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

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

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

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(B) Given the differential equation: ${e^{dy/dx}} = (x + 1)$.Taking the natural logarithm on both sides: $\frac{dy}{dx} = \log(x + 1)$.Separating the variables: $dy = \log(x + 1) dx$.Integrating both sides: $y = \int \log(x + 1) dx$.Using integration by parts,let $u = \log(x + 1)$ and $dv = dx$. Then $du = \frac{1}{x+1} dx$ and $v = x + 1$.$y = (x + 1)\log(x + 1) - \int \frac{x+1}{x+1} dx$.$y = (x + 1)\log(x + 1) - \int 1 dx$.$y = (x + 1)\log(x + 1) - x + C$.Given the initial 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 get $C = 3$.Therefore,the solution is $y = (x + 1)\log |x + 1| - x + 3$.

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

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