Finding General Solutions: Separation of Variables - AP Calculus BC
Card 1 of 30
Determine the type of equation: $\frac{dy}{dx} = x y + 1$.
Determine the type of equation: $\frac{dy}{dx} = x y + 1$.
Tap to reveal answer
Not separable. The constant term prevents separation of variables.
Not separable. The constant term prevents separation of variables.
← Didn't Know|Knew It →
Identify the integral to solve after separating variables: $\frac{dy}{dx} = x \frac{1}{y}$
Identify the integral to solve after separating variables: $\frac{dy}{dx} = x \frac{1}{y}$
Tap to reveal answer
$\text{Integrate: } \frac{1}{y} dy = x dx$. Separation gives $\frac{dy}{y} = x dx$, then integrate both sides.
$\text{Integrate: } \frac{1}{y} dy = x dx$. Separation gives $\frac{dy}{y} = x dx$, then integrate both sides.
← Didn't Know|Knew It →
State the integration result for $\frac{1}{y} dy = x dx$.
State the integration result for $\frac{1}{y} dy = x dx$.
Tap to reveal answer
$\ln|y| = \frac{x^2}{2} + C$. Antiderivatives are $\ln|y|$ and $\frac{x^2}{2}$ respectively.
$\ln|y| = \frac{x^2}{2} + C$. Antiderivatives are $\ln|y|$ and $\frac{x^2}{2}$ respectively.
← Didn't Know|Knew It →
What is the general solution for $\frac{dy}{dx} = 3xy$?
What is the general solution for $\frac{dy}{dx} = 3xy$?
Tap to reveal answer
$y = Ce^{\frac{3x^2}{2}}$. Separating gives $\frac{dy}{y} = 3x dx$, integrating yields $\ln|y| = \frac{3x^2}{2} + C$.
$y = Ce^{\frac{3x^2}{2}}$. Separating gives $\frac{dy}{y} = 3x dx$, integrating yields $\ln|y| = \frac{3x^2}{2} + C$.
← Didn't Know|Knew It →
Identify the next step: $\text{ln}|y| = \text{ln}|x| + C$.
Identify the next step: $\text{ln}|y| = \text{ln}|x| + C$.
Tap to reveal answer
Express $y$ in terms of $x$: $y = Cx$. Use properties of logarithms: $\ln|y| - \ln|x| = \ln|\frac{y}{x}| = C$.
Express $y$ in terms of $x$: $y = Cx$. Use properties of logarithms: $\ln|y| - \ln|x| = \ln|\frac{y}{x}| = C$.
← Didn't Know|Knew It →
Identify the integration: $y dy = x dx$.
Identify the integration: $y dy = x dx$.
Tap to reveal answer
$\frac{y^2}{2} = \frac{x^2}{2} + C$. Both sides integrate using the power rule.
$\frac{y^2}{2} = \frac{x^2}{2} + C$. Both sides integrate using the power rule.
← Didn't Know|Knew It →
Find the general solution: $\frac{dy}{dx} = yx$.
Find the general solution: $\frac{dy}{dx} = yx$.
Tap to reveal answer
$y = Ce^{\frac{x^2}{2}}$. Same as previous: separating $\frac{dy}{y} = x dx$ gives this result.
$y = Ce^{\frac{x^2}{2}}$. Same as previous: separating $\frac{dy}{y} = x dx$ gives this result.
← Didn't Know|Knew It →
Which method helps find a particular solution after finding a general one?
Which method helps find a particular solution after finding a general one?
Tap to reveal answer
Use initial conditions. Initial conditions determine the specific value of the constant $C$.
Use initial conditions. Initial conditions determine the specific value of the constant $C$.
← Didn't Know|Knew It →
What must be true about $C$ when using an initial condition $y(x_0) = y_0$?
What must be true about $C$ when using an initial condition $y(x_0) = y_0$?
Tap to reveal answer
$C$ is determined by substituting $x_0$ and $y_0$ into the solution. Substitute the initial condition to solve for the constant.
$C$ is determined by substituting $x_0$ and $y_0$ into the solution. Substitute the initial condition to solve for the constant.
← Didn't Know|Knew It →
Find the general solution: $\frac{dy}{dx} = y^2$.
Find the general solution: $\frac{dy}{dx} = y^2$.
Tap to reveal answer
$y = -\frac{1}{C-x}$. Separating gives $\frac{dy}{y^2} = dx$, integrating yields $-\frac{1}{y} = x + C$.
