Parallel RLC with zero input
They made a mistake in the book.

Change this equation:

$$L~C~ i'' + \frac{L}{R} i' + i = 0$$

To This equation:

$$L~C~ v'' + \frac{L}{R} v' + v = 0$$

That's why they didn't specify which current I was looking at. For example, is i(t) referring to the register, the inductor, or the capacitor..they didn't specify because the differential equation is really suppose to be a function of v instead of i.

I think they made a mistake with the initial conditions as well...

$$v(0) = 1$$

$$v'(0) = 0$$

When I make these changes, i get an answer of:

$$v(t) = e^{-t/4}\cos\bigg(\frac{\sqrt{15}}{4}t\bigg) + \frac{1}{\sqrt{15}}e^{-t/4}\sin\bigg(\frac{\sqrt{15}}{4}t\bigg)$$

well anyways... that makes the question clean.

one further note, i can rationalize the denominator of: 

$$\frac{1}{\sqrt{15}}$$

by multiplying by: 

$$\frac{\sqrt{15}}{\sqrt{15}}$$

this yields:

$$\frac{1}{\sqrt{15}} = \frac{\sqrt{15}}{15}$$

so i can rewrite the expression for v(t) as:

$$v(t) = e^{-t/4}\cos\bigg(\frac{\sqrt{15}}{4}t\bigg) + \frac{\sqrt{15}}{15} e^{-t/4}\sin\bigg(\frac{\sqrt{15}}{4}t\bigg)$$

$$v(t) = \frac{1}{15}\Bigg(15e^{-t/4}\cos\bigg(\frac{\sqrt{15}}{4}t\bigg) + \sqrt{15} e^{-t/4}\sin\bigg(\frac{\sqrt{15}}{4}t\bigg)\Bigg)$$

And that answer exactly matches the answer key in the back of the book, which confirms what i was saying about the mistakes in the question.