AP Calculus BC 7.6 Finding General Solutions Using Separation of Variables Study Notes - New Syllabus
AP Calculus BC 7.6 Finding General Solutions Using Separation of Variables Study Notes- New syllabus
AP Calculus BC 7.6 Finding General Solutions Using 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 General Solutions Using Separation of Variables
Finding General Solutions Using Separation of Variables
Finding General Solutions Using Separation of Variables
Separation of variables is a method for solving first-order differential equations where the variables can be separated so that all terms involving \(y\) are on one side of the equation and all terms involving \(x\) are on the other side. This allows us to integrate each side with respect to its own variable.
In the process of separation of variables, antidifferentiation is used to find the original function \(y\) when its derivative \(\dfrac{dy}{dx}\) is known. Antidifferentiation involves integrating both sides of the equation to obtain the general solution. Since indefinite integration introduces an arbitrary constant \(C\), the general solution represents a family of functions that satisfy the given differential equation.
General Form:
\(\dfrac{dy}{dx} = f(x) \, g(y)\)
Step-by-step:
- Rewrite the equation to separate variables:
\(\dfrac{1}{g(y)} \, dy = f(x) \, dx\) - Integrate both sides:
\(\displaystyle \int \dfrac{1}{g(y)} \, dy = \displaystyle \int f(x) \, dx\) - Find the general solution by including the constant of integration \(C\):
\(F(y) = G(x) + C\) - If an initial condition is given, substitute to find the particular solution.
Example of antidifferentiation in this context:
If we have:
\(\dfrac{dy}{dx} = f(x)\)
Then by integrating both sides:
\(y = \displaystyle \int f(x) \, dx + C\)
The constant \(C\) is determined if a point \((x_0, y_0)\) is provided.
Example:
Solve \(\dfrac{dy}{dx} = x \cdot y\)
▶️Answer/Explanation
Separate variables:
\(\dfrac{dy}{dx} = x \cdot y \quad \Rightarrow \quad \dfrac{1}{y} \, dy = x \, dx\)
Integrate both sides:
\(\displaystyle \int \dfrac{1}{y} \, dy = \displaystyle \int x \, dx\)
\(\ln|y| = \dfrac{x^2}{2} + C\)
Exponentiate to solve for \(y\):
\(y = Ae^{\frac{x^2}{2}}\), where \(A = e^C\)
Final General Solution: \(\boxed{y = Ae^{\frac{x^2}{2}}}\)
Example :
Solve \(\dfrac{dy}{dx} = (1+y^2) \cdot \cos(x)\)
▶️Answer/Explanation
Separate variables:
\(\dfrac{dy}{1+y^2} = \cos(x) \, dx\)
Integrate both sides:
\(\displaystyle \int \dfrac{dy}{1+y^2} = \displaystyle \int \cos(x) \, dx\)
\(\arctan(y) = \sin(x) + C\)
Solve for \(y\):
\(y = \tan(\sin(x) + C)\)
Final General Solution: \(\boxed{y = \tan(\sin(x) + C)}\)
Example:
Solve \(\dfrac{dy}{dx} = 2y\), given \(y(0) = 3\)
▶️Answer/Explanation
Separate variables:
\(\dfrac{1}{y} \, dy = 2 \, dx\)
Integrate both sides:
\(\displaystyle \int \dfrac{1}{y} \, dy = \displaystyle \int 2 \, dx\)
\(\ln|y| = 2x + C\)
Exponentiate:
\(y = Ae^{2x}\)
Apply initial condition \(y(0) = 3\):
\(3 = Ae^0 \quad \Rightarrow \quad A = 3\)
Final Particular Solution: \(\boxed{y = 3e^{2x}}\)