how to solve: dy/dx=e^(x-y), y(0)=ln3
Hello,\(\dfrac{dy}{dx} = \dfrac{e^x}{e^y} \\ e^y*dy=e^x*dx \\ e^y=e^x+C\\ if\ x=0\ then\ y=ln(3) ==\ \textgreater \ 3=1+C==\ \textgreater \ C=2\\ \boxed{y=ln(2+e^x)}\\\)
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 equationln 3 = ln (e^0 + C)ln 3 = ln (1 + C)3 = 1 + C C = 2 y = ln (e^x + 2)hope it helps
