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**TshingY** i need help on how to solve this differential equation

y'' = y' * e^y

initial conditions

y(0) = 0

y'(0) = 2

first i use u = y' and y'' = u*du/dy substitution to get

u*du/dy = u * e^y ---> du/dy = e^y

separating variables and integrating both sides give

u = e^y + c1 (*)

substituting y' back into u and separating variables give

dy / (e^y + c1) = dx

integrating gives

[ln(e^y) - ln(e^y + c1)] / c1 = x + c2

y - ln(e^y + c1) = c1*(x + c2) (**)

from here i am stuck. I can't get it to be in the form of y =