CUET · MATHS · PYQ PAPER 2025
The particular solution of the differential equation \(\frac{d y}{d x}=e^{x^2 / 2}+x y\), when \(x = 0\), \(y = 1\) is
- A \(y=e^{-x^2 / 2}+1\)
- B \(y=(x+1) e^{x^2 / 2}\)
- C \(y=x e^{x^2 / 2}+1\)
- D \(y=(x-1) e^{x^2 / 2}\)
Answer & Solution
Correct Answer
(B) \(y=(x+1) e^{x^2 / 2}\)
Step-by-step Solution
Detailed explanation
\( \frac{d y}{d x} - x y = e^{x^2 / 2} \) IF \( = e^{\int -x dx} = e^{-x^2 / 2} \) \( \frac{d}{dx} (y e^{-x^2 / 2}) = e^{x^2 / 2} \cdot e^{-x^2 / 2} = 1 \) \( y e^{-x^2 / 2} = x + C \) \( y = (x + C) e^{x^2 / 2} \) For \( x=0, y=1 \): \( 1 = (0 + C) e^0 \implies C=1 \)…
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