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Question about solving ODE with Complex eigenvalue

 
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Sep11-12, 10:37 AM   #1
 

Question about solving ODE with Complex eigenvalue


For example,
ODE: y'' + y = 0
solve this problem using MAPLE
f(x) = _C1*sin(x)+_C2*cos(x)

My question is Eigenvalue for D^2+1=0 is +i, -i
so general solution is f(x) = C1*exp(i*x)+C2*exp(-i*x)
according to Euler's formula f(x) = C1( cos(x)+i*sin(x) ) + C2*( cos(x)-i*sin(x) )
it is different from the general solution generated by MAPLE
why?

Thanks!
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Sep11-12, 12:23 PM   #2

Math 2012
 
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Your solution is the general solution assuming f(x) is complex, and your constants C1 and C2 are also complex. You can rearrange it as
f(x) = (C1 + C2) cos(x) + i (C1 - C2) sin(x)
or
f(x) = A1 cos(x) + A2 sin(x)
wherne A1 and A2 are complex constants.
If course if you want to restrict f(x) to be a real function, A1 and A2 must be real. That condition is the equivalent to C1 and C2 being complex conjugates, so that C1 + C2 is real and C1 - C2 is imaginary.
Sep11-12, 03:42 PM   #3
 
AlephZero, thank you! Really helpful!

Quote by AlephZero View Post
Your solution is the general solution assuming f(x) is complex, and your constants C1 and C2 are also complex. You can rearrange it as
f(x) = (C1 + C2) cos(x) + i (C1 - C2) sin(x)
or
f(x) = A1 cos(x) + A2 sin(x)
wherne A1 and A2 are complex constants.
If course if you want to restrict f(x) to be a real function, A1 and A2 must be real. That condition is the equivalent to C1 and C2 being complex conjugates, so that C1 + C2 is real and C1 - C2 is imaginary.
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