ラプラス変換 y^{\prime \prime}-6y^{\prime}+15y=2sin(3t),y(0)=-1,y^{\prime}(0)=-4

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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

simplify solve for inverse tangent line

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laplace transform y^{′ ′}−6y^′+15y=2sin(3t), y(0)=−1, y^′(0)=−4

Solution

y=110 cos(3t)+130 sin(3t)−1110 e^3tcos(√6t)−2√2e^3tsin(√6t)5√3

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2(x²+y²)²=25(x²−y²)

Treat y as y(x)

Differentiate both sides: 4(x²+y²)(2x+2ydydx )=25(2x−2ydydx )

Isolate dydx : dydx =x(25−4x²−4y²)y(4x²+4y²+25)

dydx =x(25−4x²−4y²)y(4x²+4y²+25)

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

ラプラス変換 y^{\prime \prime}-6y^{\prime}+15y=2sin(3t),y(0)=-1,y^{\prime}(0)=-4

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