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f(x)=e^{3ln(x)}

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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

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線形 4xy^{′ ′}+2y^′+y=0

Solution

y=c₁√1√x sin(√x)⁴√x+c₂√1√x cos(√x)⁴√x

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Solution steps

One step at a time

x⁵+x⁴+x+1

=(x⁵+x⁴)+(x+1)

x⁴ を x⁵+x⁴: x⁴(x+1) からくくり出す

=x⁴(x+1)+(x+1)

共通項をくくり出す (x+1): (x+1)(x⁴+1)

=(x+1)(x⁴+1)

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こんにちは, について質問する線形 4xy^{′ ′}+2y^′+y=0

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

f(x)=e^{3ln(x)}

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