second order differential equation possible solution I have a second order differential equation:
$$y''=-a^2y(t)$$
I search some possible solution such as:
$$y(t)=\sin(at)$$ or $$y(t)=\cos(at)$$
Is $y(t)= 0 $  also a possible solution or not?
 A: $y(t)=0$ is called the trivial solution. So the easy solution which doesn't really represent anything interesting. But since it solves the differential equation it is a solution. Also, when you would find possible solutions involving $\sin$ and $\cos$ (or exponentials) there will be arbitrary constants $c_1$ and $c_2$ multiplying them. You can view the trivial solution as just setting these constants equal to 0.
A: So, $D^2y+a^2y=0\implies D=\pm i\cdot a$
So if $a\ne0,y=Ae^{i\cdot ax}+Be^{-i\cdot ax}$ where $A,B$ are arbitrary constants.
So, $y=A(\cos ax+i\sin ax)+B(\cos ax-i\sin ax)=(A+B)\cos ax+i(A-B)\sin ax$
If $a=0,D=0\implies y=(A+Bx)e^{ix\cdot0}=A+Bx$ as both the are equal.
Any constant set of values of $A,B$ this will satisfy $y''+a^2y=0$
A: It depends on what you you means with the solution. For a  mathematician, second order differential equation has an infinity number of solutions because any linear combination of two particulate solutions is a solution, too. However, for physicist, there exist only one solution due to two boundary conditions: $y(x_0)=y_0$ and $y'(x_0)=y'_0$. Thus, depending on the given boundary conditions, $y=0$ could be possibly not a solution.
