How will you solve: $\frac{d y}{d x} = e^{x - y} [e^{x} - e^{y}]$ $(dy)/(dx)=e^(x-y)[e^x-e^y]$ ?

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Answer 1 · 2018-03-21T17:15:52

$y = ln (C_{0} e^{- e^{x}} + e^{x} - 1)$ $y = ln( C_0 e^(-e^x)+e^x-1)$

Making the substitution

$y = ln u$ $y = ln u$ we have the transformed differential equation

$(u'-e^(2x)+e^x u)/u = 0$ or assuming $u ne 0$

$u'+e^x u -e^(2x) = 0$

This is a linear non homogeneous differential equation easily soluble giving

$u = C_0 e^(-e^x)+e^x-1$ and finally

$y = ln( C_0 e^(-e^x)+e^x-1)$

How will you solve:(dy)/(dx)=e^(x-y)[e^x-e^y]dydx=ex−y[ex−ey](dy)/(dx)=e^(x-y)[e^x-e^y]?