What is the solution for 3x^2y^'=y^'+5xe^{-y} ?

The solution for 3x^2y^'=y^'+5xe^{-y} is y=ln(5/6 ln(3x^2-1)+c_{1})

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3x^2y^'=y^'+5xe^{-y}

3 x^{2} y^{^{'}} = y^{^{'}} + 5 x e^{- y}

y = ln (\frac{5}{6} ln (3 x^{2} - 1) + c_{1})

Solution steps

3 x^{2} y^{'} (x) = y^{'} (x) + 5 x e^{- y}

Solve separable ODE:

y = ln (\frac{5}{6} ln (3 x^{2} - 1) + c_{1})

y = ln (\frac{5}{6} ln (3 x^{2} - 1) + c_{1})

3 x^{2}, \frac{1}{3} x

\frac{d y}{d x} = (1 + e^{x}) y

\int cos^{2} (5 x) d x

(1 + e^{x} y + x e^{x} y) d x + (x e^{x} + 2) d y = 0

\int \frac{9 x ^{3} - 5 x ^{2} + 3 x}{x} d x

What is the solution for 3x^2y^'=y^'+5xe^{-y} ?
The solution for 3x^2y^'=y^'+5xe^{-y} is y=ln(5/6 ln(3x^2-1)+c_{1})