AP Calculus BC 7.5 Approximating Solutions Using Euler’s Method Study Notes - New Syllabus
AP Calculus BC 7.5 Approximating Solutions Using Euler’s Method Study Notes- New syllabus
AP Calculus BC 7.5 Approximating Solutions Using Euler’s Method 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:
- Approximating Solutions Using Euler’s Method
Approximating Solutions Using Euler’s Method
Approximating Solutions Using Euler’s Method
Euler’s Method is a numerical technique used to approximate solutions to first-order differential equations of the form:
\(\dfrac{dy}{dx} = f(x, y)\), with an initial condition \(y(x_0) = y_0\).
The method uses small steps along the slope given by the differential equation to estimate values of \(y\) for increasing \(x\).
Formula:
\(y_{n+1} = y_n + h \cdot f(x_n, y_n)\)
Where:
- \(y_{n}\) = current \(y\)-value
- \(y_{n+1}\) = next \(y\)-value
- \(x_{n}\) = current \(x\)-value
- \(h\) = step size (\(x_{n+1} – x_n\))
- \(f(x_n, y_n)\) = slope from the differential equation
Procedure:
- Start from the initial point \((x_0, y_0)\).
- Use the formula \(y_{n+1} = y_n + h \cdot f(x_n, y_n)\) to compute the next value.
- Repeat for the desired number of steps.
Example:
Use Euler’s Method with step size \(h = 0.1\) to approximate \(y(0.3)\) for the differential equation:
\(\dfrac{dy}{dx} = x + y\), with \(y(0) = 1\).
▶️Answer/Explanation
Identify components.
- Given: \(y’ = x + y\)
- Initial condition: \(x_0 = 0, y_0 = 1\)
- Step size: \(h = 0.1\)
Apply Euler’s formula.
For \(n = 0\):
\(f(x_0, y_0) = 0 + 1 = 1\)
\(y_1 = 1 + 0.1(1) = 1.1\)
For \(n = 1\):
\(x_1 = 0.1, y_1 = 1.1\)
\(f(x_1, y_1) = 0.1 + 1.1 = 1.2\)
\(y_2 = 1.1 + 0.1(1.2) = 1.22\)
For \(n = 2\):
\(x_2 = 0.2, y_2 = 1.22\)
\(f(x_2, y_2) = 0.2 + 1.22 = 1.42\)
\(y_3 = 1.22 + 0.1(1.42) = 1.362\)
Final Answer: \(y(0.3) \approx 1.362\)
Example:
Approximate \(y(0.2)\) using \(h = 0.1\) for \(\dfrac{dy}{dx} = y – x^2\), with \(y(0) = 0.5\).
▶️Answer/Explanation
Identify components.
- Given: \(y’ = y – x^2\)
- Initial condition: \(x_0 = 0, y_0 = 0.5\)
- Step size: \(h = 0.1\)
Apply Euler’s formula.
For \(n = 0\):
\(f(x_0, y_0) = 0.5 – 0^2 = 0.5\)
\(y_1 = 0.5 + 0.1(0.5) = 0.55\)
For \(n = 1\):
\(x_1 = 0.1, y_1 = 0.55\)
\(f(x_1, y_1) = 0.55 – (0.1)^2 = 0.55 – 0.01 = 0.54\)
\(y_2 = 0.55 + 0.1(0.54) = 0.604\)
Final Answer: \(y(0.2) \approx 0.604\)
Example:
Use Euler’s Method with step size \(h = 0.5\) to approximate \(y(1)\) for \(\dfrac{dy}{dx} = x + y\), given \(y(0) = 1\).
▶️Answer/Explanation
Initial values:
\(x_0 = 0\), \(y_0 = 1\), \(h = 0.5\)
Compute slopes and new \(y\)-values:
\(f(x_0, y_0) = 0 + 1 = 1\) \(y_1 = 1 + 0.5 \cdot 1 = 1.5\) \(x_1 = 0.5\), \(f(x_1, y_1) = 0.5 + 1.5 = 2\)
\(y_2 = 1.5 + 0.5 \cdot 2 = 2.5\) Thus, \(y(1) \approx 2.5\).