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The general solution of differential equation \frac{d y}{d x}=e^{\frac{x^{2}}{2}}+x y is
(a) y=C e^{-x^{2} / 2}
(b) y=C e^{x^{2} / 2}
(c) y=(x+C) e^{x^{2} / 2}
(d) y=(C-x) e^{x^{2} / 2}

Answers (1)

The answer is the option (c)

Explanation: -

\Rightarrow \quad \frac{d y}{d x}-x y=e^{\frac{x^{2}}{2}}$
This is a linear differential equation. On comparing it with \frac{d y}{d x}+P y=Q,$ we get
$$ P=-x, O=e^{x^{2} / 2} $$
\therefore \quad \mathrm{I.F.}=e^{\int -x d x}=e^{-x^{2} / 2}$
So, the general solution is:
\\\therefore \quad y \cdot e^{-x^{2} / 2}=\int e^{-x^{2} / 2} e^{x^{2} / 2} d x+C$ \\$\Rightarrow \quad y e^{-x^{2} / 2}=\int 1 d x+C$ \\$\Rightarrow \quad y e^{-x^{2} / 2}=x+C$ \\$\Rightarrow \quad y=(x+C) e^{x^{2} / 2}$

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