Question

y=y'+ln y' Find the solution of the differential equation


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Answer 1

Answer:

Y=y''(0) is The answer

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

y= y'+ ln y'

y' = ln e^ y'

y= ln e^y' + ln y'

ln ab = ln a+ ln b

y= ln( y' x e^y')

e^y = y'*e^y'

y' = W(e^y)

As per Lambert W function

x= ye^y => y = W(x)

y' =W(e^y)

dy

__ = W(e^y)

dx

dx=dy/ W(e^y)

integrating on both sides we get

x= ln W(e^y)- (1/W(e^y))+C

C is any constant

ln W(e^y) - 1/W(e^y) = x-C

W(e^y) = 1/W(e^(c-x))

y= -lnW(e^(C-x)) + 1/W(e^(C-x))