Proving a second order ODE forms a vector space?

you certainly know that if b & c are constants & y1 ,y2 are independent solutions then c1 y1 + c2 y2 is a solution { DE is linear }...so y = a[y1 + y2 ] = a y1 + a y2 is a solution , (-1) y1 is a solution { inverse }and " y + ( -y) " = 0 is a solution...closed , has a zero , an inverse , associative , commutative , scalar multiple..don't know what your 1 , 4, 5,6 are...