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

Answer 1

Charles Anctil

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

this is separable

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

#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)#

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Answer 2

Emma Aldridge

The solution to the given differential equation ( \frac{dy}{dx} = 2e^{x-y} ) with the initial condition ( y(1) = \ln(2e+1) ) is:

[ y(x) = -\ln\left(1 + \frac{2e}{e^x}\right) ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.