Posted by Colin on September 23, 2002 at 16:11:11:
In Reply to: Re: 2nd order linear DE... need help. (Prob #2) posted by Soroban on September 23, 2002 at 14:14:14:
: : I'm also at an impasse on this prob:
: : yy'' + (y')^2 = 0
: : let v = y'
: : dv/dt = y'' = (dv/dy)(dy/dt) = v(dv/dy)
: : yv(dv/dy) + v^2 = 0
: : dv/dy = -v/y
: : dv/v = - dy/y
: : ln(v) = -ln(y) + c1
: Here it is, Colin! A little mishandling of logs.
: ln(v) = -ln(y) + ln(C)
: ln(v) = ln(C/y)
: v = C/y
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