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y''+y=0

I come up with the solutions of y_{1}=c_{1}e^{ix}, y_{2}=c_{2}e^{-ix}. Now, using these I try to find the cooresponding real-valued solutions:

e^{ix}=cos(x)+isin(x)

e^{-ix}=cos(x)-isin(x)

Both of which have real-valued solutions cos(x). However, when I looked this up online, Wikipedia stated that the answer was y=c_{1}sin(x)+c_{2}cos(x). Am I doing something wrong here? Thanks.

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# Homework Help: 2[sup]nd[/sup] Order ODE

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