What is the solution for e^yy^'=e^y+4x ?

The solution for e^yy^'=e^y+4x is y=ln(c_{1}+4(-e^{-x}x-e^{-x}))+x

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e^yy^'=e^y+4x

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

y = ln (c_{1} + 4 (- e^{- x} x - e^{- x})) + x

Solution steps

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

Solve exact ODE:

y = ln (c_{1} + 4 (- e^{- x} x - e^{- x})) + x

y = ln (c_{1} + 4 (- e^{- x} x - e^{- x})) + x

sec^{2} (2 x)

y d x + (x^{2} + 4 x) d y = 0

\frac{d y}{d x} = (e^{x} + 7)^{2}

y = x^{2} cos (\frac{4}{x})

f (x) = x^{4} + 2 e^{x}, (0, 2)

What is the solution for e^yy^'=e^y+4x ?
The solution for e^yy^'=e^y+4x is y=ln(c_{1}+4(-e^{-x}x-e^{-x}))+x