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nth term a_{1}=3, a_{2}=2, a_{n}=-a_{n-1}

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laplace ngược 5x³+1x²−4

Solution

5δ′(t)+394 e^−2t+414 e^2t

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Solution steps

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L⁻¹{5x³+1x²−4 }

Lấy một phần của 5x³+1x²−4 : 5x+394(x+2) +414(x−2)

=L⁻¹{5x+394(x+2) +414(x−2) }

Sử dụng thuộc tính tuyến tính của Biến đổi Nghịch đảo Laplace:
Đối với các hàm f(s), g(s) và các hằng số a, b: L⁻¹{a·f(s)+b·g(s)}=a·L⁻¹{f(s)}+b·L⁻¹{g(s)}

=5L⁻¹{x}+L⁻¹{394(x+2) }+L⁻¹{414(x−2) }

L⁻¹{x}: δ′ (t)

L⁻¹{394(x+2) }: 394 e^−2t

L⁻¹{414(x−2) }: 414 e^2t

=5δ′ (t)+394 e^−2t+414 e^2t

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

nth term a_{1}=3, a_{2}=2, a_{n}=-a_{n-1}

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