What is the solution for y^'(e^x-1)ln(y)=y ?

The solution for y^'(e^x-1)ln(y)=y is y=e^{sqrt(2x+c_{1)}},y=e^{-sqrt(2x+c_{1)}}

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y^'(e^x-1)ln(y)=y

y^{^{'}} (e^{x} - 1) ln (y) = y

y = e^{2 x + c_{1}}, y = e^{- 2 x + c_{1}}

Solution steps

y^{'} (x) ln (y) = y

Solve separable ODE:

y = e^{2 x + c_{1}}, y = e^{- 2 x + c_{1}}

y = e^{2 x + c_{1}}, y = e^{- 2 x + c_{1}}

- \frac{16}{( 8 x + 9 ) ^{3}}

\frac{d y}{d x} = \frac{y ^{2} + y}{x ^{2} + x}

\int_{0}^{1} x e^{- 2 x} d x

y = 4 x^{2} - 1, y = 0, x = \frac{- 1}{2}, x = 1

\int \frac{x + 25}{x ^{2} + 24 x + 148} d x

What is the solution for y^'(e^x-1)ln(y)=y ?
The solution for y^'(e^x-1)ln(y)=y is y=e^{sqrt(2x+c_{1)}},y=e^{-sqrt(2x+c_{1)}}