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frobenius method 4xy''+2y'+y=0

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

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series solution 4xy^{′ ′}+2y^′+y=0

Solution

y=c₁√x(1−x6 +x²120 +… +(−1)ⁿxⁿ(2)(3)(4)(5)…(2n)(2n+1) +… )+c₂(1−x2 +x²24 +… +(−1)ⁿxⁿ(1)(2)(3)(4)…(2n−1)(2n) +… )

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4xy′ ′ +2y′ +y=0

Solve using series method: y=c₁√x(1−x6 +x²120 +… +(−1)ⁿxⁿ(2)(3)(4)(5)… (2n)(2n+1) +… )+c₂(1−x2 +x²24 +… +(−1)ⁿxⁿ(1)(2)(3)(4)… (2n−1)(2n) +… )

y=c₁√x(1−x6 +x²120 +… +(−1)ⁿxⁿ(2)(3)(4)(5)… (2n)(2n+1) +… )+c₂(1−x2 +x²24 +… +(−1)ⁿxⁿ(1)(2)(3)(4)… (2n−1)(2n) +… )

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

frobenius method 4xy''+2y'+y=0

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