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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?

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  • $\begingroup$ If $u+iv$ solves the equation, then $u$ and $v$ solve the equations. $\endgroup$ – Dave Nov 12 '17 at 17:40
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$$\begin{aligned}y&=c_1(\cos(2x) + i\sin(2x)) + c_2(\cos(2x) - i\sin(2x))\\ &=(c_1+c_2)\cos(2x)+(c_1-c_2)i\sin(2x)\\ &=C_1\cos(2x)+C_2\sin(2x)\end{aligned}$$ where $C_1=c_1+c_2,C_2=(c_1-c_2)i\in\Bbb{C}$.

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well it is usually should be written like this $$e^{x\cdot Re}(C_1\cos (x \cdot Im) + C_2\sin (x \cdot Im))$$

if your roots have no real part it is obiviously can be written as

$C_1\cos (x \cdot Im) + C_2\sin (x \cdot Im)$

As you have done. If I understood your question correctly.

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  • $\begingroup$ thank you I apreciate it. I wasnt sure that the constants could be bundled together like that but $\endgroup$ – entercaspa Nov 12 '17 at 18:12

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