If \(y = y(x)\) is the solution of the differential equation \(\displaystyle\frac{dy}{dx} = e^{2x - y}\) with \(y(0) = \log_e 2\), then \(y(\log_e 3)\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\log_e 6\)
Since \(e^{2x - y} = \frac{e^{2x}}{e^{y}}\), the variables separate:
\(e^{y}\,dy = e^{2x}\,dx\)
Integrating both sides:
\( \displaystyle e^{y} = \frac{e^{2x}}{2} + C \)
Putting x = 0, \(e^{y} = 2\):
\( \displaystyle 2 = \frac{1}{2} + C \)
\( \displaystyle C = \frac{3}{2} \)
At \(x = \log_e 3\), \(e^{2x} = 3^2 = 9\):
\( \displaystyle e^{y} = \frac{9}{2} + \frac{3}{2} = 6 \)
\(y = \log_e 6\)
Why the other options are incorrect
Option (2) \(\log_e 10\) — Incorrect. This takes \(\int e^{2x}\,dx\) as \(e^{2x}\); then C = 1 and \(e^{y} = 9 + 1 = 10\). The integral is \(\frac{e^{2x}}{2}\).
Option (3) \(\log_e 3\) — Incorrect. At \(x = \log_e 3\), \(e^{y} = 6\), not 3.
Option (4) \(\log_e 5\) — Incorrect. This uses \(e^{y} = 1\) at x = 0 (that is, y(0) = 0), giving \(C = \frac{1}{2}\) and \(e^{y} = 5\); but \(e^{y(0)} = e^{\log_e 2} = 2\).