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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{\partial}{\partial x} (W \cdot L \cdot H (W, L, x))

\frac{\partial}{\partial y} (\frac{y}{x})

\frac{d}{d x} (\frac{x ^{2} + 3}{x})

y = x^{2} sin^{6} (x) + x cos^{- 2} (x)

x \to 4 lim ((3 x - 9)^{4})

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