Posted by Subhotosh Khan on September 29, 2002 at 16:11:57:
In Reply to: ANYONE??? I'm still stuck on this... posted by Colin on September 29, 2002 at 00:32:10:
: : the book's answer is:
: : e^y = (t + c2)^2 + c1
: : Any clue how they arrived at that?
: : thx.
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Knowing the answer
Assume:
y = ln[f(t)]..............e^y = f(t)
y' = f'/f
y" = [f"*f -(f')^2]/f^2
D.E
y" + (y')^2 = 2e^-y
[f"*f -(f')^2]/f^2 + (f'/f)^2 = 2/f
f" = 2
f' = 2t + c1
f = t^2 + t*c1 + c2 = (t+c3)^2 + c4 = e^y