Clarification of a DE example

**sixstringartist** · October 23rd 2007, 04:46 PM

My book states that:
$y'' +4y = 0$
has a general solution of:
$y(t) = c_1\cos {2t} + c_2\sin{2t}$

I dont see how they achieve this since the initial equation has a characteristic form of:
$r^2 + 4 = 0$
$r_{1,2} = \pm 2$
$\therefore$
$y(t) = c_1e^{2t} + c_2e^{-2t}$

**Jhevon** · October 23rd 2007, 05:01 PM

Originally Posted by sixstringartist

My book states that:
$y'' +4y = 0$
has a general solution of:
$y(t) = c_1\cos {2t} + c_2\sin{2t}$

I dont see how they achieve this since the initial equation has a characteristic form of:
$r^2 + 4 = 0$
$r_{1,2} = \pm 2$
$\therefore$
$y(t) = c_1e^{2t} + c_2e^{-2t}$

no, you solutions are not correct, we have complex solutions here. your solutions work for the difference of two squares, you have the sum of two squares here. use the quadratic formula

**sixstringartist** · October 23rd 2007, 05:29 PM

Thank you. In my haste I was taking the sqrt of 4 and not -4. Thanks again.

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