What is the solution for y^'-2yy^'+e^x=0 ?

The solution for y^'-2yy^'+e^x=0 is y=-(-1+sqrt(1+4(-c_{1)+e^x)})/2 ,y=-(-1-sqrt(1+4(-c_{1)+e^x)})/2

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y^'-2yy^'+e^x=0

y^{^{'}} - 2 y y^{^{'}} + e^{x} = 0

y = - \frac{- 1 + 1 + 4 ( - c _{1} + e ^{x} )}{2}, y = - \frac{- 1 - 1 + 4 ( - c _{1} + e ^{x} )}{2}

Solution steps

y^{'} (x) - 2 y y^{'} (x) + e^{x} = 0

Solve separable ODE:

y = - \frac{- 1 + 1 + 4 ( - c _{1} + e ^{x} )}{2}, y = - \frac{- 1 - 1 + 4 ( - c _{1} + e ^{x} )}{2}

y = - \frac{- 1 + 1 + 4 ( - c _{1} + e ^{x} )}{2}, y = - \frac{- 1 - 1 + 4 ( - c _{1} + e ^{x} )}{2}

x e^{(x^{2})} + y y^{^{'}} = 0

y^{^{'}} = e^{- 5 y} (2 7 x - 4), y (0) = 1

y^{^{'}} = \frac{y ^{2} + 3}{( 2 x y )}

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

ln (x) y^{^{'}} - (\frac{2 \cdot y}{x}) = 0

What is the solution for y^'-2yy^'+e^x=0 ?
The solution for y^'-2yy^'+e^x=0 is y=-(-1+sqrt(1+4(-c_{1)+e^x)})/2 ,y=-(-1-sqrt(1+4(-c_{1)+e^x)})/2