what is the solution of the associated homogeneous equation y'' + y = 0

Answer: [tex]y(t)=Acos(t)+Bsin(t)[/tex]Step-by-step explanation:To find the solution of a given differential equation ay''+by'+cy=0, a≠0, you have to consider the quadratic polynomial ax²+bx+c=0, called the characteristic polynomial.  Using the quadratic formula, this polynomial will always have one or two roots, for example r and s. The general solution of the differential equation is:[tex]y(t)= Ae^{rt}+Be^{st}[/tex] , if the roots r and s are real numbers and r≠s.[tex]y(t)= A e^{rt}+B*t*e^{rt}[/tex] , if r=s is real.[tex]y(t)=Acos(\beta t)e^{\alpha t} +Bsin(\beta t)e^{\alpha t}[/tex] , if the roots r and s are complex numbers α+βi and  α−βi.In this case, the characteristic polynomial is:[tex]x^{2} +1=0\\x^{2} =-1\\x1=i; x2=-i[/tex]Since the roots are complex numbers, with α=0 and β=1, then the answer is: [tex]y(t)=Acos(t)+Bsin(t)[/tex]