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(B) Given the differential equation: $e^{\frac{dy}{dx}} = (x+1)$.Taking the natural logarithm on both sides,we get: $\frac{dy}{dx} = \log(x+1)$.Integr

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$y = (x+1) \log (x+1) - x + 3$

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(B) Given the differential equation: $e^{\frac{dy}{dx}} = (x+1)$.Taking the natural logarithm on both sides,we get: $\frac{dy}{dx} = \log(x+1)$.Integrating both sides with respect to $x$: $\int dy = \int \log(x+1) dx + C$.Using integration by parts for $\int \log(x+1) dx$ (let $u = \log(x+1)$ and $dv = dx$):$y = (x+1) \log(x+1) - \int (x+1) \cdot \frac{1}{x+1} dx + C$.$y = (x+1) \log(x+1) - \int 1 dx + C$.$y = (x+1) \log(x+1) - x + C$.Given the condition $y(0) = 3$,substitute $x = 0$ and $y = 3$:$3 = (0+1) \log(0+1) - 0 + C$.$3 = 1 \cdot \log(1) + C$.Since $\log(1) = 0$,we have $3 = 0 + C$,so $C = 3$.Therefore,the particular solution is $y = (x+1) \log(x+1) - x + 3$.

The particular solution of the differential equation $e^{\frac{dy}{dx}} = (x+1)$ with the condition $y(0) = 3$ is

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