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ty'+2y=t^2-t+1

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

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ty^′+2y=t²−t+1

Solution

y=t²4 −t3 +12 +c₁t²

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ty′ +2y=t²−t+1

Solve linear ODE: y=t²4 −t3 +12 +c₁t²

y=t²4 −t3 +12 +c₁t²

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

ty'+2y=t^2-t+1

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