The solution of e^{dy/dx} = x+1, y(0) = 3 is | vedclass

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

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

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(D) Given the differential equation $e^{dy/dx} = x+1$. Taking the natural logarithm on both sides,we get $\frac{dy}{dx} = \log(x+1)$. Integrating both sides with respect to $x$,we have $\int dy = \int \log(x+1) dx$. Using integration by parts,$\int \log(x+1) dx = (x+1) \log(x+1) - (x+1) + C$. Alternatively,$y = x \log(x+1) - \int \frac{x}{x+1} dx = x \log(x+1) - \int (1 - \frac{1}{x+1}) dx = x \log(x+1) - x + \log(x+1) + C$. Thus,$y = (x+1) \log(x+1) - x + C$. Given the initial condition $y(0) = 3$,we substitute $x=0$ and $y=3$: $3 = (0+1) \log(1) - 0 + C \Rightarrow 3 = 0 - 0 + C \Rightarrow C = 3$. Substituting $C=3$ into the general solution,we get $y = (x+1) \log(x+1) - x + 3$. Rearranging the terms,we obtain $y+x-3 = (x+1) \log(x+1)$.

The solution of $e^{dy/dx} = x+1, y(0) = 3$ is

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