What is the solution for e^{-y}y^'=e^x+4x^2 ?

The solution for e^{-y}y^'=e^x+4x^2 is y=-ln(-e^x-(4x^3)/3-c_{1})

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e^{-y}y^'=e^x+4x^2

e^{- y} y^{^{'}} = e^{x} + 4 x^{2}

y = - ln (- e^{x} - \frac{4 x ^{3}}{3} - c_{1})

Solution steps

e^{- y} y^{'} (x) = e^{x} + 4 x^{2}

Solve separable ODE:

y = - ln (- e^{x} - \frac{4 x ^{3}}{3} - c_{1})

y = - ln (- e^{x} - \frac{4 x ^{3}}{3} - c_{1})

y^{^{'}} = \frac{x ^{2} - 8}{y ^{2}}, y (0) = 2

y^{^{'}} = - (sin (x)) (cos^{2} (y))

y^{^{'}} = - \frac{9 x}{y}

6 (y^{2} + 2 x y \frac{d y}{d x}) + 18 x = - 4 y^{3} \frac{d y}{d x}

y^{^{'}} (x) = \frac{y}{x ( y - 3 )}, y (2) = 6

What is the solution for e^{-y}y^'=e^x+4x^2 ?
The solution for e^{-y}y^'=e^x+4x^2 is y=-ln(-e^x-(4x^3)/3-c_{1})