If \(y = y(x)\) is the solution of the differential equation \((x + 1)\displaystyle\frac{dy}{dx} = 2xy\), \(x \gt -1\), satisfying \(y(0) = 3\), then \(y(1)\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{3e^2}{4}\)
The equation is variables separable:
\( \displaystyle \frac{dy}{y} = \frac{2x}{x + 1}\,dx \)
Writing \(2x = 2(x + 1) - 2\) splits the right side:
\( \displaystyle \frac{2x}{x + 1} = 2 - \frac{2}{x + 1} \)
Integrating both sides (here \(x + 1 \gt 0\)):
\(\log|y| = 2x - 2\log(x + 1) + \log|C|\)
\( \displaystyle y = \frac{C\,e^{2x}}{(x + 1)^2} \)
Putting x = 0, y = 3:
\(3 = C\)
At x = 1:
\( \displaystyle y(1) = \frac{3e^{2}}{(1 + 1)^2} = \frac{3e^2}{4} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3e^2}{2}\) — Incorrect. This uses \(\log(x + 1)\) instead of \(2\log(x + 1)\), dropping the 2 in \(\frac{2}{x + 1}\).
Option (2) \(3e^2\) — Incorrect. This keeps only \(\int 2\,dx\) and leaves out the term \(-\frac{2}{x + 1}\).
Option (4) \(\displaystyle\frac{e^2}{4}\) — Incorrect. This takes C = 1 and ignores the condition y(0) = 3.