What is the solution for 9x^2y^'=y^'+6xe^{-y} ?

The solution for 9x^2y^'=y^'+6xe^{-y} is y=ln(1/3 ln(9x^2-1)+c_{1})

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9x^2y^'=y^'+6xe^{-y}

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

y = ln (\frac{1}{3} ln (9 x^{2} - 1) + c_{1})

Solution steps

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

Solve separable ODE:

y = ln (\frac{1}{3} ln (9 x^{2} - 1) + c_{1})

y = ln (\frac{1}{3} ln (9 x^{2} - 1) + c_{1})

y^{^{'}} = y (x y^{5} + 1)

y = ∣ x ∣, y = x^{2} - 1

(cos (x))^{\frac{1}{x}}

x = y^{2} - 4 y, x = 0

y^{^{'}} + (tan (t)) y = tan (t)

What is the solution for 9x^2y^'=y^'+6xe^{-y} ?
The solution for 9x^2y^'=y^'+6xe^{-y} is y=ln(1/3 ln(9x^2-1)+c_{1})