sostituzione y^{\prime \prime}-y=0

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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 inverse tangent line

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substitution y^{′ ′}−y=0

Solution

y=−√c₁2e^t+c₂ +√c₁e^t+c₂2 , y=−√c₁2e^−t+c₂ +√c₁e^−t+c₂2

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y′ ′ −y=0

Solve using substitution: y=−√c₁2e^t+c₂ +√c₁e^t+c₂2 , y=−√c₁2e^−t+c₂ +√c₁e^−t+c₂2

y=−√c₁2e^t+c₂ +√c₁e^t+c₂2 , y=−√c₁2e^−t+c₂ +√c₁e^−t+c₂2

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

sostituzione y^{\prime \prime}-y=0

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