$y = -\frac{1}{C-x}$. Separating gives $\frac{dy}{y^2} = dx$, integrating yields $-\frac{1}{y} = x + C$.
← Didn't Know|Knew It →
What is the role of the constant $C$ in the solution of a differential equation?
What is the role of the constant $C$ in the solution of a differential equation?
Tap to reveal answer
$C$ represents an arbitrary constant, allowing for all solutions. The constant encompasses all possible initial conditions.
$C$ represents an arbitrary constant, allowing for all solutions. The constant encompasses all possible initial conditions.
← Didn't Know|Knew It →
What is the result of integrating $x dx = y dy$?
What is the result of integrating $x dx = y dy$?
Tap to reveal answer
$\frac{x^2}{2} = \frac{y^2}{2} + C$. Both sides integrate to $\frac{x^2}{2}$ and $\frac{y^2}{2}$ respectively.
$\frac{x^2}{2} = \frac{y^2}{2} + C$. Both sides integrate to $\frac{x^2}{2}$ and $\frac{y^2}{2}$ respectively.
← Didn't Know|Knew It →
What technique is used to verify the solution of a separable differential equation?
What technique is used to verify the solution of a separable differential equation?
Tap to reveal answer
Differentiate the solution and compare with the original equation. Substitution back into the original equation confirms correctness.
Differentiate the solution and compare with the original equation. Substitution back into the original equation confirms correctness.
← Didn't Know|Knew It →
What must be checked after finding a general solution?
What must be checked after finding a general solution?
Tap to reveal answer
Verify the solution by differentiating and substituting. Solutions must satisfy the original differential equation.
Verify the solution by differentiating and substituting. Solutions must satisfy the original differential equation.
← Didn't Know|Knew It →
State the form of a separable differential equation.
State the form of a separable differential equation.
Tap to reveal answer
$\frac{dy}{dx} = g(x)h(y)$. This form allows variables to be separated into $\frac{dy}{h(y)} = g(x)dx$.
$\frac{dy}{dx} = g(x)h(y)$. This form allows variables to be separated into $\frac{dy}{h(y)} = g(x)dx$.
← Didn't Know|Knew It →
Find the general solution: $\frac{dy}{dx} = y\cos(x)$.
Find the general solution: $\frac{dy}{dx} = y\cos(x)$.
Tap to reveal answer
$y = Ce^{\sin(x)}$. Separating gives $\frac{dy}{y} = \cos(x) dx$, integrating yields $\ln|y| = \sin(x) + C$.
$y = Ce^{\sin(x)}$. Separating gives $\frac{dy}{y} = \cos(x) dx$, integrating yields $\ln|y| = \sin(x) + C$.
← Didn't Know|Knew It →
Identify the step after integration: $\text{ln}|y| = x^2 + C$.
Identify the step after integration: $\text{ln}|y| = x^2 + C$.
Tap to reveal answer
Solve for $y$: $y = e^{x^2 + C}$. Exponentiate both sides to isolate $y$ from the logarithm.
Solve for $y$: $y = e^{x^2 + C}$. Exponentiate both sides to isolate $y$ from the logarithm.
← Didn't Know|Knew It →
Find the general solution: $\frac{dy}{dx} = xy$.
Find the general solution: $\frac{dy}{dx} = xy$.
Tap to reveal answer
$y = Ce^{\frac{x^2}{2}}$. Separating gives $\frac{dy}{y} = x dx$, integrating yields $\ln|y| = \frac{x^2}{2} + C$.
$y = Ce^{\frac{x^2}{2}}$. Separating gives $\frac{dy}{y} = x dx$, integrating yields $\ln|y| = \frac{x^2}{2} + C$.
← Didn't Know|Knew It →
Why is verifying a solution important?
Why is verifying a solution important?
Tap to reveal answer
To ensure it satisfies the original equation. Verification confirms the solution is mathematically correct.
To ensure it satisfies the original equation. Verification confirms the solution is mathematically correct.
← Didn't Know|Knew It →
What is the next step after separating variables in a differential equation?
What is the next step after separating variables in a differential equation?
Tap to reveal answer
Integrate both sides with respect to their variables. Integration produces antiderivatives on each side of the equation.
Integrate both sides with respect to their variables. Integration produces antiderivatives on each side of the equation.
← Didn't Know|Knew It →
How is a particular solution found from a general solution?
How is a particular solution found from a general solution?
Tap to reveal answer
Use initial conditions to solve for $C$. Initial conditions determine the specific constant value.
