# Second order ODE: finding solution.

1. Nov 27, 2017

### knockout_artist

1. The problem statement, all variables and given/known data

d2u/d2x + 1/2Lu = 0 where L is function of x

2. Relevant equations

I am try to find solutions y1 and y2 of this equation.

3. 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 ?

2. Nov 27, 2017

### phyzguy

No, not if L is a function of x. When you took the derivatives of your function y, you treated L as a constant, not as a function of x. If L is a function of x, there will be additional terms in the derivatives due to the chain rule. The solution will depend on the functional form of the function L(x).

3. Nov 27, 2017

### Ray Vickson

Correct if $L > 0$ is constant, but wrong if $L$ is a function of $x$, as you stated in the original question. For a general $L = L(x) > 0$ there may be no known formula for the solution, in which case you would need to solve the DE numerically.

4. Nov 27, 2017

### Delta²

Your solutions are correct only if L is a constant. IF L is not a constant then you have to apply the chain and product rule in order to calculate correctly the derivatives.
For example for the first derivative it would be $y'(x)=-(\frac{L'(x)x}{2\sqrt{2L(x)}}+\sqrt{\frac{L(x)}{2}})sin(\sqrt{\frac{L(x)}{2}}x)+(\frac{L'(x)x}{2\sqrt{2L(x)}}+\sqrt{\frac{L(x)}{2}})cos(\sqrt{\frac{L(x)}{2}}x)$.

Last edited: Nov 27, 2017