I read this as y"+4y=0 where y" is d^2y/dx^2 and y' is dy/dx. We start with the simple fact that the differential of sine is cosine and the differential of cosine is minus sine, so we have a repeating pattern in the differential. Furthermore, when we differentiate twice we also get a change of sign from which we can see that the double differential of sine or cosine is minus sine or cosine. This is the main clue in solving the differential equation. Let y=bsin(ax), y'=abcos(ax) and y"=-a^2bsin(ax)=-a^2y. So y"+a^2y=0. (Notice that the constant b is carried through unaffected by the differentiation.) Comparing this with y"+4y=0, a^2=4 so a is -2 or +2.
Check:
a=2
y=bsin(2x)
y'=2bcos(2x)
y"=-4bsin(2x)=-4y, y"+4y=0
a=-2
y=bsin(-2x)=-bsin(2x)
y'=-2bcos(2x)
y"=4bsin(2x)=-4y, y"+4y=0
However, the same result applies if y=bcos(2x). If we have two functions y1 and y2 we can put y=y1+y2. It's true that y'=y1'+y2' and y"=y1"+y2". So we can let y1=bsin(2x) and y2=ccos(2x), making y=bsin(2x)+ccos(2x); y"=-4bsin(2x)-4ccos(2x)=-4y, so the general answer is y=Asin(2x)+Bcos(2x), where A and B are constants.