AP Calculus BC 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables Study Notes - New Syllabus
AP Calculus BC 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables Study Notes- New syllabus
AP Calculus BC 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Solving differential equations allows us to determine functions and develop models.
Key Concepts:
- Finding Particular Solutions, Using Initial Conditions, and Separation of Variables
Finding Particular Solutions, Using Initial Conditions, and Separation of Variables
Finding Particular Solutions, Using Initial Conditions, and Separation of Variables
A general solution to a differential equation represents a family of functions that satisfy the equation. These solutions typically include an arbitrary constant \( C \), meaning infinitely many curves fit the general form. A particular solution is a specific member of this family that passes through a given point, determined by applying initial conditions.
- A general solution may describe infinitely many solutions to a differential equation. There is only one particular solution passing through a given point.
- The function \( F(x) = y_0 + \displaystyle \int_{a}^{x} f(t)\,dt \) is a particular solution to the differential equation \(\dfrac{dy}{dx} = f(x)\), satisfying \( F(a) = y_0 \).
- Solutions to differential equations may be subject to domain restrictions due to the nature of the function or its derivation (for example, restrictions from square roots or division by zero).
Procedure for Finding Particular Solutions Using Initial Conditions:
- Solve the differential equation to find the general solution, which will include the constant \( C \).
- Substitute the given initial condition (e.g., \( y(x_0) = y_0 \)) into the general solution.
- Solve for \( C \) to determine the unique particular solution.
Separation of Variables Method:
This is a common technique for solving first-order differential equations that can be written in the form:
\(\dfrac{dy}{dx} = g(x)h(y)\)
Steps:
- Rewrite the equation as \(\dfrac{1}{h(y)}\,dy = g(x)\,dx\).
- Integrate both sides with respect to their respective variables.
- Simplify and include the constant \( C \) in the solution.
- Apply the initial condition to find the particular solution.
Example:
The differential equation \(\dfrac{dy}{dx} = 3x^2\) has the initial condition \(y(1) = 4\). Find the particular solution.
▶️Answer/Explanation
We first integrate: \( y = \displaystyle \int 3x^2\,dx = x^3 + C \)
Using the initial condition \(y(1) = 4\): \( 4 = (1)^3 + C \quad \Rightarrow \quad C = 3 \)
So the particular solution is: \( y = x^3 + 3 \)
Example:
Solve \(\dfrac{dy}{dx} = xy\) given that \(y(0) = 5\).
▶️Answer/Explanation
We write: \( \frac{1}{y} \, dy = x \, dx \)
Integrating both sides: \( \ln|y| = \frac{x^2}{2} + C \)
Exponentiating: \( y = Ae^{x^2/2} \)
Applying \(y(0) = 5\): \( 5 = Ae^{0} \quad \Rightarrow \quad A = 5 \) So: \( y = 5e^{x^2/2} \)
Example:
Solve \(\dfrac{dy}{dx} = \dfrac{1}{x-2}\) given that \(y(3) = 4\), and state the domain of the solution.
▶️Answer/Explanation
Integrating: \( y = \displaystyle \int \frac{1}{x-2} \, dx = \ln|x-2| + C \)
Applying \(y(3) = 4\): \( 4 = \ln|3-2| + C \quad \Rightarrow \quad 4 = \ln(1) + C \quad \Rightarrow \quad C = 4 \)
So the particular solution is: \( y = \ln|x-2| + 4 \)
Domain restriction: Since \(\ln|x-2|\) is defined only for \(x \neq 2\), the solution is valid either for \(x > 2\) or \(x < 2\), but not across \(x = 2\).
Given \(x = 3\) in the initial condition, the domain is \(x > 2\).
Example: Solve \(\dfrac{dy}{dx} = 2x\), given that \(y(0) = 3\), using the integral form.
▶️Answer/Explanation
We use: \( F(x) = y_0 + \displaystyle \int_a^x f(t)\,dt \)
Here, \(a = 0\), \(y_0 = 3\), and \(f(t) = 2t\).
\( F(x) = 3 + \displaystyle \int_0^x 2t \, dt \) \( F(x) = 3 + [t^2]_0^x = 3 + x^2 \)
Thus, the particular solution is: \( y = x^2 + 3 \)
Example:
Solve \(\dfrac{dy}{dx} = \dfrac{y}{x}\) given that \(y(1) = 2\), and state the domain.
▶️Answer/Explanation
Separate variables: \( \frac{dy}{y} = \frac{dx}{x} \)
Integrate: \( \ln|y| = \ln|x| + C \)
Exponentiate: \( |y| = k|x| \quad \text{where } k = e^C \)
Using \(y(1) = 2\): \( 2 = k(1) \implies k = 2 \) So \(y = 2x\).
Since \(\ln|x|\) was used, \(x > 0\).
Domain: \(x > 0\).