The solution of the differential equation \(\displaystyle\frac{dy}{dx} + 2y = 4x\) that satisfies \(y(0) = 0\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(y = 2x - 1 + e^{-2x}\)
This is a linear equation with \(P = 2\) and \(Q = 4x\). The integrating factor is \(e^{2x}\), so
\( \displaystyle y\,e^{2x} = \int 4x\,e^{2x}\,dx \)
Integrating by parts:
\( \displaystyle \int 4x\,e^{2x}\,dx = 2x\,e^{2x} - \int 2e^{2x}\,dx = (2x - 1)e^{2x} + C \)
So
\( y = 2x - 1 + Ce^{-2x} \)
Using \(y(0) = 0\):
\( 0 = -1 + C \Rightarrow C = 1 \)
\( y = 2x - 1 + e^{-2x} \)
Why the other options are incorrect
Option (2) \(y = 2x + 1 - e^{-2x}\) — Incorrect. It has \(y(0) = 0\), but \(y' + 2y = 4x + 4\), not 4x.
Option (3) \(y = 2x - 1 + e^{2x}\) — Incorrect. Dividing by \(e^{2x}\) gives \(e^{-2x}\), not \(e^{2x}\); here \(y' + 2y = 4x + 4e^{2x}\).
Option (4) \(y = 2x\left(1 - e^{-2x}\right)\) — Incorrect. It has \(y(0) = 0\), but \(y' + 2y = 4x + 2 - 2e^{-2x}\), not 4x.