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4(1+u)=-8

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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преобразование Лапласа dydt +2y=12sin(2t), y(0)=5

Solution

y=−3cos(2t)+3sin(2t)+8e^−2t

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dydt +2y=12sin(2t)

Решите с помощью преобразования Лапласа: y=−3cos(2t)+3sin(2t)+8e^−2t

y=−3cos(2t)+3sin(2t)+8e^−2t

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

4(1+u)=-8

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