Discussion
Let \(y_{1}(x)\) and \(y_{2}(x)\) be two solutions of :
\[ (r(x)y')' + p(x)y = 0 \]
such that \(y_{1}(x)\) and \(y_{2}(x)\) have a common zero in \([a, b]\). Then :
Answer: B
The Wronskian vanishes: \(W(y_{1}, y_{2}) = 0\).
For the self-adjoint equation \((r y')' + p y = 0\), Abel's formula gives \(r(x)\,W(y_{1}, y_{2})(x) = \text{constant}\). So, with \(r\) nonzero on [a, b], the Wronskian is either identically zero or never zero.
At the common zero \(x_{0}\), \(W = y_{1}(x_{0}) y_{2}'(x_{0}) - y_{1}'(x_{0}) y_{2}(x_{0}) = 0 - 0 = 0\). Hence \(W \equiv 0\) on [a, b].
That makes \(W \neq 0\) false, and it also means \(y_{1}\) and \(y_{2}\) are linearly dependent, so 'linearly independent' is false. (Two independent solutions of such an equation can never share a zero; this is the starting point of the Sturm separation theorem.)
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development