Verifying Solutions for Differential Equations - AP Calculus BC
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What is the integrating factor for $\frac{dy}{dx} + Py = Q$?
What is the integrating factor for $\frac{dy}{dx} + Py = Q$?
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$e^{\textstyle\int P , dx}$. Factor to make coefficient of $\frac{dy}{dx}$ equal to $1$.
$e^{\textstyle\int P , dx}$. Factor to make coefficient of $\frac{dy}{dx}$ equal to $1$.
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Is $y = 2x^2 + 3$ a solution to $y' = 4x$?
Is $y = 2x^2 + 3$ a solution to $y' = 4x$?
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Yes, it is a solution. Derivative: $\frac{dy}{dx} = 4x$ matches the equation.
Yes, it is a solution. Derivative: $\frac{dy}{dx} = 4x$ matches the equation.
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Identify the particular solution for $y' = 2y$ given $y(0) = 5$.
Identify the particular solution for $y' = 2y$ given $y(0) = 5$.
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$y = 5e^{2x}$. Using initial condition $y(0) = 5$ with $y = Ce^{2x}$.
$y = 5e^{2x}$. Using initial condition $y(0) = 5$ with $y = Ce^{2x}$.
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What is the general solution for $y'' + y = 0$?
What is the general solution for $y'' + y = 0$?
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$y = C_1 \text{cos}(x) + C_2 \text{sin}(x)$. Solution using characteristic equation $r^2 + 1 = 0$.
$y = C_1 \text{cos}(x) + C_2 \text{sin}(x)$. Solution using characteristic equation $r^2 + 1 = 0$.
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What is the general form of a first-order differential equation?
What is the general form of a first-order differential equation?
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$\frac{dy}{dx} = f(x, y)$. Standard form where $f$ depends on both $x$ and $y$.
$\frac{dy}{dx} = f(x, y)$. Standard form where $f$ depends on both $x$ and $y$.
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What type of differential equation is $y' + 3y = 6$?
What type of differential equation is $y' + 3y = 6$?
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First-order linear. Has form $y' + Py = Q$ with constant coefficient.
First-order linear. Has form $y' + Py = Q$ with constant coefficient.
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What is the characteristic equation for $y'' - 4y = 0$?
What is the characteristic equation for $y'' - 4y = 0$?
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$r^2 - 4 = 0$. Replace $y''$ with $r^2$ and set equal to zero.
$r^2 - 4 = 0$. Replace $y''$ with $r^2$ and set equal to zero.
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Verify if $y = 1 + \text{ln}(x)$ satisfies $xy' = 1$.
Verify if $y = 1 + \text{ln}(x)$ satisfies $xy' = 1$.
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Yes, it satisfies the equation. $y' = \frac{1}{x}$, so $x \cdot \frac{1}{x} = 1$ ✓
Yes, it satisfies the equation. $y' = \frac{1}{x}$, so $x \cdot \frac{1}{x} = 1$ ✓
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Verify if $y = \frac{1}{1-x}$ is a solution to $y' = y^2$.
Verify if $y = \frac{1}{1-x}$ is a solution to $y' = y^2$.
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Yes, it is a solution. $y' = \frac{1}{(1-x)^2}$ and $y^2 = \frac{1}{(1-x)^2}$ ✓
Yes, it is a solution. $y' = \frac{1}{(1-x)^2}$ and $y^2 = \frac{1}{(1-x)^2}$ ✓
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Verify if $y = \frac{1}{x}$ is a solution to $xy' + 2y = 0$.
Verify if $y = \frac{1}{x}$ is a solution to $xy' + 2y = 0$.
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No, it is not a solution. $y' = -\frac{1}{x^2}$, so $x(-\frac{1}{x^2}) + 2(\frac{1}{x}) = \frac{1}{x} \neq 0$
No, it is not a solution. $y' = -\frac{1}{x^2}$, so $x(-\frac{1}{x^2}) + 2(\frac{1}{x}) = \frac{1}{x} \neq 0$
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What is the form of a second-order linear homogeneous differential equation?
What is the form of a second-order linear homogeneous differential equation?
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$a y'' + b y' + c y = 0$. Standard form with constant coefficients and zero right side.
$a y'' + b y' + c y = 0$. Standard form with constant coefficients and zero right side.
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What is the form of a non-homogeneous differential equation?
What is the form of a non-homogeneous differential equation?
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$a y'' + b y' + c y = g(x)$. Same as homogeneous but with non-zero right side.
$a y'' + b y' + c y = g(x)$. Same as homogeneous but with non-zero right side.
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Determine if $y = \frac{e^x}{x}$ is a solution to $x^2y' - xy = e^x$.
Determine if $y = \frac{e^x}{x}$ is a solution to $x^2y' - xy = e^x$.
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Yes, it is a solution. Computing $y'$ using quotient rule verifies the equation.
Yes, it is a solution. Computing $y'$ using quotient rule verifies the equation.
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What is the general solution to $\frac{dy}{dx} = y$?
What is the general solution to $\frac{dy}{dx} = y$?
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$y = Ce^x$, where $C$ is a constant. Solution to the simplest exponential differential equation.
$y = Ce^x$, where $C$ is a constant. Solution to the simplest exponential differential equation.
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What is the characteristic equation for $y'' + 3y' + 2y = 0$?
What is the characteristic equation for $y'' + 3y' + 2y = 0$?
