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How do you solve #y=lny'# given y(2)=0?

Answer 1

Elijah Abram

# y = ln(1/(3-x)) #

A am assuming that #y# is a function of #x#, ie #y=f(x)#

# y=lny' # # :. ln(dy/dx)=y # # :. dy/dx=e^y #

We now have a first order separable differential equation, so we can separate the variables to get:

# \ \ \ \ \ \ int 1/e^ydy = int dx # # :. int e^-ydy = int dx #

We can now integrate to get:

# -e^-y = x+c # # :. e^-y = -x-c #

Using #y(2)=0 => #

#e^0=-2+c# # :. x=1+2=3 #

# :. e^-y = 3-x # # :. -y = ln(3-x) # # :. y = -ln(3-x) # # :. y = ln(1/(3-x)) #

CHECK

# y = ln(1/(3-x)) # # :. y(2) = ln(1/1)=0#, so the initial condition is met

And

# dy/dx = 1/((1/(3-x))) * (-(3-x)^-2) (-1) # # :. dy/dx = (3-x) * 1/(3-x)^2# # :. dy/dx = 1/(3-x)# # :. ln(dy/dx) = y#, so the DE condition is met

Confirming our solution is correct.

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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.