Discussion
The solution of the linear partial differential equation
\[ \begin{aligned} &\frac{\partial^{2} z}{\partial x^{2}} - 2\frac{\partial^{2} z}{\partial y \partial x} \\ &+ \frac{\partial^{2} z}{\partial y^{2}} = 0 \end{aligned} \]
is :
Answer: C
The solution is \(z = \phi_{1}(x + y) + x\phi_{2}(x + y)\).
With \(D = \partial/\partial x\) and \(D' = \partial/\partial y\), the equation is \((D^{2} - 2DD' + D'^{2})z = (D - D')^{2}z = 0\). For \((D - mD')^{2}z = 0\) the general solution is \(\phi_{1}(y + mx) + x\phi_{2}(y + mx)\); here \(m = 1\), so the argument is \(x + y\) in both terms.
Check: if \(z = \phi(x + y)\), then \(z_{xx} = z_{xy} = z_{yy} = \phi''\) and \(\phi'' - 2\phi'' + \phi'' = 0\).
Option A has \(x - y\), which belongs to \(D + D'\), a different factor. Option B mixes the two arguments, and option D's \(x^{2}\) would arise only from a cubed factor.
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