Question

Let \( L[y] = x^2\dfrac{d^2y}{dx^2} + px\dfrac{dy}{dx} + qy, \) where \( p, q \) are real constants. Let \( y_1(x) \) and \( y_2(x) \) be two solutions of \( L[y] = 0, \, x > 0, \) that satisfy \( y_1(x_0) = 1, y_1'(x_0) = 0, y_2(x_0) = 0, y_2'(x_0) = 1 \) for some \( x_0 > 0. \) Then,

• A. \( y_1(x) \) is not a constant multiple of \( y_2(x) \)
• B. \( y_1(x) \) is a constant multiple of \( y_2(x) \)
• C. \( 1, \ln x \) are solutions of \( L[y] = 0 \) when \( p = 1, q = 0 \)
• D. \( x, \ln x \) are solutions of \( L[y] = 0 \) when \( p + q \neq 0 \)

Hint:

Check linear independence using the Wronskian. For Euler-type equations, trial solutions like \( y = x^m \) help identify parameters \( p, q \).

Answer 1

The Correct Option is A, C

Step 1: Independence of \( y_1 \) and \( y_2 \). 
The Wronskian of \( y_1 \) and \( y_2 \) is \[ W(y_1, y_2)(x_0) = \begin{vmatrix} y_1(x_0) & y_2(x_0) \\ y_1'(x_0) & y_2'(x_0) \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1 \neq 0. \] Thus, \( y_1 \) and \( y_2 \) are linearly independent, and hence \( y_1(x) \) is not a constant multiple of \( y_2(x) \). Therefore, (A) is true. 
Step 2: Check for \( 1 \) and \( \ln x \) as solutions. 
For \( p = 1, q = 0 \), \[ L[y] = x^2 y'' + x y' = 0. \] Let \( y = 1 \): then \( y' = y'' = 0 \Rightarrow L[1] = 0. \) Let \( y = \ln x \): \( y' = \frac{1}{x}, y'' = -\frac{1}{x^2} \Rightarrow L[\ln x] = x^2(-\frac{1}{x^2}) + x(\frac{1}{x}) = -1 + 1 = 0. \) Hence both \( 1 \) and \( \ln x \) are solutions, so (C) is true. 

Step 3: Verify (D). 
For \( y = x, \ln x \), we get: \[ L[x] = x^2(0) + p x(1) + qx = x(p + q). \] This equals zero only when \( p + q = 0 \). Hence, (D) is false as stated. 
 

Final Answer: \[ \boxed{\text{(A) and (C)}} \]