Discussion
Let \(f(x) = x^{2}\) and \(g(x) = x^{2} \log x\) on \(|x| \gt 0\). Then :
Answer: C
The correct statement is that \(f\) and \(g\) can be solutions of an ordinary differential equation.
They are linearly independent, since \(g/f = \log x\) is not constant, which rules out 'linearly dependent'. Their Wronskian is
\[ \begin{aligned} W &= x^{2}(2x\log x + x) - 2x \cdot x^{2}\log x \\ &= x^{3} \end{aligned} \]
so it is not \(x^{2}\). Since \(W \neq 0\) for \(x \neq 0\), they form a fundamental pair for a second-order linear equation. Indeed \(x^{2}\) and \(x^{2}\log x\) are the solutions of the Euler equation \(x^{2}y'' - 3xy' + 4y = 0\), whose indicial equation \(m^{2} - 4m + 4 = 0\) has the repeated root 2. So 'cannot be solutions of an ordinary differential equation' is false.
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