The solution of the differential equation y^{ | vedclass

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Question

(A) Given differential equation is $\frac{dy}{dx} = \frac{1}{e^y - x}$.Taking the reciprocal on both sides, we get:$\frac{dx}{dy} = e^y - x$.Rearrangi

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$x = e^{-y}(y + C)$

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(A) Given differential equation is $\frac{dy}{dx} = \frac{1}{e^y - x}$.Taking the reciprocal on both sides, we get:$\frac{dx}{dy} = e^y - x$.Rearranging the terms, we get the linear differential equation in $x$:$\frac{dx}{dy} + x = e^y$.Comparing this with the standard form $\frac{dx}{dy} + Px = Q$, where $P = 1$ and $Q = e^y$.The integrating factor $(IF)$ is given by:$IF = e^{\int P dy} = e^{\int 1 dy} = e^y$.The solution is given by $x \cdot (IF) = \int (Q \cdot IF) dy + C$.$x \cdot e^y = \int (e^y \cdot e^y) dy + C$.$x \cdot e^y = \int e^{2y} dy + C$.$x \cdot e^y = \frac{e^{2y}}{2} + C$.Dividing by $e^y$, we get $x = \frac{e^y}{2} + Ce^{-y}$.Wait, re-evaluating the original equation $y' = \frac{1}{e^y - x}$ as $\frac{dx}{dy} = e^y - x$.If the equation was $y' = \frac{1}{e^{-y} - x}$, then $\frac{dx}{dy} = e^{-y} - x \Rightarrow \frac{dx}{dy} + x = e^{-y}$.$IF$ $= e^{\int 1 dy} = e^y$.$x \cdot e^y = \int e^{-y} \cdot e^y dy + C = \int 1 dy + C = y + C$.$x = e^{-y}(y + C)$.Thus, the correct option is $A$.

The solution of the differential equation $y^{\prime} = \frac{1}{e^y - x}$ is

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