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Let L = (1+t²)D² – 4tD + 6. Check that yi =1– 3t2 and y2 = t - are two solutions to the differential equation L(y) = 0. Check that y =t is a solution to the differential equation L(y) = 2t. Use parts (a) and (b) to solve the IVP (1+t²)y/" – 4ty + 6y = 2t, y(0) = 1, y'(0) = 0. %3D

Let L = (1+t²)D² – 4tD + 6. Check that yi =1– 3t2 and y2 = t - are two solutions to the differential equation L(y) = 0. Check that y =t is a solution to the differential equation L(y) = 2t. Use parts (a) and (b) to solve the IVP (1+t²)y/" – 4ty + 6y = 2t, y(0) = 1, y'(0) = 0. %3D

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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Let L = (1+t²)D² – 4tD +6. OCheck that yı =1– 3t2 and Y2 =t – are two solutions to the differential equation L(y) = 0. OCheck that y = Use parts (a) and (b) to solve the IVP (1+t)y" – 4ty + 6y = 2t, y(0) = 1, 3/(0) = 0. %3D t is a solution to the differential equation L(y) = 2t.
Transcribed Image Text:Let L = (1+t²)D² – 4tD +6. OCheck that yı =1– 3t2 and Y2 =t – are two solutions to the differential equation L(y) = 0. OCheck that y = Use parts (a) and (b) to solve the IVP (1+t)y" – 4ty + 6y = 2t, y(0) = 1, 3/(0) = 0. %3D t is a solution to the differential equation L(y) = 2t.
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