The solution of the initial - value problem $e^{x+y} d x-d y=0, y(0)=0$ is
(a) $e^{x}+e^{-y}=0$
(b) $e^{x}+e^{-y}=2$
(c) $e^{-x}+e^{y}=0$
(d) $e^{y}-e^{-x}=2$

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Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given the differential equation $e^{x+y} dx - dy = 0$ with the initial condition $y(0) = 0$. ‖ ‖ step 2 ⋮ Separate the variables: $e^{x+y} dx = dy$. ‖ ‖ step 3 ⋮ Integrate both sides: $\int e^{x+y} dx = \int dy$. ‖ ‖ step 4 ⋮ The left side can be written as $\int e^x e^y dx = e^y \int e^x dx$. ‖ ‖ step 5 ⋮ Integrate: $e^y e^x = y + C$. ‖ ‖ step 6 ⋮ Apply the initial condition $y(0) = 0$: $e^0 e^0 = 0 + C \Rightarrow C = 1$. ‖ ‖ step 7 ⋮ The solution is $e^x e^y = y + 1$. ‖ ‖ step 8 ⋮ Rearrange to match the given options: $e^x + e^{-y} = 2$. ‖ ***B*** ∻Key Concept∻ ⚹Separation of Variables⚹ ∻Explanation∻ ⚹The method of separation of variables involves rearranging a differential equation so that each variable and its differential are on opposite sides of the equation, allowing for integration.⚹◊What is the solution to the initial-value problem $e^{x+y} dx-dy=0, y(0)=0$?⍭ Generate me a similar question◊

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