Verifying Solutions for Differential Equations - AP Calculus AB
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Verify if $y = e^x$ satisfies $y' = y$.
Verify if $y = e^x$ satisfies $y' = y$.
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Yes, $y' = e^x = y$. Derivative of $e^x$ equals itself.
Yes, $y' = e^x = y$. Derivative of $e^x$ equals itself.
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What is the general solution for $y' = 2y$?
What is the general solution for $y' = 2y$?
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$y = Ce^{2x}$. Exponential growth with rate constant 2.
$y = Ce^{2x}$. Exponential growth with rate constant 2.
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What is the first step in verifying a solution?
What is the first step in verifying a solution?
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Substitute the function into the differential equation. Direct substitution method for verification.
Substitute the function into the differential equation. Direct substitution method for verification.
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What is a particular solution?
What is a particular solution?
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A solution with specified initial conditions. General solution evaluated at given conditions.
A solution with specified initial conditions. General solution evaluated at given conditions.
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What is a solution to a differential equation?
What is a solution to a differential equation?
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A function satisfying the differential equation. When substituted, makes the equation true.
A function satisfying the differential equation. When substituted, makes the equation true.
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Identify the solution form for $y'' + y = 0$.
Identify the solution form for $y'' + y = 0$.
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General solution: $y = C_1 \text{cos}(x) + C_2 \text{sin}(x)$. Characteristic equation $r^2 + 1 = 0$ gives $r = \pm i$.
General solution: $y = C_1 \text{cos}(x) + C_2 \text{sin}(x)$. Characteristic equation $r^2 + 1 = 0$ gives $r = \pm i$.
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State the general solution for $y' = k$.
State the general solution for $y' = k$.
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$y = kt + C$. Integrate the constant $k$ with respect to time.
$y = kt + C$. Integrate the constant $k$ with respect to time.
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What is the general solution for $y'' = 0$?
What is the general solution for $y'' = 0$?
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$y = C_1x + C_2$. Double integration of zero gives linear function.
$y = C_1x + C_2$. Double integration of zero gives linear function.
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Is $y = x^2 + C$ a solution to $y'' = 2$?
Is $y = x^2 + C$ a solution to $y'' = 2$?
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Yes, $y'' = 2$. Second derivative of $x^2 + C$ is constant 2.
Yes, $y'' = 2$. Second derivative of $x^2 + C$ is constant 2.
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Verify $y = x^3$ for $y'' = 6x$.
Verify $y = x^3$ for $y'' = 6x$.
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Yes, $y'' = 6x$. Second derivative of $x^3$ is $6x$.
Yes, $y'' = 6x$. Second derivative of $x^3$ is $6x$.
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Verify if $y = e^{3x}$ solves $y' = 3y$.
Verify if $y = e^{3x}$ solves $y' = 3y$.
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Yes, $y' = 3e^{3x}$. Derivative matches 3 times the function.
Yes, $y' = 3e^{3x}$. Derivative matches 3 times the function.
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Which function is a solution to $y' = 3x^2$?
Which function is a solution to $y' = 3x^2$?
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Any antiderivative of $3x^2$, e.g., $y = x^3 + C$. Integrate $3x^2$ to get $x^3 + C$.
Any antiderivative of $3x^2$, e.g., $y = x^3 + C$. Integrate $3x^2$ to get $x^3 + C$.
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Determine if $y = \frac{1}{x}$ solves $xy' = -y^2$.
Determine if $y = \frac{1}{x}$ solves $xy' = -y^2$.
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Yes, $y' = -\frac{1}{x^2}$. Check: $x(-\frac{1}{x^2}) = -\frac{1}{x} = -(\frac{1}{x})^2$.
Yes, $y' = -\frac{1}{x^2}$. Check: $x(-\frac{1}{x^2}) = -\frac{1}{x} = -(\frac{1}{x})^2$.
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Identify the order of the differential equation: $y'' + 3y' - 5y = 0$.
Identify the order of the differential equation: $y'' + 3y' - 5y = 0$.
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Order 2. Highest derivative is the second, so order is 2.
Order 2. Highest derivative is the second, so order is 2.
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What is a linear differential equation?
What is a linear differential equation?
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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 $y$.
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 $y$.
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Verify if $y = \frac{1}{2}e^{2x}$ solves $y' = y$.
Verify if $y = \frac{1}{2}e^{2x}$ solves $y' = y$.
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No, $y' = e^{2x} \neq y$. $y' = e^{2x} \neq \frac{1}{2}e^{2x} = y$.
No, $y' = e^{2x} \neq y$. $y' = e^{2x} \neq \frac{1}{2}e^{2x} = y$.
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Verify $y = 2x - 1$ for $y' = 2$.
Verify $y = 2x - 1$ for $y' = 2$.
