Explanation Theorem used: Variation of parameters: If y 1 and y 2 are two linearly independent solutions to the homogenous equation y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = 0 on an interval I where p ( t ) , q ( t ) and g ( t ) are continuous, then a particular solution to y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) is y p = v 1 y 1 + v 2 y 2 , where v 1 and v 2 are determined up to a constant by the pairs of equations y 1 v ′ 1 + y 2 v ′ 2 = 0 and y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = g , which have the solution: v 1 ( t ) = ∫ − g ( t ) y 2 ( t ) W [ y 1 , y 2 ] ( t ) d t v 2 ( t ) = ∫ g ( t ) y 1 ( t ) W [ y 1 , y 2 ] ( t ) d t Procedure used: Consider the differential equation a t 2 y ″ ( t ) + b t y ′ ( t ) + c y = 0 . The associated auxiliary equation is a r 2 + ( b − a ) r + c = 0 . Solving the auxiliary equation gives the roots r 1 and r 2 . (a) If the auxiliary equation has distinct real roots r 1 and r 2 , then y ( t ) = c 1 t r 1 + c 2 t r 2 is a general solution. (b) If the auxiliary equation has repeated root r , then y ( t ) = c 1 t r + c 2 t r ln t is a general solution. (c) If the auxiliary equation has complex root r = α ± i β , then y ( t ) = c 1 t α cos ( β ln t ) + c 2 t α sin ( β ln t ) is a general solution. Calculation: The given differential equation is, t 2 y ″ + 3 t y ′ + y = t − 1 . Compare t 2 y ″ + 3 t y ′ + y = t − 1 with the general equation form a t 2 y ″ ( t ) + b t y ′ ( t ) + c y = 0 to obtain a = 1 , b = 3 and c = 1 . The associated auxiliary equation is, a r 2 + ( b − a ) r + c = 0 becomes, 1 r 2 + ( 3 − 1 ) r + ( 1 ) = 0 r 2 + 2 r + 1 = 0 Simplify r 2 + 2 r + 1 = 0 as, r 2 + r + r + 1 = 0 r ( r + 1 ) + 1 ( r + 1 ) = 0 ( r + 1 ) ( r + 1 ) = 0 r = − 1 , − 1 That is, the repeated roots are r 1 = − 1 and r 2 = − 1 . Thus, the given differential equation has solution of the form, y ( t ) = c 1 t − 1 + c 2 t − 1 ln t . That is, the two independent solutions to the homogenous equation t 2 y ″ + 3 t y ′ + y = 0 are t − 1 and t − 1 ln t . Let y 1 ( t ) = t − 1 and y 2 ( t ) = t − 1 ln t . Then, y ′ 1 ( t ) = − 1 t 2 and y ′ 2 ( t ) = 1 − ln t t 2 . Set the particular solution y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) as, y p ( t ) = v 1 ( t ) t − 1 + v 2 ( t ) t − 1 ln t . For applying the variation of parameters, the equation is to be simplified in the form y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) . Divide the equation t 2 y ″ + 3 t y ′ + y = t − 1 by t 2 as, t 2 t 2 y ″ + 3 t t 2 y ′ + 1 t 2 y = t − 1 t 2 y ″ + 3 t y ′ + 1 t 2 y = 1 t 3 Compare y ″ + 3 t y ′ + 1 t 2 y = 1 t 3 with the general equation form y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) to obtain p ( t ) = 3 t , q ( t ) = 1 t 2 and g ( t ) = 1 t 3

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 4.7, Problem 38E

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The general solution to the differential equation t 2 y ″ + 3 t y ′ + y = t − 1 .

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