IB Mathematics AHL 5.15 Slope fields and their diagrams - AI HL Paper 2- Exam Style Questions- New Syllabus
Question
Examine the differential equation
\( \frac{dy}{dx} = \frac{x}{e^{2y}}. \)
(a) Determine which among diagrams A, B, or C depicts the slope field for this differential equation. Justify your selection. [2]
Given that for a specific solution, the coordinates are \( x = 0 \) and \( y = 0 \).
(b) Develop an expression for \( y \) as a function of \( x \) for this particular solution. [7]
(c) Calculate \( \frac{dy}{dx} \) in terms of \( x \) by applying differentiation to your result from part (b). [2]
(d) Confirm that your result from part (b) fulfills the differential equation \( \frac{dy}{dx} = \frac{x}{e^{2y}}. \) [2]
▶️ Answer/Explanation
(a) Diagram C.
Justification: Based on \( \frac{dy}{dx} = \frac{x}{e^{2y}} \):
- At \( x = 0 \), \( \frac{dy}{dx} = 0 \), resulting in horizontal tangents along the \( y \)-axis.
- Since \( e^{2y} > 0 \) for all \( y \), the slope’s sign depends on \( x \): positive when \( x > 0 \) and negative when \( x < 0 \).
These characteristics align with Diagram C.
(b) Separate variables and perform integration:
\( e^{2y} \, dy = x \, dx \)
\( \int e^{2y} \, dy = \int x \, dx \quad \Longrightarrow \quad \frac{1}{2} e^{2y} = \frac{1}{2} x^2 + C. \)
Apply the given point \( (x, y) = (0, 0) \):
\( \frac{1}{2} e^0 = \frac{1}{2} \cdot 0^2 + C \quad \Rightarrow \quad \frac{1}{2} = C. \)
Thus, \( \frac{1}{2} e^{2y} = \frac{1}{2} x^2 + \frac{1}{2} \), so
\( e^{2y} = x^2 + 1. \)
Taking the natural logarithm,
\( y = \frac{1}{2} \ln(x^2 + 1). \)
(c) Differentiate \( y = \frac{1}{2} \ln(x^2 + 1) \):
\( \frac{dy}{dx} = \frac{1}{2} \times \frac{1}{x^2 + 1} \times 2x = \frac{x}{x^2 + 1}. \)
(d) From part (b), \( e^{2y} = x^2 + 1 \). Therefore,
\( \frac{x}{e^{2y}} = \frac{x}{x^2 + 1}. \)
This matches the derivative from part (c), confirming that
\( y = \frac{1}{2} \ln(x^2 + 1) \) satisfies \( \frac{dy}{dx} = \frac{x}{e^{2y}}. \)