What is a solution to the differential equation #dy/dx=2e^(x-y)# with the initial condition #y(1)=ln(2e+1)#?

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Answer 1 · 2016-07-31T11:46:47

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

this is separable

#dy/dx=2e^(x-y) = 2e^(x)e^(-y)#

So

#e^y dy/dx=2e^(x)#

#int \ e^y dy/dx \dx =int \ 2e^(x) \ dx#

#int \d/dx( e^y) \dx =int \ 2e^(x) \ dx#

#e^y = 2e^(x) + C#

#y = ln ( 2e^(x) + C)#

applying IV: #y(1)=ln(2e+1)#

#ln(2e+1) = ln ( 2e + C) implies C = 1#

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