What is the solution for e^{2x+y}y^'=e^{x-y} ?

The solution for e^{2x+y}y^'=e^{x-y} is y=(ln(-2e^{-x}+c_{1}))/2

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e^{2x+y}y^'=e^{x-y}

e^{2 x + y} y^{^{'}} = e^{x - y}

y = \frac{ln ( - 2 e ^{- x} + c _{1} )}{2}

Solution steps

e^{2 x + y} y^{'} (x) = e^{x - y}

Solve separable ODE:

y = \frac{ln ( - 2 e ^{- x} + c _{1} )}{2}

y = \frac{ln ( - 2 e ^{- x} + c _{1} )}{2}

\frac{d}{d x} (x^{5} - 5 x^{3})

\frac{e ^{x}}{x ^{2} + 3}

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\frac{x}{7}

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What is the solution for e^{2x+y}y^'=e^{x-y} ?
The solution for e^{2x+y}y^'=e^{x-y} is y=(ln(-2e^{-x}+c_{1}))/2