Use initial conditions to solve for $C$. Initial conditions determine the specific constant value.
← Didn't Know|Knew It →
Identify the final form: $\frac{dy}{y} = x dx$.
Identify the final form: $\frac{dy}{y} = x dx$.
Tap to reveal answer
$\text{ln}|y| = \frac{x^2}{2} + C$. Integration of separated form $\frac{dy}{y} = x dx$.
$\text{ln}|y| = \frac{x^2}{2} + C$. Integration of separated form $\frac{dy}{y} = x dx$.
← Didn't Know|Knew It →
What is the general solution for $\frac{dy}{dx} = 2x$?
What is the general solution for $\frac{dy}{dx} = 2x$?
Tap to reveal answer
$y = x^2 + C$. Direct integration of $\frac{dy}{dx} = 2x$ gives $y = x^2 + C$.
$y = x^2 + C$. Direct integration of $\frac{dy}{dx} = 2x$ gives $y = x^2 + C$.
← Didn't Know|Knew It →
Find the general solution: $\frac{dy}{dx} = 3x^2 y$.
Find the general solution: $\frac{dy}{dx} = 3x^2 y$.
Tap to reveal answer
$y = Ce^{x^3}$. Separating gives $\frac{dy}{y} = 3x^2 dx$, integrating yields $\ln|y| = x^3 + C$.
$y = Ce^{x^3}$. Separating gives $\frac{dy}{y} = 3x^2 dx$, integrating yields $\ln|y| = x^3 + C$.
← Didn't Know|Knew It →
Identify the error: $\frac{dy}{dx} = x^2y^2$ rewritten as $y^2 dy = x^2 dx$.
Identify the error: $\frac{dy}{dx} = x^2y^2$ rewritten as $y^2 dy = x^2 dx$.
Tap to reveal answer
Correct: $\frac{1}{y^2} dy = x^2 dx$. The $y^2$ term should move to the left with reciprocal: $\frac{1}{y^2}$.
Correct: $\frac{1}{y^2} dy = x^2 dx$. The $y^2$ term should move to the left with reciprocal: $\frac{1}{y^2}$.
← Didn't Know|Knew It →
What is the general solution for $\frac{dy}{dx} = y$?
What is the general solution for $\frac{dy}{dx} = y$?
Tap to reveal answer
$y = Ce^x$. This is the standard exponential growth/decay solution.
$y = Ce^x$. This is the standard exponential growth/decay solution.
← Didn't Know|Knew It →
Find the general solution: $\frac{dy}{dx} = y\text{e}^x$.
Find the general solution: $\frac{dy}{dx} = y\text{e}^x$.
Tap to reveal answer
$y = Ce^{\text{e}^x}$. Separating gives $\frac{dy}{y} = e^x dx$, integrating yields $\ln|y| = e^x + C$.
$y = Ce^{\text{e}^x}$. Separating gives $\frac{dy}{y} = e^x dx$, integrating yields $\ln|y| = e^x + C$.
← Didn't Know|Knew It →
Identify the integration: $\frac{dy}{y} = x dx$.
Identify the integration: $\frac{dy}{y} = x dx$.
Tap to reveal answer
$\text{ln}|y| = \frac{x^2}{2} + C$. Standard result from separating $\frac{dy}{y} = x dx$.
$\text{ln}|y| = \frac{x^2}{2} + C$. Standard result from separating $\frac{dy}{y} = x dx$.
← Didn't Know|Knew It →
Determine the type of equation: $\frac{dy}{dx} = x^2 + y^2$.
Determine the type of equation: $\frac{dy}{dx} = x^2 + y^2$.
Tap to reveal answer
Not separable. Cannot separate $x$ and $y$ terms to opposite sides.
Not separable. Cannot separate $x$ and $y$ terms to opposite sides.
← Didn't Know|Knew It →
Find the general solution: $\frac{dy}{dx} = y^2 \text{sin}(x)$.
Find the general solution: $\frac{dy}{dx} = y^2 \text{sin}(x)$.
Tap to reveal answer
$y = \frac{1}{C - \text{cos}(x)}$. Separating gives $\frac{dy}{y^2} = \sin(x) dx$, integrating yields $-\frac{1}{y} = -\cos(x) + C$.
$y = \frac{1}{C - \text{cos}(x)}$. Separating gives $\frac{dy}{y^2} = \sin(x) dx$, integrating yields $-\frac{1}{y} = -\cos(x) + C$.
← Didn't Know|Knew It →