Question

URGENT!!! Just realized my answer was incorrect.

if I have integral 1/(sqrt(1+v^2)dv = integral1/x dx
where v = y/x
I simplified the equation to
arcsinhv = lnx +c

With the initial condition at point (2,0), how do I find arbitrary constant and get y explicitly in terms of x?

Answer 1

my bad. its sinh not sin.

modifying answer.

Answer 2

plug in (x, y) = (2, 0) and solve for C

so arcsinh(0/2) = ln(2) + C

C = - ln(2)

and you can plug that C in your solution:

you get:

\[\sinh^{-1}\left( \frac{ y }{ x } \right) = \ln(x) - \ln(2)\]\[\frac{ y }{ x } = \sinh[\ln(x) - \ln(2)]\]\[y = xsinh[\ln(x) - \ln(2)]\]

Answer 3

THANKS! LIFESAVER!

Answer 4

happy to help ^_^