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nghịch đảo sqrt(x+1)-1

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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laplace ngược x⁴−x²+2x²(x−1)

Solution

δ′(t)+δ(t)−2H(t)−2t+2e^t

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L⁻¹{x⁴−x²+2x²(x−1) }

Lấy một phần của x⁴−x²+2x²(x−1) : x+1−2x −2x² +2x−1

=L⁻¹{x+1−2x −2x² +2x−1 }

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)}

=L⁻¹{x}+L⁻¹{1}−L⁻¹{2x }−L⁻¹{2x² }+L⁻¹{2x−1 }

L⁻¹{x}: δ′ (t)

L⁻¹{1}: δ(t)

L⁻¹{2x }: 2H(t)

L⁻¹{2x² }: 2t

L⁻¹{2x−1 }: 2e^t

=δ′ (t)+δ(t)−2H(t)−2t+2e^t

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

nghịch đảo sqrt(x+1)-1

Solution

Solution steps

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