Let $ y_1(x) $ and $ y_2(x) $ be two linearly independent solutions of the differential equation: $$ x^2 \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + 2y = 0, \quad x0. $$ Let $ W(y_1, y_2)(x) $ denote the Wronskian of $ y_1(x) $ and $ y_2(x) $ at $ x $. If $ W(y_1, y_2)(1) = 1 $, then $ W(y_1, y_2)(2) $ is equal to ________ .

Question

Let $ y_1(x) $ and $ y_2(x) $ be two linearly independent solutions of the differential equation: $$ x^2 \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + 2y = 0, \quad x>0. $$ Let $ W(y_1, y_2)(x) $ denote the Wronskian of $ y_1(x) $ and $ y_2(x) $ at $ x $. If $ W(y_1, y_2)(1) = 1 $, then $ W(y_1, y_2)(2) $ is equal to ________ .

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The given differential equation is: $$ x^2 \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + 2y = 0 $$ This is a Cauchy-Euler equation, and for this type of equation, the Wronskian $ W(y_1, y_2)(x) $ satisfies the following property: $$ \frac{dW}{dx} = \frac{1}{x} W $$ The solution to this equation is: $$ W(y_1, y_2)(x) = Cx $$ where $ C $ is a constant. We are given that $ W(y_1, y_2)(1) = 1 $, so: $$ C \times 1 = 1 \quad \Rightarrow \quad C = 1 $$ Thus, the Wronskian is: $$ W(y_1, y_2)(x) = x $$ Now, we can calculate $ W(y_1, y_2)(2) $: $$ W(y_1, y_2)(2) = 2 $$ Thus, the value of $ W(y_1, y_2)(2) $ is: $$ \boxed{2} $$

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Correct Answer: 4

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