The general solution of the differential equation \(\displaystyle\frac{dy}{dx} + 2xy = 2x\) is (C is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(y = 1 + Ce^{-x^2}\)
This is a linear equation with \(P = 2x\) and \(Q = 2x\). The integrating factor is
\( e^{\int 2x\,dx} = e^{x^2} \)
So
\( \displaystyle y\,e^{x^2} = \int 2x\,e^{x^2}\,dx \)
With \(t = x^2\), \(dt = 2x\,dx\):
\( y\,e^{x^2} = e^{x^2} + C \)
\( y = 1 + Ce^{-x^2} \)
Why the other options are incorrect
Option (2) \(y = 1 + Ce^{x^2}\) — Incorrect. Dividing by the integrating factor \(e^{x^2}\) gives \(e^{-x^2}\), not \(e^{x^2}\). Check: \(y' + 2xy = 2x + 4Cxe^{x^2}\), not \(2x\).
Option (3) \(y = x + Ce^{-x^2}\) — Incorrect. Check with \(C = 0\): \(y = x\) gives \(y' + 2xy = 1 + 2x^2\), not \(2x\).
Option (4) \(y = 2x + Ce^{-x^2}\) — Incorrect. Check with \(C = 0\): \(y = 2x\) gives \(y' + 2xy = 2 + 4x^2\), not \(2x\).