Homogeneous differential equations of second order I can not find other solutions of the equation
$$
y'' + 4y = 0
$$
addition of the solutions $y = 0$ and $y = \sin 2x$. There are positive solutions? Or solutions in terms of exponential function? Thanks.
 A: Hint: Note that $7\cos 2x$ is a solution, and so. is  $\sin 2x-\cos 2x$.
Remark: The characteristic equation $r^2+4=0$ has the solutions $r=\pm 2 i$. So the general solution is $Ae^{2ix} +Be^{-2ix}$, where $A$ and $B$ are arbitrary (not necessarily reaal) constants. Thus the solutions can be expressed in terms of complex exponentials. You may not have covered this material yet. 
A: So for these equations we take a guess that the solution is exponential form($e^{rt}$).
$$y''+4y=0$$
$$\frac {d^2} {dt^2} e^{rt}+4e^{rt}$$
$$r^2e^{rt}+4e^{rt}=0$$
$$e^{rt}(r^2+4)=0$$ 
$$r^2=-4$$
$$r=\pm2i$$
So the solution to your equation is $Ae^{\pm2ix}$.  If you covered this in class, you know that you can expand the exponential:
$$e^{i\alpha}=\cos(\alpha)+i\sin(\alpha)$$
So your general solution is thus
$$y(x)=A\cos(2x)+B\sin(2x)$$
The 2 is because of the exponential, and there is no i in front of the sin term because it gets  absorbed into the constant B.
You can then check that this works:
$$y''+4y=0$$
$$\frac {d^2} {dx^2} y +4y$$
$$-4A\cos(2x)-4B\sin(2x)+4A\cos(2x)+4B\sin(2x)=0$$
A: The characteristic equation is 
$$r^2+4=0$$
so its roots are $r_1=2i$ and $r_2=-2i$ and then the general solution is 
$$y(x)=A\cos(2x)+B\sin(2x)$$
