Since you have ln y = (e^x)+c, you need to free the y from the ln. How you do that raise everything to the power of e.
From that, you obtain e^(ln y) = e^[(e^x)+c].
y = e^[(e^x)+c]
The last step before substitution is understanding that e^(a+b) can be written as (e^a)*(e^b) and that when you have e^c is just another constant and can be written just as C
I made sure to mention how exponents work in my post but here is it again.
For example if we have 2^(2+x), then we could write (2^2)*(2^x). This works because 2*2=(2^1)*(2^1)=2^(1+1)=2^2=2*2.
Therefore, e^[e^x + c]= e^(e^x)*e^c and e^c is just another constant so e^c=C. Now we have (e^(e^x))*c=C*(e^(e^x).
y= C*(e^(e^x))
y(0)=C*(e^(e^0))=2*e
e^0=1
I think you should be able to obtain the correct solution from there.