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(D) Given differential equation is $(x+1) \frac{dy}{dx} - xy = 1$. Dividing by $(x+1)$,we get $\frac{dy}{dx} - \frac{x}{x+1}y = \frac{1}{x+1}$. This i

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$y = \frac{1}{1+x}(2e^x - 1)$

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(D) Given differential equation is $(x+1) \frac{dy}{dx} - xy = 1$. Dividing by $(x+1)$,we get $\frac{dy}{dx} - \frac{x}{x+1}y = \frac{1}{x+1}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{x}{x+1} = -(\frac{x+1-1}{x+1}) = -1 + \frac{1}{x+1}$ and $Q = \frac{1}{x+1}$. The integrating factor $IF = e^{\int P dx} = e^{\int (-1 + \frac{1}{x+1}) dx} = e^{-x + \log(x+1)} = e^{-x} \cdot (x+1)$. The general solution is $y \cdot IF = \int (Q \cdot IF) dx + C$. $y \cdot (x+1)e^{-x} = \int (\frac{1}{x+1} \cdot (x+1)e^{-x}) dx + C$. $y(x+1)e^{-x} = \int e^{-x} dx + C = -e^{-x} + C$. Dividing by $e^{-x}$,we get $y(x+1) = -1 + Ce^x$. Given $y(0) = 1$,substituting $x=0$ and $y=1$: $1(0+1) = -1 + Ce^0 \implies 1 = -1 + C \implies C = 2$. Thus,$y(x+1) = -1 + 2e^x$,which gives $y = \frac{2e^x - 1}{x+1}$.

The solution of the differential equation $(x+1) \frac{dy}{dx} - xy = 1$,satisfying $y(0) = 1$ is

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