If y=y(x) is the solution of the differential | vedclass

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(D) Given the differential equation: $e^{y}\left(\frac{dy}{dx}-1\right)=e^{x}$.This can be rewritten as $e^{y} \frac{dy}{dx} - e^{y} = e^{x}$.Let $e^{

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$1+\log _{e} 2$

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(D) Given the differential equation: $e^{y}\left(\frac{dy}{dx}-1\right)=e^{x}$.This can be rewritten as $e^{y} \frac{dy}{dx} - e^{y} = e^{x}$.Let $e^{y} = t$. Then,differentiating with respect to $x$,we get $e^{y} \frac{dy}{dx} = \frac{dt}{dx}$.Substituting these into the equation,we get the linear differential equation: $\frac{dt}{dx} - t = e^{x}$.The integrating factor ($I$.$F$.) is $e^{\int -1 dx} = e^{-x}$.Multiplying both sides by the $I$.$F$.,we get $\frac{d}{dx}(t e^{-x}) = e^{x} \cdot e^{-x} = 1$.Integrating both sides with respect to $x$,we get $t e^{-x} = x + c$.Substituting $t = e^{y}$,we have $e^{y} e^{-x} = x + c$,which simplifies to $e^{y-x} = x + c$.Given $y(0) = 0$,we substitute $x = 0$ and $y = 0$: $e^{0-0} = 0 + c \Rightarrow 1 = c$.So,the solution is $e^{y-x} = x + 1$.To find $y(1)$,substitute $x = 1$: $e^{y-1} = 1 + 1 = 2$.Taking the natural logarithm on both sides,$y - 1 = \log_{e} 2$.Therefore,$y(1) = 1 + \log_{e} 2$.

If $y=y(x)$ is the solution of the differential equation $e^{y}\left(\frac{dy}{dx}-1\right)=e^{x}$ such that $y(0)=0,$ then $y(1)$ is equal to

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