What is the solution for (1+e^x)*y*y^'=e^x,y(0)=1 ?

The solution for (1+e^x)*y*y^'=e^x,y(0)=1 is y=sqrt(2ln(1+e^x)+1-2ln(2))

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(1+e^x)yy^'=e^x,y(0)=1

(1 + e^{x}) \cdot y \cdot y^{^{'}} = e^{x}, y (0) = 1

y = 2 ln (1 + e^{x}) + 1 - 2 ln (2)

Solution steps

(1 + e^{x}) \cdot y \cdot y^{'} (x) = e^{x}

Solve separable ODE:

y = 2 ln (1 + e^{x}) + 1 - 2 ln (2)

y = 2 ln (1 + e^{x}) + 1 - 2 ln (2)

y^{^{'}} = 0.05 y (1 - \frac{y}{4000})

- \frac{d h}{d t} = kh (t)^{n}

(x + \frac{x}{y}) d x - d y = 0

\frac{d y}{d x} y = sin^{3} (cos (9 x))

\frac{d y}{d t} 400 = - \frac{π}{9} \cdot 8 y

What is the solution for (1+e^x)*y*y^'=e^x,y(0)=1 ?
The solution for (1+e^x)*y*y^'=e^x,y(0)=1 is y=sqrt(2ln(1+e^x)+1-2ln(2))