Why is it important to find both solutions to a second order linear differential equation? Given the equation $$y'' + y=0$$
A solution is $y=\sin(t)$
Why can't we stop there since we know a way to solve the system? Why should we consider all of the ways to solve the system? 
I would really like to see a real world example when having a single solution is inadequate. I know this is asking a lot, but I often find mathematics only becomes easier to understand once I need to use it to solve something and it becomes relate-able to real things. 
 A: Consider a point mass $m= 1 \ \mathrm{kg}$ attached to one end of a spring with spring constant $k = 1 \ \mathrm{N/m}$. Suppose that the spring is suspended vertically from an immovable support. The oscillations of the mass around its equilibrium position can then be described by the equation
$$y''+y = 0$$
where $'$ denotes time derivative, $y$ the displacement and where we measure distance in $\mathrm{m}$ and time in $\mathrm{s}$.
As you have said, $y = \sin t$ is a solution to this equation. For this solution, we have $y(0) = 0$ and $y'(0) = 1$. This means that the mass starts from rest with unit speed. Furthermore, the maximum displacement of the mass is $1$.
But what if the mass doesn't start from rest and what if it doesn't have unit speed? What if we release the mass $1.5 \ \mathrm{m}$ from its equilibrium at $t=0$ and with zero initial speed? Then your solution would be $y = 1.5 \cos t$. This is physically different from $y = \sin t$.
In order to be able to account for different initial conditions, you need the general solution, which can be written in many ways, one of which is $y = A\sin (t) + B\sin (t)$. We could also write this as $y = C\sin(t+ \phi)$, as $C\sin(t+\phi) = C\cos(\phi)\sin(t) + C\sin(\phi)\cos(t)$. However, this is still very different from simply $y = \sin t$.  
