- #1
knockout_artist
- 70
- 2
Homework Statement
d2u/d2x + 1/2Lu = 0 where L is function of x
Homework Equations
I am try to find solutions y1 and y2 of this equation.
The Attempt at a Solution
y = [cos √(L/2) x] + [sin √(L/2) x]
y' = - [√(L/2) sin √(L/2) x] + [ √(L/2) cos √(L/2) x]
y'' = -[(L/2) cos √(L/2) x] - [(L/2) sin √(L/2) x ]so now we have
y = [cos √(L/2) x] + [sin √(L/2) x]
and if we multiply it with (1/2 L ) we get negative y'' which will satisfy original equation.
so solutions are
y1 = cos √(L/2) x
y2 = sin √(L/2) x
is that correct ?