The functions y1(t) = t² and y2(t) = t³ are solutions of the homogeneous differential equation ty" – 4ty' + on (0, o0). To find a solution of the nonhomogeneous differential equation t2y" – 4ty' + 6y = 4t³ on (0, 00) the function u(t) of yp = u1(t)t² + uz(t)t³? %3D %3D Lütfen birini seçin: O 4 -4t -4t2

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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The functions y(t) = t² and y2(t) = t³ are solutions of the homogeneous differential equation t2y" – 4ty' +
on (0, 0). To find a solution of the nonhomogeneous differential equation ty" – 4ty'+ 6y = 4t³ on (0, ∞0)
the function u(t) of y, = u1(t)t² + u2(t)t³ ?
Lütfen birini seçin:
-4t
-4
-4t2
Transcribed Image Text:The functions y(t) = t² and y2(t) = t³ are solutions of the homogeneous differential equation t2y" – 4ty' + on (0, 0). To find a solution of the nonhomogeneous differential equation ty" – 4ty'+ 6y = 4t³ on (0, ∞0) the function u(t) of y, = u1(t)t² + u2(t)t³ ? Lütfen birini seçin: -4t -4 -4t2
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