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satisfies (2, 2), y>=-1/3 x+2, y<2x+3

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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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lời giải chuỗi xy^{′ ′}+y^′−y=0

Solution

y=c₁(1+x+x²4 +… +xⁿ4… n² +… )+c₂(ln(x)(1+x+x²4 +… +xⁿ4… n² +… )−2x−3x²4 −11x³108 +… )

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xy′ ′ +y′ −y=0

Giải bằng phương pháp chuỗi: y=c₁(1+x+x²4 +… +xⁿ4… n² +… )+c₂(ln(x)(1+x+x²4 +… +xⁿ4… n² +… )−2x−3x²4 −11x³108 +… )

y=c₁(1+x+x²4 +… +xⁿ4… n² +… )+c₂(ln(x)(1+x+x²4 +… +xⁿ4… n² +… )−2x−3x²4 −11x³108 +… )

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

satisfies (2, 2), y>=-1/3 x+2, y<2x+3

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