What is a particular solution to the differential equation #dy/dx=e^(x-y)# with #y(0)=2#?

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Answer 1 · 2016-07-18T19:10:30

# y =ln( e^x + e^2 - 1 )#

this is separable

#dy/dx=e^(x-y) = e^x e^(-y)#

#e^(y)dy/dx= e^x #

#int \ e^(y)dy/dx \ dx=int \ e^x \ dx#

#int \ e^(y) \ dy =int \ e^x \ dx#

# e^(y) = e^x + C#

#y(0) = 2 implies e^2 = 1 + C implies C = e^2 - 1#

# e^(y) = e^x + e^2 - 1#

# y =ln( e^x + e^2 - 1 )#