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inverse of sin^{65}(x)

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

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

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substituição y^{′ ′}−4y=0

Solution

y=−√c₁4e^2t+2c₂ +√c₁e^2t+2c₂4 , y=−√c₁4e^−2t+2c₂ +√c₁e^−2t+2c₂4

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y′ ′ −4y=0

Solve using substitution: y=−√c₁4e^2t+2c₂ +√c₁e^2t+2c₂4 , y=−√c₁4e^−2t+2c₂ +√c₁e^−2t+2c₂4

y=−√c₁4e^2t+2c₂ +√c₁e^2t+2c₂4 , y=−√c₁4e^−2t+2c₂ +√c₁e^−2t+2c₂4

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

inverse of sin^{65}(x)

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inverse of sin^{65}(x)

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