The functions y1(t) = t² and y2(t) = t³ are solutions of the homogeneous differential equation t’y" – 4ty' + 6y = 0 %3! on (0, 0). When using variation of parameters with y = u1(t)t² +u2(t)t³ to find a solution of the nonhomogeneous differential equation t*y" – 4ty' + 6y = 4t%, %3D what is the function u(t) ? (Submit the corresponding number without parentheses.)

Algebra and Trigonometry (6th Edition)
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ISBN:9780134463216
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ChapterP: Prerequisites: Fundamental Concepts Of Algebra
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The functions yı(t) = t² and y2(t) = t³ are solutions of the homogeneous differential equation
%3D
%3D
t’y" – 4ty' + 6y = 0 on (0, ∞).
When using variation of parameters with y = u1(t)t² +U2(t)t³ to find a solution of the nonhomogeneous
differential equation
t²y" – 4ty' + 6y = 4t°,
what is the function u(t) ? (Submit the corresponding number without parentheses.)
(1) 4
(2) –4
(3)
(5) –4t
(6) –4t²
Transcribed Image Text:The functions yı(t) = t² and y2(t) = t³ are solutions of the homogeneous differential equation %3D %3D t’y" – 4ty' + 6y = 0 on (0, ∞). When using variation of parameters with y = u1(t)t² +U2(t)t³ to find a solution of the nonhomogeneous differential equation t²y" – 4ty' + 6y = 4t°, what is the function u(t) ? (Submit the corresponding number without parentheses.) (1) 4 (2) –4 (3) (5) –4t (6) –4t²
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