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**2nd Order ODE**

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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