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interval notation-5=4

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for expand factor rationalize

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series solution y^{′ ′}−y=0

Solution

y=c₁(1+t²2 +t⁴24 +… +t²ⁿ(1)(2)(3)(4)…(2n−1)· 2n +… )+c₂(t+t³6 +t⁵120 +… +t²ⁿ⁺¹(2)(3)(4)(5)… 2n(2n+1) +… )

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y′ ′ −y=0

Solve using series method: y=c₁(1+t²2 +t⁴24 +… +t²ⁿ(1)(2)(3)(4)… (2n−1)· 2n +… )+c₂(t+t³6 +t⁵120 +… +t²ⁿ⁺¹(2)(3)(4)(5)… 2n(2n+1) +… )

y=c₁(1+t²2 +t⁴24 +… +t²ⁿ(1)(2)(3)(4)… (2n−1)· 2n +… )+c₂(t+t³6 +t⁵120 +… +t²ⁿ⁺¹(2)(3)(4)(5)… 2n(2n+1) +… )

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Notebook

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

interval notation-5=4

Solution

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