Differential Equation - general solution

$$36y'' - y = 0$$ $$y(-6) = 1, y'(-6) = -1$$

This can be solved by characteristic equation $$36r^2 - 1 = 0$$ which yields $r = \frac{1}{6} , -\frac{1}{6}$

The general solution of this is : $$y = C_1e^{-\frac{1}{6}t} + C_2e^{\frac{1}{6}t}$$

By my understanding, I can then use the initial conditions to solve the constants.

However, according to the solution in my textbook, the general solution is rewritten (from step above) as :

$$y = \frac{d_1e^{-\frac{t+6}{6}}}{6} + \frac{d_2e^{\frac{t+6}{6}}}{6}$$

and this is where I don't understand.

1. Why do I need to rewrite my general solution in a different form?
2. I cannot see how the second equation is derived.
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Let us first check that the rewritten equation is correct.

If we simplify, for example, $\frac{1}{6}d_1e^{-\frac{t+6}{6}}$, we get $\frac{1}{6}d_1e^{-1}e^{-t/6}$. And we can think of $d_1$ as being defined by $C_1=\frac{1}{6}d_1e^{-1}$.

Remark: The rewriting was done to make the substitutions of $t=-6$ more pleasant. It succeeds, but is certainly not necessary. So the answer to your first question is that you do not need to rewrite the equation in that style.

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@stackplasm They are technically equal. If you let your constants $d_1$ and $d_2$ absorb $\frac{1}{6e}$ and $\frac{e}{6}$ respectively you can retrieve your equation $y = C_1 e^{-t/6} + C_2 e^{t/6}$.
Edit: As per Andre's comment, it seems that the rewrite does make the substitution for $t = -6$ more convenient by letting both exponentials to simplify to zero such that we have $e^0$ for both the term preceded by $C_1$ and by $C_2$.