Discussion
The Cauchy problem \(uu_{x} + yu_{y} = x\) with \(u(x, 1) = 2x\), when solved using its characteristic equations with an independent variable \(t\), is found to admit of a solution in the form
\[ \begin{aligned} & x = \frac{3}{2}se^{t} - \frac{1}{2}se^{-t}, \\ & y = e^{t}, \quad u = f(s, t). \end{aligned} \]
Then \(f(s, t) =\)
Answer: A
\(f(s, t) = \frac{3}{2}se^{t} + \frac{1}{2}se^{-t}\).
The characteristic equations of \(uu_{x} + yu_{y} = x\) are \(\frac{dx}{dt} = u\), \(\frac{dy}{dt} = y\), \(\frac{du}{dt} = x\), started from the initial curve \(x = s\), \(y = 1\), \(u = 2s\) at \(t = 0\).
Since \(\frac{dx}{dt} = u\), the solution is simply the derivative of the given \(x\):
\[ \begin{aligned} u &= \frac{dx}{dt} = \frac{d}{dt}\left(\frac{3}{2}se^{t} - \frac{1}{2}se^{-t}\right) \\ &= \frac{3}{2}se^{t} + \frac{1}{2}se^{-t} \end{aligned} \]
Check: at \(t = 0\), \(u = \frac{3}{2}s + \frac{1}{2}s = 2s\), matching \(u(x, 1) = 2x\), and \(\frac{du}{dt} = \frac{3}{2}se^{t} - \frac{1}{2}se^{-t} = x\). The option equal to \(x\) itself gives \(u = s\) at \(t = 0\), and the other two fail the initial condition.
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