To determine The intervals in which solutions are sure to exist for the differential equation t ( t − 1 ) y ( 4 ) + e t y ″ + 4 t 2 y = 0 . Explanation Theorem used: “If the functions p 1 , p 2 , … , p n , and g are continuous on the open interval I , then there exist exactly on solution y = ϕ ( t ) of the differential equation L [ y ] = d n y d t n + p 1 ( t ) d n − 1 y d t n − 1 + ⋯ + p n − 1 ( t ) d y d t + p n ( t ) y = g ( t ) that also satisfies the initial conditions y ( t 0 ) = y 0 , y ′ ( t 0 ) = y ′ 0 , … , y n − 1 ( t 0 ) = y 0 ( n − 1 ) , where t 0 is any point in I . this solution exists throughout the interval I .” Calculation: Consider the differential equation t ( t − 1 ) y ( 4 ) + e t y ″ + 4 t 2 y = 0 . Divide by t ( t − 1 ) on both sides to the above equation. t ( t − 1 ) y ( 4 ) t ( t − 1 ) + e t y ″ t ( t − 1 ) + 7 t 2 y t ( t − 1 ) = 0 t ( t − 1 ) y ( 4 ) + e t t ( t − 1 ) y ″ + 7 t 2 t ( t − 1 ) y = 0 By comparing the given differential equation y ( 4 ) + e t t ( t −

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 4.1, Problem 3P

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The intervals in which solutions are sure to exist for the differential equation t(t−1)y(4)+ety″+4t2y=0.

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Chapter 4.1, Problem 2P

Chapter 4.1, Problem 4P

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Elementary Differential Equations

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