Second order ODE: finding solution.

[knockout_artist]

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 ?

 

[phyzguy]

is that correct ?

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).

 

[Ray Vickson]

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 ?

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.

 

[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 $y'(x)=-(\frac{L'(x)x}{2\sqrt{2L(x)}}+\sqrt{\frac{L(x)}{2}})sin(\sqrt{\frac{L(x)}{2}}x)+(\frac{L'(x)x}{2\sqrt{2L(x)}}+\sqrt{\frac{L(x)}{2}})cos(\sqrt{\frac{L(x)}{2}}x)$.