Discussion
The general solution of
\[ \begin{aligned} &\frac{d^{4}y}{dx^{4}} - 2\frac{d^{3}y}{dx^{3}} + 2\frac{d^{2}y}{dx^{2}} \\ &\quad - 2\frac{dy}{dx} + y = 0 \end{aligned} \]
is
Answer: C
The general solution is \(c_{1}e^{x} + c_{2}xe^{x} + c_{3}\cos(x) + c_{4}\sin(x)\).
The characteristic equation factors neatly:
\[ \begin{aligned} r^{4} - 2r^{3} + 2r^{2} - 2r + 1 &= 0 \\ (r^{2} - 2r + 1)(r^{2} + 1) &= 0 \\ (r - 1)^{2}(r^{2} + 1) &= 0 \end{aligned} \]
The double root r = 1 gives \(e^{x}\) and \(xe^{x}\), and the roots \(\pm i\) give \(\cos x\) and \(\sin x\).
The cosh/sinh option treats \(\pm 1\) as roots; the options with \(e^{-x}\) or \(xe^{-x}\) also assume a root at -1, which does not exist. Expanding the factors confirms the coefficients -2, 2, -2, 1.
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