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

cubanwongux 2022-09-13 Answered
What is a solution to the differential equation d y d x = 2 e x - y with the initial condition y ( 1 ) = ln ( 2 e + 1 ) ?
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Answers (1)

Conner Singleton
Answered 2022-09-14 Author has 13 answers
this is separable
d y d x = 2 e x - y = 2 e x e - y
So
e y d y d x = 2 e x
  e y d y d x d x =   2 e x   d x
d d x ( e y ) d x =   2 e x   d x
e y = 2 e x + C
y = ln ( 2 e x + C )
applying IV: y ( 1 ) = ln ( 2 e + 1 )
ln ( 2 e + 1 ) = ln ( 2 e + C ) C = 1
y = ln ( 2 e x + 1 )

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