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lecture_26 (dragged) 2

lecture_26 (dragged) 2 - A t 3 e t 4 e t = Y(3 − 3 Y(2 3...

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MA 36600 LECTURE NOTES: MONDAY, MARCH 30 3 This is known as the Method of Undetermined Coe cients . (Compare with Lecture #19 from Monday, March 2.) Example. We wish to find the general solution to the di ff erential equation y (3) 3 y (2) + 3 y (1) y = 4 e t . First, we find a fundamental set of solutions { y 1 , y 2 , y 3 } to the homogeneous equation y (3) 3 y (2) + 3 y (1) y = 0 . Since this has constant coe cients, we guess that there are solutions in the form y = e rt . The characteristic polynomial is Z ( r ) = r 3 3 r 2 + 3 r 1 = r 1 3 . Hence there is only p = 1 distinct real root and q = 0 pairs complex roots, so that we may choose y 1 ( t ) = e t y 2 ( t ) = t e t y 3 ( t ) = t 2 e t = y c ( t ) = C 1 y 1 ( t ) + C 2 y 2 ( t ) + C 3 y 3 ( t ) = C 1 + C 2 t + C 3 t 2 e t . Second, we find a particular solution Y = Y ( t ) to the nonhomogeneous equation. We may guess a solution Y ( t ) = A 0 + A 1 t + A 2 t 2 + A 3 t 3 e t . Instead, we try guess of a monomial, say Y ( t ) = A e t : 4 e t = Y (3) 3 Y (2) + 3 Y (1) Y = A e t 3 A e t + 3 A e t A e t = 0 . It is easy to see that this does not work because A e t = A y 1 ( t ). Similarly, the guesses A t e t = A y 2 ( t ) and A t 2 e t = A y 3 ( t ) will not work, so we guess instead
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Unformatted text preview: A t 3 e t : 4 e t = Y (3) − 3 Y (2) + 3 Y (1) − Y = ° A t 3 e t + 9 A t 2 e t + 18 A t e t + 6 A e t ± − 3 ° A t 3 e t + 6 A t 2 e t + 6 A t e t ± + 3 ° A t 3 e t + 3 A t 2 e t ± − ° A t 3 e t ± = 6 A e t = ⇒ A = 2 3 . Hence the general solution to the nonhomogeneous equation is y ( t ) = y c ( t ) + Y ( t ) = ² C 1 + C 2 t + C 3 t 2 ³ e t + 2 3 t 3 e t . We mention in passing that where is another way to Fnd this general solution. We express the di±erential equation using operator notation: y (3) − 3 y (2) + 3 y (1) − y = 4 e t = ⇒ ´ µ d dt − 1 ¶ 3 · y = 4 e t . Hence we Fnd that ´ µ d dt − 1 ¶ 4 · y = 4 µ d dt − 1 ¶ e t = 0 = ⇒ yt ) = ² C 1 + C 2 t + C 3 t 2 + C 4 t 3 ³ e t . Hence we may as well have guessed Y ( t ) = C 4 t 3 e t , so that C 4 = 2 / 3....
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