2nd order linear DE

Posted by colin on September 26, 2002 at 16:52:14:

This is one of those 2nd order linear DEs where the independent variable is
missing.
After making a transformation using v = y' the eq is supposed to be a
Bernoulli eq.

y'' + (y')^2 = 2e^-y

v' + v^2 = 2e^-y

v' = dv/dt = (dv/dy)(dy/dt) = v(dv/dy)

v(dv/dy) + v^2 = 2e^-y

dv/dy + v = 2(v^-1)(e^-y)

Is this a Bernoulli eq. with n = -1 ?

If so, then I'm letting r = v^(1-n) = r = v^2
dr/dy = 2v(dv/dy)

v(dv/dy) + v^2 = 2e^-y
1/2(dr/dy) + r = 2e^-y
r' + 2r = 4e^-y

integrating factor = e^2y
re^2y = 4e^y + c1
r = 4e^-y +c1*e^-2y
v = 4e^-y/2 + c1*e^-y
4e^y/2 + c1*e^y dy = dt
t + c2 = 8e^y/2 + c1*e^y

right about this time I realize that I'm nowhere near the book answer...
can some help point out my screw up?? thx...

Follow Ups:

• Re: 2nd order linear DE Subhotosh Khan 08:32:08 09/27/02 (3)
  • Re: book answer quite different... colin 11:00:18 09/27/02 (2)
    • ANYONE??? I'm still stuck on this... Colin 00:32:10 09/29/02 (1)
      • Re: ANYONE??? I'm still stuck on this... Subhotosh Khan 16:11:57 09/29/02 (0)

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