ComplexRoots

# Lets focus on r 1 and rewrite it as r 1 α iβ in

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are both complex numbers. Let’s focus on r 1 and rewrite it as r 1 = α + , in terms of its real and imaginary parts. We can easily check that e r 1 x = e ( α + ) x = e αx e iβx = e αx (cos( βx ) + i sin( βx ) = e αx cos( βx ) + i e αx sin( βx ) is a complex-valued solution to our differential equation (just take derivatives and plug them in). It follows that both the real and imaginary parts are solutions. This gives us the two real-valued solutions: y 1 ( x ) = e αx cos( βx ) and y 2 ( x ) = e αx sin( βx ) . One of your homework problems is to show W [ y 1 , y 2 ] = βe 2 αx . This is nonzero for all x since e 2 αx > 0 and β 6 = 0 (since otherwise our roots would be real and we are assuming they are not).

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Hence by a previous theorem, the general solution in the case of complex roots is y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) = c 1 e αx cos( βx ) + c 2 e αx sin( βx ) or after factoring, y ( x ) = e αx c 1 cos( βx ) + c 2 sin( βx ) where r 1 = α + is a complex root of the characteristic equation (it doesn’t matter which one). I Example Solve the IVP: 2 y 00 - y 0 + 3 y = 0; y (1) = 1 , y 0 (1) = 1 Solution We first note that b 2 - 4 ac = 1 - 4(2)(3) = - 23 < 0 , so we are in the case of complex roots. Using the quadratic formula we compute the roots of the characteristic equation 2 r 2 - r + 3 = 0 to be r = 1 ± - 23 4 = 1 4 ± i 23 4 Thus we have α = 1 / 4 and β = 23 / 4 . Then our general solution is y ( x ) = e x/ 4 c 1 cos 23 4 x !
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• Fall '08
• staff
• Complex Numbers, Cos, Complex number, RHS, general solution, characteristic equation

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