# Lecture 8 - Second Order Equations Complex Roots of the...

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Second Order Equations Complex Roots of the Characteristic Equation

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Recall the Characteristic Equation Remember, to solve y 00 y 0 y + + = 0 A B C We find roots of the characteristic equation ( A r 2 + B r + C ) Find Write the general solution: r 1 and r 2 y = C 1 e r 1 t + C 2 e r 2 t
Recall the Characteristic Equation We find roots of the characteristic equation ( A r 2 + B r + C ) r = - B ± p B 2 - 4 AC 2 A In general, using quadratic formula This is fine if the discriminant B 2 - 4 AC > 0

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Recall the Characteristic Equation This is fine if the discriminant B 2 - 4 AC > 0 But what if B 2 - 4 AC < 0 ? Then p B 2 - 4 AC is imaginary! This results in imaginary or complex roots.
Example y 00 - 2 y 0 + 2 y = 0 r = 2 ± p 4 - 4 (1) (2) 2 (1) r 2 - 2 r + 2 = 0 r 1 = 1 + 1 i Has the characteristic equation And roots r 2 = 1 - 1 i Which are:

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But what to do? y 00 - 2 y 0 + 2 y = 0 r 1 = 1 + 1 i r 2 = 1 - 1 i Well, let’s just plug it in y = C 1 e ( 1+1 i ) t + C 2 e ( 1 - 1 i ) t That’s pretty ugly! How do we get rid of those imaginary numbers?
Euler’s Magic Formula e i = cos ( ) + i sin ( ) Remember Euler’s Magic Formula y 00 - 2 y 0 + 2 y = 0 So for Our Problem y = C 1 e ( 1+1 i ) t + C 2 e ( 1 - 1 i ) t e ( 1+1 i ) t = e t e i t = e t (cos ( t ) + i sin ( t )) We can rewrite: e ( 1 - 1 i ) t = e t e - i t = e t (cos ( - t ) + i sin ( - t )) = e t (cos ( t ) - i sin ( t ))

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Euler’s Magic Formula y 00 - 2 y 0 + 2 y = 0 So for Our Problem y = C 1 e ( 1+1 i ) t + C 2 e ( 1 - 1 i ) t e ( 1+1 i ) t = e t e i t = e t (cos ( t ) + i sin ( t )) We can rewrite: e ( 1 - 1 i ) t = e t e - i t = e t (cos ( - t ) + i sin ( - t )) = e t (cos ( t ) - i sin ( t )) which gives us y = + e (1+1 i ) t C 1 C 2 e (1 - 1 i ) t
Euler’s Magic Formula y 00 - 2 y 0 + 2 y = 0 So for Our Problem y = C 1 e ( 1+1 i ) t + C 2 e ( 1 - 1 i ) t e ( 1+1 i ) t = e t e i t = e t (cos ( t ) + i sin ( t )) We can rewrite: e ( 1 - 1 i ) t = e t e - i t = e t (cos ( - t ) + i sin ( - t )) = e t (cos ( t ) - i sin ( t )) which gives us y = C 1 e t e i t + C 2 e t e - i t

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Euler’s Magic Formula y 00 - 2 y 0 + 2 y = 0 So for Our Problem y = C 1 e ( 1+1 i ) t + C 2 e ( 1 - 1 i ) t e ( 1+1 i ) t = e t e i t = e t (cos ( t ) + i sin ( t )) We can rewrite: e ( 1 - 1 i ) t = e t e - i
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