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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

2 sec (2 t) 2 t \cdot y^{^{'}} = 1

y = 3 x^{3}, y = 75 x

x - 3

7 π r^{2}

x^{2} - 7

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