Posted by Soroban on September 23, 2002 at 14:02:37:
In Reply to: 2nd order linear DE... need help. (Prob #1) posted by colin on September 23, 2002 at 11:12:12:
: I'm stuck on these 2nd order linear DEs.
: 1. Using substitution v = y', v' = y'' to get an FODE
: y'' + t(y')^2 = 0
: v' + tv^2 = 0
: dv/dt = -tv^2
: dv/v^-2 = -t dt
: -v^-1 = -1/2(t^2) + c1
: v = 2/(t^2 +c1)
: let k^2 = c1
: v = 2/(t^2 + k^2)
: v = 2(k^2)/[(t^2/k^2) + 1]
You factored out k^2 in the denominator.
v = 2/(k^2)(t^2/k^2 + 1)
: v = dy/dt = 2(k^2)/[(t^2/k^2) + 1]
Here it is, Colin!
u = t/k. You forgot the du.
: y = 2(k^2)arctan(t/k) + c2
: The book says ans: y = 2k*arctan(t/k) + c2
: How did they get 2k and I got 2k^2?
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