# Thread: Clarification of a DE example

1. ## Clarification of a DE example

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

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

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

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

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