The particular solution of the differential equation \(\displaystyle\frac{dy}{dx} = e^{x^2/2} + xy\), when \(x = 0, y = 1\), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(y = (x+1)e^{x^2/2}\)
The equation is linear in y:
\( \displaystyle \frac{dy}{dx} - xy = e^{x^2/2} \)
Here \(P = -x\), so the integrating factor is
\( \displaystyle \text{I.F.} = e^{\int -x\,dx} = e^{-x^2/2} \)
Multiplying the equation by the I.F.:
\( \displaystyle \frac{d}{dx}\left(y\,e^{-x^2/2}\right) = e^{x^2/2} \cdot e^{-x^2/2} = 1 \)
\(y\,e^{-x^2/2} = x + C\)
Putting \(x = 0, y = 1\):
\(1 \cdot e^{0} = 0 + C,\quad C = 1\)
Hence
\(y = (x + 1)e^{x^2/2}\)