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$r^2 + 3r + 2 = 0$. Replace $y''$ with $r^2$, $y'$ with $r$, and $y$ with $1$.
$r^2 + 3r + 2 = 0$. Replace $y''$ with $r^2$, $y'$ with $r$, and $y$ with $1$.
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What is the general solution to $y'' - 5y' + 6y = 0$?
What is the general solution to $y'' - 5y' + 6y = 0$?
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$y = C_1 e^{2x} + C_2 e^{3x}$. Using characteristic roots $r = 2, 3$ from factoring.
$y = C_1 e^{2x} + C_2 e^{3x}$. Using characteristic roots $r = 2, 3$ from factoring.
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Determine if $y = x^3$ is a solution for $y'' = 6x$.
Determine if $y = x^3$ is a solution for $y'' = 6x$.
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Yes, it is a solution. $y' = 3x^2$, $y'' = 6x$, so equation holds.
Yes, it is a solution. $y' = 3x^2$, $y'' = 6x$, so equation holds.
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What is the order of $y'' + 2xy' + y = 0$?
What is the order of $y'' + 2xy' + y = 0$?
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Second order. Highest derivative is the second derivative.
Second order. Highest derivative is the second derivative.
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Determine if $y = x + 1$ is a solution for $y' = 1$.
Determine if $y = x + 1$ is a solution for $y' = 1$.
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Yes, it is a solution. Derivative of $x + 1$ is constant $1$.
Yes, it is a solution. Derivative of $x + 1$ is constant $1$.
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Verify if $y = x^3 + x$ is a solution to $y' = 3x^2 + 1$.
Verify if $y = x^3 + x$ is a solution to $y' = 3x^2 + 1$.
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Yes, it is a solution. Derivative matches: $\frac{d}{dx}(x^3 + x) = 3x^2 + 1$.
Yes, it is a solution. Derivative matches: $\frac{d}{dx}(x^3 + x) = 3x^2 + 1$.
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Verify if $y = \text{sin}(x)$ satisfies $y'' + y = 0$.
Verify if $y = \text{sin}(x)$ satisfies $y'' + y = 0$.
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Yes, it satisfies the equation. $y' = \cos(x)$, $y'' = -\sin(x)$, so $y'' + y = 0$ ✓
Yes, it satisfies the equation. $y' = \cos(x)$, $y'' = -\sin(x)$, so $y'' + y = 0$ ✓
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What is the order of the differential equation $y''' + 2y' = 0$?
What is the order of the differential equation $y''' + 2y' = 0$?
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Third order. Highest derivative is the third derivative.
Third order. Highest derivative is the third derivative.
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Does $y = 3x + 2$ solve $y' = 3$?
Does $y = 3x + 2$ solve $y' = 3$?
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Yes, it solves the equation. Derivative of $3x + 2$ is constant $3$.
Yes, it solves the equation. Derivative of $3x + 2$ is constant $3$.
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Verify if $y = e^{-x}$ is a solution to $y' = -y$.
Verify if $y = e^{-x}$ is a solution to $y' = -y$.
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Yes, it is a solution. Derivative: $\frac{d}{dx}(e^{-x}) = -e^{-x} = -y$ ✓
Yes, it is a solution. Derivative: $\frac{d}{dx}(e^{-x}) = -e^{-x} = -y$ ✓
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What is the characteristic equation for $y'' + 9y = 0$?
What is the characteristic equation for $y'' + 9y = 0$?
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$r^2 + 9 = 0$. Replace $y''$ with $r^2$ and $y$ with $1$.
$r^2 + 9 = 0$. Replace $y''$ with $r^2$ and $y$ with $1$.
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What is the order of $y^{(4)} + 2y'' + y = 0$?
What is the order of $y^{(4)} + 2y'' + y = 0$?
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Fourth order. Highest derivative is the fourth derivative.
Fourth order. Highest derivative is the fourth derivative.
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Verify if $y = \frac{x^2}{2}$ satisfies $y'' = 1$.
Verify if $y = \frac{x^2}{2}$ satisfies $y'' = 1$.
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Yes, it satisfies the equation. $y' = x$, $y'' = 1$, so equation is satisfied.
Yes, it satisfies the equation. $y' = x$, $y'' = 1$, so equation is satisfied.
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What is the general solution of $\frac{dy}{dx} = ky$?
What is the general solution of $\frac{dy}{dx} = ky$?
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$y = Ce^{kx}$, where $C$ is a constant. Solution to exponential growth/decay differential equation.
$y = Ce^{kx}$, where $C$ is a constant. Solution to exponential growth/decay differential equation.
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Identify the particular solution for $y' = 5y$ given $y(0) = 2$.
Identify the particular solution for $y' = 5y$ given $y(0) = 2$.
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$y = 2e^{5x}$. Using initial condition $y(0) = 2$ with $y = Ce^{5x}$.
$y = 2e^{5x}$. Using initial condition $y(0) = 2$ with $y = Ce^{5x}$.
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Verify if $y = e^{2x}$ is a solution to $\frac{dy}{dx} = 2y$.
Verify if $y = e^{2x}$ is a solution to $\frac{dy}{dx} = 2y$.
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Yes, it is a solution. Taking derivative: $\frac{dy}{dx} = 2e^{2x} = 2y$ ✓
Yes, it is a solution. Taking derivative: $\frac{dy}{dx} = 2e^{2x} = 2y$ ✓
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