# Roots of a second order ODE

Trying to find all solutions on (-infinity,+infinity) for : $y''+4y = 0$

I know that the discriminant of the characteristic equation is -16 so the roots are complex. so $k=0.5 \cdot \sqrt{-16} = 2i$

$f_1(x) = e^{(2ix)} = \cos(2x) + i\sin(2x)$

$f_2(x) = e^{(-2ix)} = \cos(2x) - i\sin(2x)$

and so the general solution therefore is

$y=c_1(\cos(2x) + i\sin(2x)) + c_2(\cos(2x) - i\sin(2x))$

but the answers say that it is

$y=C_1\cos(2x) + C_2\sin(2x)$

So I am having trouble interpreting the real parts of the complex roots. Could someone please explain how to get to the answer from here?

• If $u+iv$ solves the equation, then $u$ and $v$ solve the equations. – Dave Nov 12 '17 at 17:40

\begin{aligned}y&=c_1(\cos(2x) + i\sin(2x)) + c_2(\cos(2x) - i\sin(2x))\\ &=(c_1+c_2)\cos(2x)+(c_1-c_2)i\sin(2x)\\ &=C_1\cos(2x)+C_2\sin(2x)\end{aligned} where $C_1=c_1+c_2,C_2=(c_1-c_2)i\in\Bbb{C}$.
well it is usually should be written like this $$e^{x\cdot Re}(C_1\cos (x \cdot Im) + C_2\sin (x \cdot Im))$$
$C_1\cos (x \cdot Im) + C_2\sin (x \cdot Im)$