Respuesta :

caylus
Hello,

[tex] \dfrac{dy}{dx} = \dfrac{e^x}{e^y} \\ e^y*dy=e^x*dx \\ e^y=e^x+C\\ if\ x=0\ then\ y=ln(3) ==\ \textgreater \ 3=1+C==\ \textgreater \ C=2\\ \boxed{y=ln(2+e^x)}\\ [/tex]
 dy/dx=e^(x-y) 
dy/dx = e^(x)/e^(y) 
e^(y)dy = e^(x)dx 
e^(y) = e^(x) + C 
y = ln[C + e^(x)]
 y(0) = ln 3 to solve for C in that equation
ln 3 = ln (e^0 + C) ln 3
= ln (1 + C) 3
 = 1 + C
    C = 2
y = ln (e^x + 2)
hope it helps

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