What is the solution for 4x^2y^'=y^'+6xe^{-y} ?

The solution for 4x^2y^'=y^'+6xe^{-y} is y=ln(3/4 ln(4x^2-1)+c_{1})

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4x^2y^'=y^'+6xe^{-y}

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

y = ln (\frac{3}{4} ln (4 x^{2} - 1) + c_{1})

Solution steps

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

Solve separable ODE:

y = ln (\frac{3}{4} ln (4 x^{2} - 1) + c_{1})

y = ln (\frac{3}{4} ln (4 x^{2} - 1) + c_{1})

\frac{d}{d x} (6 x e^{x})

f (x) = x^{3} - 3 x^{2} + 2 x - 2, at x = 0

\frac{\partial}{\partial y} (\frac{3 x - 4 y}{3 x + 4 y})

x \to 9 lim (cos (\frac{1}{x - 9}))

What is the solution for 4x^2y^'=y^'+6xe^{-y} ?
The solution for 4x^2y^'=y^'+6xe^{-y} is y=ln(3/4 ln(4x^2-1)+c_{1})