If \(y = y(x)\) is the solution of the differential equation \(\displaystyle\frac{dy}{dx} + y = 1\) satisfying \(y(0) = 3\), then \(y(\log_e 2)\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2
Writing the equation as \(\frac{dy}{dx} = 1 - y\), it is variables separable. Since y(0) = 3, y stays greater than 1, so
\( \displaystyle \frac{dy}{y - 1} = -\,dx \)
Integrating both sides:
\(\log_e (y - 1) = -x + C_1\)
\(y - 1 = Ce^{-x}\)
Putting x = 0, y = 3:
\(C = 2\)
\(y = 1 + 2e^{-x}\)
At \(x = \log_e 2\), where \(e^{-x} = \frac{1}{2}\):
\( \displaystyle y = 1 + 2 \times \frac{1}{2} = 2 \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{2}\) — Incorrect. This is the value of \(3e^{-x}\), which solves \(\frac{dy}{dx} = -y\) and leaves out the 1 on the right side.
Option (2) 5 — Incorrect. This is the value of \(1 + 2e^{x}\), which has the wrong sign in the exponent.
Option (3) 4 — Incorrect. It would need \(2e^{-x} = 3\) at \(x = \log_e 2\), but there \(2e^{-x} = 2 \times \frac{1}{2} = 1\).