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

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

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

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

y = ln (ln (2 x^{2} - 1) + c_{1})

Solution steps

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

Solve separable ODE:

y = ln (ln (2 x^{2} - 1) + c_{1})

y = ln (ln (2 x^{2} - 1) + c_{1})

x \to 0 lim (\frac{1}{2 x ^{2} + x})

\int \frac{9}{4 + 9 x} d x

\int \frac{y}{a ^{2} - y ^{2}} d y

x \to 0 lim (x^{2} csc (x))

\frac{\partial}{\partial y} (3 x + 4 y)

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