Write is as: . Now, integrate that with respect to y' on the left. I get:
Square it then isolate with long division and solve for :
You can finish that right? Just take square roots, get two equations, then solve those.
hello,
I am trying to solve
K = y''/{[1+(y')^2]}^(3/2), K=const.
i see that there is no y term so i make the substitution y'=p => y''=pp'
K = pp'/[1+(p^2)]^(3/2)
then i'm thinking maybe... 1+(p^2)=u => 2pp'=u'
K/2 = u'/ [u^(3/2)] => (Kx/2)+C = -2/[u^(1/2)] =>
u = 4/[(Kx/2)+C]^2 = 1+(p^2) => p = {{4/[(Kx/2)+C]^2}-1}^(1/2) = y' =>
y = integr of {{4/[(Kx/2)+C]^2}-1}^(1/2)
which is honestly quite unwieldy