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]
v = dy/dt = 2(k^2)/[(t^2/k^2) + 1]
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?