$y' = \int \sqrt{1 + (y')^{2}}$

By Yatir Halevi on Monday, February 18, 2002 - 06:09 pm:

Please help me solve these (describe your steps):
1)
$y' = \int \sqrt{1 + (y')^{2}}$
2)
$\int y = \int \sqrt{1 + (y')^{2}}$
Thanks,
Yatir

P.s.
I already found out that:
$y = \int \sqrt{1 + (y')^{2}}$ Is not solvable

By Andre Rzym on Tuesday, February 19, 2002 - 09:19 am:

If I read (2) correctly, we can differentiate and square to get:

y² =1+(y')²

This would appear to have the solution y=cosh(x).

Andre

By David Loeffler on Tuesday, February 19, 2002 - 01:31 pm:

And for the first, we get

y'' = \sqrt{1 + y < quote / >^{2}}

, so if

v = y'

dv / dx = \sqrt{1 + v^{2}}

. So

\int dv / (1 + v^{2})^{1 / 2} = \int dx

\sinh^{- 1} v = x + c

, so

v = \sinh (x + c)

y = \cosh (x + c) + d

for arbitrary

c

d