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 z ″ + t z ′ + 9 z = − tan ( 3 ln t ) . For easier calculation, assume the differential equation to be of the form t 2 y ″ + t y ′ + 9 y = − tan ( 3 ln t ) . Compare t 2 y ″ + t y ′ + 9 y = − tan ( 3 ln t ) with the general equation form a t 2 y ″ ( t ) + b t y ′ ( t ) + c y = 0 to obtain a = 1 , b = 1 and c = 9 . The associated auxiliary equation is, a r 2 + ( b − a ) r + c = 0 becomes, 1 r 2 + ( 1 − 1 ) r + ( 9 ) = 0 r 2 + 0 r + 9 = 0 r 2 + 9 = 0 Simplify r 2 + 9 = 0 as, r 2 = − 9 r = ± − 9 r = ± 3 i ( ∵ i = − 1 ) That is, the roots are r 1 = 3 i and r 2 = − 3 i . Here, α = 0 and β = 3 . Thus, the given differential equation has solution of the form, y ( t ) = c 1 cos [ 3 ln ( t ) ] + c 2 sin [ 3 ln ( t ) ] . That is, the two independent solutions to the homogenous equation t 2 y ″ + t y ′ + 9 y = 0 are cos [ 3 ln t ] and sin [ 3 ln t ] . Let y 1 ( t ) = cos [ 3 ln t ] and y 2 ( t ) = sin [ 3 ln t ] . Then, y ′ 1 ( t ) = − 3 t sin [ 3 ln t ] and y ′ 2 ( t ) = 3 t cos [ 3 ln t ] . 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 ) cos [ 3 ln t ] + v 2 ( t ) sin [ 3 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 ″ + t y ′ + 9 y = − tan ( 3 ln t ) by t 2 as, t 2 t 2 y ″ + t t 2 y ′ + 9 t 2 y = − tan ( 3 ln t ) t 2 y ″ + 1 t y ′ + 9 t 2 y = − tan ( 3 ln t ) t 2 Compare y ″ + 1 t y ′ + 9 t 2 y = − tan ( 3 ln t ) t 2 with the general equation form y ″ ( t ) + p ( t ) y ′ ( t ) + q ( t ) y = g ( t ) to obtain p ( t ) = 1 t , q ( t ) = 9 t 2 and g ( t ) = − tan ( 3 ln t ) t 2 . Obtain the value of W [ y 1 , y 2 ] ( t ) as, W [ y 1 , y 2 ] ( t ) = | y 1 ( t ) y 2 ( t ) y 1 ′ ( t ) y 2 ′ ( t ) | = | cos [ 3 ln t ] sin [ 3 ln t ] − 3 t sin [ 3 ln t ] 3 t cos [ 3 ln t ] | = 3 t cos [ 3 ln t ] cos [ 3 ln t ] − sin [ 3 ln t ] ( − 3 t sin [ 3 ln t ] ) = 3 t cos 2 [ 3 ln t ] + 3 t sin 2 [ 3 ln t ] Simplify further as, W [ y 1 , y 2 ] ( t ) = 3 t ( cos 2 [ 3 ln t ] + sin 2 [ 3 ln t ] ) = 3 t ( 1 ) ( ∵ cos 2 x + sin 2 x = 1 ) = 3 t The equation v 1 ( t ) = ∫ − g ( t ) y 2 ( t ) W [ y 1 , y 2 ] ( t ) d t for g ( t ) = − tan ( 3 ln t ) t 2 , y 2 ( t ) = sin [ 3 ln t ] and W [ y 1 ,

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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The general solution to the differential equation t2z″+tz′+9z=−tan(3lnt).

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Chapter 4.7, Problem 36E

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The general solution to the differential equation t 2 z ″ + t z ′ + 9 z = − tan ( 3 ln t ) .

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The general solution to the differential equation t2z″+tz′+9z=−tan(3lnt).

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