Question

The value of \( \alpha \) for which the Euler method with step size \( h = 0.1 \) provides \( y(1.1) = 1.2 \) for the following initial value problem is \[ \frac{dy}{dx} = -2xy^\alpha, \quad y(1) = 2 \]

• A. \( -2 \)
• B. \( -1 \)
• C. \( 1 \)
• D. \( 2 \)

Hint:

When using Euler’s method for an ODE involving powers of \( y \), plug in the initial values and solve algebraically to determine unknown constants like \( \alpha \).

Answer 1

The Correct Option is D

Step 1: Apply Euler’s Method. Euler’s method updates the value of \( y \) using: \[ y_{n+1} = y_n + h \cdot f(x_n, y_n) \] Given: \[ x_0 = 1, \quad y_0 = 2, \quad h = 0.1, \quad f(x, y) = -2xy^\alpha \] Step 2: Plug into the update formula. \[ y_1 = y_0 + h \cdot (-2x_0 y_0^\alpha) = 2 + 0.1 \cdot (-2 \cdot 1 \cdot 2^\alpha) = 2 - 0.2 \cdot 2^\alpha \] We are told that \( y_1 = y(1.1) = 1.2 \), so: \[ 1.2 = 2 - 0.2 \cdot 2^\alpha \quad \Rightarrow \quad 0.2 \cdot 2^\alpha = 0.8 \Rightarrow 2^\alpha = \frac{0.8}{0.2} = 4 \Rightarrow \alpha = 2 \]