.
$y = e^{2x}$. From $y = Ce^{2x}$ with initial condition $y(0) = 1$.
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What is the general solution to $\frac{dy}{dx} = ky$ using separation of variables?
What is the general solution to $\frac{dy}{dx} = ky$ using separation of variables?
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$y = Ce^{kx}$. Standard form for exponential growth/decay differential equations.
$y = Ce^{kx}$. Standard form for exponential growth/decay differential equations.
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How do you integrate $y' = \frac{dy}{dx} = x^2 y$ using separation of variables?
How do you integrate $y' = \frac{dy}{dx} = x^2 y$ using separation of variables?
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$\frac{dy}{y} = x^2 dx$. Divide both sides by $y$ and multiply by $dx$.
$\frac{dy}{y} = x^2 dx$. Divide both sides by $y$ and multiply by $dx$.
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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. This allows variables to be on different sides for integration.
Separate the variables. This allows variables to be on different sides for integration.
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What is the solution to $\frac{dy}{dx} = 4y$ using separation of variables?
What is the solution to $\frac{dy}{dx} = 4y$ using separation of variables?
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$y = Ce^{4x}$. Standard exponential solution with growth rate $k = 4$.
$y = Ce^{4x}$. Standard exponential solution with growth rate $k = 4$.
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