Question

Consider the ordinary differential equation \( x^2 \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + 2y = 0 \) with \( y(x) \) as a general solution. Given the values \( y(1) = 1 \), \( y(2) = 5 \), the value of \( y(3) \) is equal to ..........

• A. 9
• B. 12
• C. 15
• D. -15

Hint:

For Cauchy-Euler differential equations, use a trial solution of the form \( y = x^m \) to reduce the equation to an algebraic characteristic equation.

Answer 1

The Correct Option is B

Given the differential equation: \[ x^2 \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + 2y = 0 \] We identify this as a Cauchy-Euler (or equidimensional) differential equation. To solve, we substitute: \[ y = x^m ⇒ \frac{dy}{dx} = mx^{m-1}, \frac{d^2y}{dx^2} = m(m-1)x^{m-2} \] Substitute into the equation: \[ x^2 \cdot m(m-1)x^{m-2} - 2x \cdot m x^{m-1} + 2x^m = 0 \] \[ m(m-1)x^m - 2m x^m + 2x^m = 0 ⇒ x^m [m(m-1) - 2m + 2] = 0 \] \[ x^m [m^2 - m - 2m + 2] = x^m [m^2 - 3m + 2] = 0 \] Solve the auxiliary equation: \[ m^2 - 3m + 2 = 0 ⇒ (m - 1)(m - 2) = 0 ⇒ m = 1, 2 \] Thus, the general solution is: \[ y(x) = A x + B x^2 \] Use the initial conditions:
1. \( y(1) = 1 ⇒ A(1) + B(1)^2 = A + B = 1 \)
2. \( y(2) = 5 ⇒ A(2) + B(4) = 2A + 4B = 5 \)
Solve the system:
From (1): \( A + B = 1 ⇒ A = 1 - B \)
Substitute into (2): \[ 2(1 - B) + 4B = 5 ⇒ 2 - 2B + 4B = 5 ⇒ 2 + 2B = 5 ⇒ 2B = 3 ⇒ B = \frac{3}{2} ⇒ A = 1 - \frac{3}{2} = -\frac{1}{2} \] Now compute \( y(3) \): \[ y(3) = A(3) + B(9) = -\frac{1}{2}(3) + \frac{3}{2}(9) = -\frac{3}{2} + \frac{27}{2} = \frac{24}{2} = 12 \]