What is the solution for e^x3yy^'=e^{-y}+e^{-4x-y} ?

The solution for e^x3yy^'=e^{-y}+e^{-4x-y} is y=\W((-e^{-5x}(5e^{4x}+1)+c_{1})/(15e))+1

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e^x3yy^'=e^{-y}+e^{-4x-y}

e^{x} 3 y y^{^{'}} = e^{- y} + e^{- 4 x - y}

y = W (\frac{- e ^{- 5 x} ( 5 e ^{4 x} + 1 ) + c _{1}}{15 e}) + 1

Solution steps

e^{x} 3 y y^{'} (x) = e^{- y} + e^{- 4 x - y}

Solve separable ODE:

y = W (\frac{- e ^{- 5 x} ( 5 e ^{4 x} + 1 ) + c _{1}}{15 e}) + 1

y = W (\frac{- e ^{- 5 x} ( 5 e ^{4 x} + 1 ) + c _{1}}{15 e}) + 1

y^{^{'}} = \frac{9 x}{y + x ^{2} y}, y (0) = - 7

\frac{d y}{d x} = - \frac{4}{7} x^{3} y^{3}

\frac{d x}{d y} - \frac{1}{y} = \frac{2 x}{2 y}, y (1) = 2

\frac{d y}{d x} = \frac{- 2 x y}{x ^{2} + 1}

a (x d y + 2 y d x) = x y d x

What is the solution for e^x3yy^'=e^{-y}+e^{-4x-y} ?
The solution for e^x3yy^'=e^{-y}+e^{-4x-y} is y=\W((-e^{-5x}(5e^{4x}+1)+c_{1})/(15e))+1