Discussion
Consider the differential equation \(\frac{dy}{dx} + 4x = 0\). The solution of this equation for the particular case of \(y = 1\), at \(x = 0\) is
Answer: C
The solution is \(y = 1 - 2x^{2}\).
Separate and integrate, then fix the constant from \(y = 1\) at \(x = 0\):
\[ \begin{aligned} \frac{dy}{dx} &= -4x \\ y &= \int -4x\,dx = -2x^{2} + C \\ y(0) = 1 &\Rightarrow C = 1 \\ y &= 1 - 2x^{2} \end{aligned} \]
Check: differentiating gives \(-4x\), and \(-4x + 4x = 0\). The option \(y = -4x^{2} + 1\) forgets to halve on integrating, so its derivative is \(-8x\). \(y = 4x\) gives \(y = 0\) at \(x = 0\), and \(\log(x)\) is not even defined at \(x = 0\).
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