What is the solution for (y^')/(2yln(y))=-1/(e^x) ?

The solution for (y^')/(2yln(y))=-1/(e^x) is y=e^{e^{2/(e^x)+c_{1}}}

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(y^')/(2yln(y))=-1/(e^x)

\frac{y ^{^{'}}}{2 y ln ( y )} = - \frac{1}{e ^{x}}

y = e^{e^{\frac{2}{e ^{x}} + c_{1}}}

Solution steps

\frac{y ^{'} ( x )}{2 y ln ( y )} = - \frac{1}{e ^{x}}

Solve separable ODE:

y = e^{e^{\frac{2}{e ^{x}} + c_{1}}}

y = e^{e^{\frac{2}{e ^{x}} + c_{1}}}

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