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MissTica
3 years ago
14

How to solve: dy/dx=e^(x-y), y(0)=ln3

Mathematics
2 answers:
ella [17]3 years ago
8 0
Hello,

\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)}\\


OverLord2011 [107]3 years ago
8 0
 <span>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)]
</span><span> y(0) = ln 3 to solve for C in that equation
</span><span>ln 3 = ln (e^0 + C) ln 3
= ln (1 + C) 3
 = 1 + C
    C = 2
</span><span> y = ln (e^x + 2)
</span>hope it helps
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