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Yes, $y' = 2$. Derivative of linear function $2x - 1$ is 2.
Yes, $y' = 2$. Derivative of linear function $2x - 1$ is 2.
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Determine if $y = \text{sin}(x)$ solves $y'' + y = 0$.
Determine if $y = \text{sin}(x)$ solves $y'' + y = 0$.
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Yes, $y'' + y = 0$. $y'' = -\sin x$, so $y'' + y = 0$.
Yes, $y'' + y = 0$. $y'' = -\sin x$, so $y'' + y = 0$.
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What is a separable differential equation?
What is a separable differential equation?
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An equation where variables can be separated on opposite sides. Variables can be moved to opposite sides.
An equation where variables can be separated on opposite sides. Variables can be moved to opposite sides.
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Verify if $y = Cx^{-1}$ is a solution for $xy' + y = 0$.
Verify if $y = Cx^{-1}$ is a solution for $xy' + y = 0$.
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Yes, it satisfies $xy' + y = 0$. $y' = -Cx^{-2}$, and $xy' + y = 0$ checks out.
Yes, it satisfies $xy' + y = 0$. $y' = -Cx^{-2}$, and $xy' + y = 0$ checks out.
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What is the next step after substitution in solution verification?
What is the next step after substitution in solution verification?
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Check if the equation holds true. Verify both sides are equal after substitution.
Check if the equation holds true. Verify both sides are equal after substitution.
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Determine if $y = \text{cos}(x)$ satisfies $y'' + y = 0$.
Determine if $y = \text{cos}(x)$ satisfies $y'' + y = 0$.
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Yes, $y'' + y = 0$. $y'' = -\cos x$, so $y'' + y = 0$.
Yes, $y'' + y = 0$. $y'' = -\cos x$, so $y'' + y = 0$.
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What is the general solution for $y'' - 4y = 0$?
What is the general solution for $y'' - 4y = 0$?
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$y = C_1 e^{2x} + C_2 e^{-2x}$. Characteristic equation $r^2 - 4 = 0$ gives $r = \pm 2$.
$y = C_1 e^{2x} + C_2 e^{-2x}$. Characteristic equation $r^2 - 4 = 0$ gives $r = \pm 2$.
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Verify if $y = C \text{cos}(2x)$ solves $y'' + 4y = 0$.
Verify if $y = C \text{cos}(2x)$ solves $y'' + 4y = 0$.
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Yes, it satisfies $y'' + 4y = 0$. $y'' = -4C\cos(2x)$, so $y'' + 4y = 0$.
Yes, it satisfies $y'' + 4y = 0$. $y'' = -4C\cos(2x)$, so $y'' + 4y = 0$.
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What is the solution form for $y'' + 9y = 0$?
What is the solution form for $y'' + 9y = 0$?
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General solution: $y = C_1 \text{cos}(3x) + C_2 \text{sin}(3x)$. Characteristic equation $r^2 + 9 = 0$ gives $r = \pm 3i$.
General solution: $y = C_1 \text{cos}(3x) + C_2 \text{sin}(3x)$. Characteristic equation $r^2 + 9 = 0$ gives $r = \pm 3i$.
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Is $y = Ce^{3x}$ a solution for $y' = 3y$?
Is $y = Ce^{3x}$ a solution for $y' = 3y$?
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Yes, $y' = 3y$. Derivative of $Ce^{3x}$ is $3Ce^{3x} = 3y$.
Yes, $y' = 3y$. Derivative of $Ce^{3x}$ is $3Ce^{3x} = 3y$.
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Verify if $y = x^2$ satisfies $y'' = 2$.
Verify if $y = x^2$ satisfies $y'' = 2$.
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Yes, $y'' = 2$. Second derivative of $x^2$ is constant 2.
Yes, $y'' = 2$. Second derivative of $x^2$ is constant 2.
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Verify if $y = Ce^{-2x}$ solves $y' + 2y = 0$.
Verify if $y = Ce^{-2x}$ solves $y' + 2y = 0$.
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Yes, it satisfies $y' + 2y = 0$. $y' = -2Ce^{-2x} = -2y$, so $y' + 2y = 0$.
Yes, it satisfies $y' + 2y = 0$. $y' = -2Ce^{-2x} = -2y$, so $y' + 2y = 0$.
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Determine if $y = e^{-x}$ solves $y' = -y$.
Determine if $y = e^{-x}$ solves $y' = -y$.
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Yes, $y' = -e^{-x}$. Derivative equals negative of the function.
Yes, $y' = -e^{-x}$. Derivative equals negative of the function.
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What is the general solution for $y' = -y$?
What is the general solution for $y' = -y$?
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$y = Ce^{-x}$. Exponential decay with rate constant 1.
$y = Ce^{-x}$. Exponential decay with rate constant 1.
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