What is the solution for 8x^2y^'=y^'+6xe^{-y} ?

The solution for 8x^2y^'=y^'+6xe^{-y} is y=ln(3/8 ln(8x^2-1)+c_{1})

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8x^2y^'=y^'+6xe^{-y}

8 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

y = ln (\frac{3}{8} ln (8 x^{2} - 1) + c_{1})

Solution steps

8 x^{2} y^{'} (x) = y^{'} (x) + 6 x e^{- y}

Solve separable ODE:

y = ln (\frac{3}{8} ln (8 x^{2} - 1) + c_{1})

y = ln (\frac{3}{8} ln (8 x^{2} - 1) + c_{1})

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

\int 9.8 \cdot e^{- 0.4 x} d x

\int \frac{( x + 8 ) ^{3}}{2 x} d x

f (x) = 10 x^{2}

x \to 1 lim (0)

What is the solution for 8x^2y^'=y^'+6xe^{-y} ?
The solution for 8x^2y^'=y^'+6xe^{-y} is y=ln(3/8 ln(8x^2-1)+c_{1})