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The technique works for equations of any order, for example, first order. If is a solution of the differential equation ak+2 then its coefficients a satisfy the following recurrence relation. For k ≥ 0, We also have y = 2 2 k+2 2 a₁ = -2 ao help (numbers) Therefore, write the first few terms of a solution where y(0) = 2 48 ak+1+ 1 k+2 +-4 x²+-2 + help (numbers) x+ y=[akz* k=0 2³+ y + (2x + 2)y=0, a help (formulas)

The technique works for equations of any order, for example, first order. If is a solution of the differential equation ak+2 then its coefficients a satisfy the following recurrence relation. For k ≥ 0, We also have y = 2 2 k+2 2 a₁ = -2 ao help (numbers) Therefore, write the first few terms of a solution where y(0) = 2 48 ak+1+ 1 k+2 +-4 x²+-2 + help (numbers) x+ y=[akz* k=0 2³+ y + (2x + 2)y=0, a help (formulas)

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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The technique works for equations of any order, for example, first order. If is a solution of the differential equation ak+2 then its coefficients a satisfy the following recurrence relation. For k ≥ 0, We also have 2 k+2 2 48 a₁ = -2 ao. help (numbers) Therefore, write the first few terms of a solution where y(0) = 2 y = 2 ak+1+ 4. 1 k+2 +-4 x² +-2 + help (numbers) x+ y=[akak k=0 2³+ y + (2x + 2)y=0, a. help (formulas)
Transcribed Image Text:The technique works for equations of any order, for example, first order. If is a solution of the differential equation ak+2 then its coefficients a satisfy the following recurrence relation. For k ≥ 0, We also have 2 k+2 2 48 a₁ = -2 ao. help (numbers) Therefore, write the first few terms of a solution where y(0) = 2 y = 2 ak+1+ 4. 1 k+2 +-4 x² +-2 + help (numbers) x+ y=[akak k=0 2³+ y + (2x + 2)y=0, a. help (formulas)
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