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

1 Answer
Jul 31, 2016

y = ln ( 2e^x + 1)

Explanation:

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)