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lecture_27 (dragged)

lecture_27 (dragged) - n τ y n − 1 1 τ y n − 1 2 τ ...

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MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 1 Higher Order Linear Equations Variation of Parameters. Recall that when n = 2, a particular solution to the diferential equation P 0 ( t ) Y °° + P 1 ( t ) Y ° + P 2 ( t ) Y = G ( t ) is given by the integral Y ( t )= ° t G ( τ ) P 0 ( τ ) K ( τ,t ) in terms oF the kernel K ( τ,t )= y 1 ( τ ) y 2 ( t ) y 1 ( t ) y 2 ( τ ) y 1 ( τ ) y ° 2 ( τ ) y ° 1 ( τ ) y 2 ( τ ) = ± ± ± ± ± y 1 ( τ ) y 2 ( τ ) y 1 ( t ) y 2 ( t ) ± ± ± ± ± ²± ± ± ± ± y 1 ( τ ) y 2 ( τ ) y ° 1 ( τ ) y ° 2 ( τ ) ± ± ± ± ± . In general, consider the n th order nonhomogeneous linear diferential equation P 0 ( t ) d n Y dt n + P 1 ( t ) d n 1 Y dt n 1 + ··· + P n 1 ( t ) dY dt + P n ( t ) Y = G ( t ) . Say that { y 1 ,y 2 ,...,y n } is a Fundamental set oF solutions to the homogeneous equation: n ³ m =0 P n m ( t ) y ( m ) i = 0 For i =1 , 2 ,...,n . We will show that a particular solution to the nonhomogeneous equation is given by the integral Y ( t )= ° t G ( τ ) P 0 ( τ ) K ( τ,t ) in terms oF the kernel K ( τ,t )= ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± y 1 ( τ ) y 2 ( τ ) ··· y n ( τ ) y (1) 1 ( τ ) y (1) 2 ( τ ) ··· y (1) n ( τ ) . . . . . . . . . . . . y ( n 2) 1 ( τ ) y ( n 2) 2 ( τ ) ··· y ( n 2) n ( τ ) y 1 ( t ) y 2 ( t ) ··· y n ( t ) ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ² ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± y 1 ( τ ) y 2 ( τ ) ··· y n ( τ ) y (1) 1 ( τ ) y (1) 2 ( τ ) ··· y (1) n ( τ ) . . . . . . . . . . . . y ( n 2) 1 ( τ ) y ( n 2) 2 ( τ ) ··· y ( n 2)
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Unformatted text preview: n ( τ ) y ( n − 1) 1 ( τ ) y ( n − 1) 2 ( τ ) · · · y ( n − 1) n ( τ ) ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± . We explain why. Write Y ( t ) = u 1 ( t ) y 1 ( t ) + u 2 ( t ) y 2 ( t ) + · · · + u n ( t ) y n ( t ) For some Functions u i ( t ) to be determined. It is easy to show by induction that Y = ´ n ³ i =1 u i ( t ) y i ( t ) µ dY dt = ´ n ³ i =1 u i ( t ) dy i dt µ + ¶ n ³ i =1 du i dt y i · d 2 Y dt 2 = ´ n ³ i =1 u i ( t ) d 2 y i dt 2 µ + ´ n ³ i =1 du i dt y i ( t ) µ (1) + ´ n ³ i =1 du i dt y (1) i µ . . . d m Y dt m = ´ n ³ i =1 u i ( t ) d m y i dt m µ + m ³ j =1 ´ n ³ i =1 du i dt y ( j − 1) i ( t ) µ ( m − j ) . 1...
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