What is the solution for (1+e^x)yy^'=e^x ?

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

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(1+e^x)yy^'=e^x

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

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

Solution steps

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

Solve separable ODE:

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

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

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