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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 7, y (0) = 7, y^{^{'}} (0) = 7

\frac{\partial}{\partial x} (2 y (y^{2} + e^{2 x}))

\int_{0}^{N} e^{(- s t)} sin (2 t) d t

h \to 3 lim (\frac{3 ( x + h ) ^{3} - 3 x ^{3}}{h})

\int_{- 1}^{2} \frac{2 x - 7}{x ^{3}} d x

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