What is the solution for 8x^2y^'=y^'+2xe^{-y} ?

The solution for 8x^2y^'=y^'+2xe^{-y} is y=ln(1/8 ln(8x^2-1)+c_{1})

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8x^2y^'=y^'+2xe^{-y}

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

y = ln (\frac{1}{8} ln (8 x^{2} - 1) + c_{1})

Solution steps

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

Solve separable ODE:

y = ln (\frac{1}{8} ln (8 x^{2} - 1) + c_{1})

y = ln (\frac{1}{8} ln (8 x^{2} - 1) + c_{1})

\frac{s + 3}{s ^{2} + s + 3}

\int x^{3} (x^{4} - 1)^{2} d x

\int 8 cos (4 x) d x

\frac{d}{d x} (\frac{x ^{3}}{x + 1})

x \to 0 lim (x^{3 s i n (x)})

What is the solution for 8x^2y^'=y^'+2xe^{-y} ?
The solution for 8x^2y^'=y^'+2xe^{-y} is y=ln(1/8 ln(8x^2-1)+c_{1})