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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)

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 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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