The general solution of the differential equation \(\displaystyle\frac{dy}{dx} - 2y = 3e^{-x}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(y = -e^{-x} + Ce^{2x}\)
This is a linear equation with \(P = -2\) and \(Q = 3e^{-x}\). The integrating factor is
\( e^{\int -2\,dx} = e^{-2x} \)
So
\( \displaystyle y\,e^{-2x} = \int 3e^{-x} \cdot e^{-2x}\,dx = \int 3e^{-3x}\,dx \)
\( y\,e^{-2x} = -e^{-3x} + C \)
Multiplying by \(e^{2x}\):
\( y = -e^{-x} + Ce^{2x} \)
Why the other options are incorrect
Option (2) \(y = e^{-x} + Ce^{2x}\) — Incorrect. The integral of \(3e^{-3x}\) is \(-e^{-3x}\); the minus sign is lost here. Check: \(y' - 2y = -3e^{-x}\), not \(3e^{-x}\).
Option (3) \(y = -e^{-x} + Ce^{-2x}\) — Incorrect. Dividing by the integrating factor \(e^{-2x}\) multiplies C by \(e^{2x}\), not \(e^{-2x}\).
Option (4) \(y = -3e^{-x} + Ce^{2x}\) — Incorrect. The factor \(\frac{1}{3}\) from integrating \(e^{-3x}\) is missed. Check: \(y' - 2y = 9e^{-x}\).