Kashmir said:
I don't know. I tried searching on the internet but didn't find it. Can you tell me what that means?
Let's look at the general case. We have a function ##u(\theta)## which satisfies a differential equation. We then look at a substitution, which in general is of the form ##\theta = f(\alpha)## or ##\alpha = g(\theta)##. Typical examples might be ##\theta = a\alpha + b## or ##\alpha = \cos(\theta)##. In this case we have the substitution ##\alpha = -\theta##.
Note that if you don't "see" what's happening, then it can be confusing to try to work with variables ##\theta## and ##-\theta##, so it's conceptually clearer to let ##\alpha = -\theta##.
A second point is that physicists especially tend to use the same function for both the original ##u(\theta)## and the new ##u(\alpha)##. This is a notational shortcut, but in fact we really want to look at a new function, which we'll call ##v##, where: $$v(\alpha) = u(\theta) = u(f(\alpha))$$ This shows that ##v## is not the same mathematically defined function as ##u##, but the composition of ##u \circ f##. That said, it's seen by physicists as the same "physical" function, so usually they don't bother with this subtlety.
Then, we have to look for the equation that is satisfied by the function ##v(\alpha)## - and it may or may not take the same form as the original equation. Often these substitutions are used to try to change the equation to one that is easier to solve. In this case, however, it is to show that the system has some sort of symmetry in the variable ##\theta##.
The full, rigorous way to tackle this, therefore, is to use the substitution ##\alpha = -\theta## and see what becomes of the original equation. We can start with the first term and look at the second derivative:
$$\frac{dv}{d\alpha} = \frac{dv(\alpha)}{d\alpha} = \frac{dv(\alpha)}{d\theta}\frac{d\theta}{d\alpha} = \frac{du(\theta)}{d\theta}(-1) = -\frac{du}{d\theta} $$
And, if we take the second derivative, we get another factor of ##-1##, which cancels out and we find that:
$$\frac{d^2v}{d\alpha^2} = \frac{d^2u}{d\theta^2} $$
I'll leave you to work on the terms on the RHS.
As I said, this is the full rigorous way to see what's happening. It's certainly quicker just to say that it's obvious!