What is the solution for 3x^2y^'=y^'+6xe^{-y} ?

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

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3x^2y^'=y^'+6xe^{-y}

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

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

Solution steps

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

Solve separable ODE:

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

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

(e^{4 x^{2}})^{7}

\frac{d x}{d t} = 5 x - 3

12 x

y^{^{'}} = e^{3 y} sin (4 x)

\frac{\partial}{\partial x} (6 - x^{2} - y^{2})

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