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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d z} (- z)

\int \frac{1}{x ^{2} + 4 x + 6} d x

ln (x + 6)

f (x) = 2 x^{2} - 2 x - 4, at x = 1

y = (ln (4) x^{2}) (- x^{3} - 4)

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