The solution of the differential equation e^{ | vedclass

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Question

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

Vedclass · Accepted Answer

$y = (x+1) \log(x+1) - x + 5$

Vedclass · Answer

(D) Given the differential equation $e^{\frac{dy}{dx}} = x+1$. Taking the natural logarithm on both sides,we get $\frac{dy}{dx} = \log_{e}(x+1)$. Integrating both sides with respect to $x$: $\int dy = \int \log_{e}(x+1) dx$. Using integration by parts,let $u = \log_{e}(x+1)$ and $dv = dx$. Then $du = \frac{1}{x+1} dx$ and $v = x+1$. $\int \log_{e}(x+1) dx = (x+1) \log_{e}(x+1) - \int \frac{x+1}{x+1} dx = (x+1) \log_{e}(x+1) - x + C$. So,$y = (x+1) \log_{e}(x+1) - x + C$. Given the initial condition $y(0) = 5$,substitute $x=0$ and $y=5$: $5 = (0+1) \log_{e}(0+1) - 0 + C \Rightarrow 5 = 1 \cdot \log_{e}(1) - 0 + C \Rightarrow 5 = 0 - 0 + C \Rightarrow C = 5$. Therefore,the solution is $y = (x+1) \log_{e}(x+1) - x + 5$.

The solution of the differential equation $e^{\frac{dy}{dx}} = x+1$ with the initial condition $y(0) = 5$ for $x \in (-1, \infty)$ is:

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