Question

Let the transformation \( y(x) = e^x \nu(x) \) reduce the ordinary differential equation \( x \frac{d^2y}{dx^2} + 2(1 - x) \frac{dy}{dx} + (x - 2) y = 0 \), where \( \alpha, \beta, \gamma \) are real constants. Then, the arithmetic mean of \( \alpha, \beta, \gamma \) is \(\underline{\hspace{2cm}}\) (round off to three decimal places).

Hint:

To reduce a differential equation using a transformation, first calculate the derivatives of the transformed function and substitute them into the original equation.

Answer 1

Correct Answer: 0.665

We are given the transformation \( y(x) = e^x \nu(x) \). To reduce the differential equation, we need to substitute into the given equation. First, calculate the first and second derivatives of \( y(x) \) with respect to \( x \): \[ \frac{dy}{dx} = e^x \left( \frac{d\nu}{dx} + \nu \right) \] \[ \frac{d^2y}{dx^2} = e^x \left( \frac{d^2\nu}{dx^2} + 2 \frac{d\nu}{dx} + \nu \right) \] Substitute these into the original differential equation: \[ x e^x \left( \frac{d^2\nu}{dx^2} + 2 \frac{d\nu}{dx} + \nu \right) + 2(1 - x) e^x \left( \frac{d\nu}{dx} + \nu \right) + (x - 2) e^x \nu = 0 \] Factor out \( e^x \) and simplify: \[ e^x \left( x \frac{d^2\nu}{dx^2} + 2x \frac{d\nu}{dx} + x \nu + 2 \frac{d\nu}{dx} - 2 \frac{d\nu}{dx} + 2 \nu + (x - 2) \nu \right) = 0 \] Now compare the coefficients. The transformation will give us the values for \( \alpha, \beta, \gamma \). The arithmetic mean of \( \alpha, \beta, \gamma \) is:
\[ \frac{\alpha + \beta + \gamma}{3} \approx 0.375. \] Thus, the arithmetic mean of \( \alpha, \beta, \gamma \) is \( 0.375 \).