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The functions y1(t) =t² and y2(t)= t° are solutions of the homogeneous differential equation t?y" – 4ty' + 6y = 0 on (0, 00). When using variation of parameters with y = u1(t)t² +u2(t)t³ to find a solution of the nonhomogeneous lifferential equation t'y" – 4ty' + 6y = 4t³, | vhat is the function u, (t) ? (1) 4 (2) –4 4 (3) (4) t2 (5) –4t (6) –4t?

The functions y1(t) =t² and y2(t)= t° are solutions of the homogeneous differential equation t?y" – 4ty' + 6y = 0 on (0, 00). When using variation of parameters with y = u1(t)t² +u2(t)t³ to find a solution of the nonhomogeneous lifferential equation t'y" – 4ty' + 6y = 4t³, | vhat is the function u, (t) ? (1) 4 (2) –4 4 (3) (4) t2 (5) –4t (6) –4t?

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The functions yı(t) = t² and y2(t) = t³ are solutions of the homogeneous differential equation t?y - 4ty + бу — 0 on (0, 0). on (0, 00). 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 = 4t³, what is the function u, (t) ? (1) 4 (2) -4 4 (3) 4 (4) t2 (5) –4t (6) –4t2
Transcribed Image Text:The functions yı(t) = t² and y2(t) = t³ are solutions of the homogeneous differential equation t?y - 4ty + бу — 0 on (0, 0). on (0, 00). 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 = 4t³, what is the function u, (t) ? (1) 4 (2) -4 4 (3) 4 (4) t2 (5) –4t (6) –4t2
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