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

2. Write is as: $\frac{dy'}{\left(1+(y')^2\right)^{3/2}}=k$. Now, integrate that with respect to y' on the left. I get:

$\frac{y'}{\sqrt{1+(y')^2}}=kx+c_1$

Square it then isolate $(y')^2$ with long division and solve for $(y')^2$:

$(y')^2=\frac{1}{1+(kx+c_1)^2}-1$

You can finish that right? Just take square roots, get two equations, then solve those.

3. Originally Posted by shawsend
Write is as: $\frac{dy'}{\left(1+(y')^2\right)^{3/2}}=k$. Now, integrate that with respect to y' on the left. I get:

$\frac{y'}{\sqrt{1+(y')^2}}=kx+c_1$

Square it then isolate $(y')^2$ with long division and solve for $(y')^2$:

$(y')^2=\frac{1}{1+(kx+c_1)^2}-1$

You can finish that right? Just take square roots, get two equations, then solve those.
thanks for the help