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You can verify that the differential equation: 2ty" – 4t(t + 1)y + 4(t + 1)y= 0, t> 0 has solutions y1 = 2t and y2 = 3t exp (2t). a. Compute the Wronskian W between y, and y2. W (t) = 12?e2 b. The solutions yı and y2 form a fundamental set of solutions because there is a point to where W(to) # 0: (D W #0.

You can verify that the differential equation: 2ty" – 4t(t + 1)y + 4(t + 1)y= 0, t> 0 has solutions y1 = 2t and y2 = 3t exp (2t). a. Compute the Wronskian W between y, and y2. W (t) = 12?e2 b. The solutions yı and y2 form a fundamental set of solutions because there is a point to where W(to) # 0: (D W #0.

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 12CR
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You can verify that the differential equation: 2t°y" - 4t(t + 1)y + 4(t +1)y= 0, t> 0 has solutions y1 = 2t and y2 = 3t exp (2t). a. Compute the Wronskian W between y1 and y2. W (t) = | 12-e2t b. The solutions y1 and y2 form a fundamental set of solutions because there is a point to where W(to) + 0: W # 0. Note: It simply wants you to find some number to to put into the first answer blank and plug into the Wronskian to get something nonzero, which goes into the second answer blank.
Transcribed Image Text:You can verify that the differential equation: 2t°y" - 4t(t + 1)y + 4(t +1)y= 0, t> 0 has solutions y1 = 2t and y2 = 3t exp (2t). a. Compute the Wronskian W between y1 and y2. W (t) = | 12-e2t b. The solutions y1 and y2 form a fundamental set of solutions because there is a point to where W(to) + 0: W # 0. Note: It simply wants you to find some number to to put into the first answer blank and plug into the Wronskian to get something nonzero, which goes into the second answer blank.
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