Discussion
The solution of the linear differential equation \(\frac{dy}{dx} + y = e^{-x}\), when \(x = 0, y = 1\),is
Answer: C
The solution is \(ye^{x} = x + 1\).
The equation is linear with P = 1, so the integrating factor is \(e^{\int 1\,dx} = e^{x}\):
\[ \begin{aligned} \frac{d}{dx}(ye^{x}) &= e^{-x} \cdot e^{x} = 1 \\ ye^{x} &= x + C \end{aligned} \]
At x = 0, y = 1 gives C = 1, so \(ye^{x} = x + 1\). \(ye^{x} = x - 1\) has the wrong constant (it gives y = -1 at x = 0), and the options with \(e^{-x}\) use the reciprocal of the integrating factor.
No comment is present. Be the first to comment.
Loading…
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