382_pdfsam_math 54 differential equation solutions odd

382_pdfsam_math 54 differential equation solutions odd - 1)...

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Chapter 6 5. Since the nonhomogeneous term, g ( x )=tan x , is not a solution to a homogeneous linear diferential equation with constant coefficients, we will Fnd a particular solution by the method o± variation o± parameters. To do this, we must Frst Fnd a ±undamental solution set ±or the corresponding homogeneous equation, y 0 + y 0 = 0. Its auxiliary equation is r 3 + r =0 , which ±actors as r 3 + r = r ( r 2 + 1). Thus, the roots to this auxiliary equation are r =0 , ± i . There±ore, a ±undamental solution set to the homogeneous equation is { 1 , cos x, sin x } and y p ( x )= v 1 ( x )+ v 2 ( x )cos x + v 3 ( x )sin x. To accomplish this, we must Fnd the ±our determinants W [1 , cos x, sin x ]( x ), W 1 ( x ), W 2 ( x ), W 3 ( x ). That is, we calculate W [1 , cos x, sin x ]( x )= ± ± ± ± ± ± ± ± 1c o s x sin x 0 sin x cos x 0 cos x sin x ± ± ± ± ± ± ± ± =s in 2 x +cos 2 x =1 , W 1 ( x )=( 1) 3 1 W [cos x, sin x ]( x )= ± ± ± ± ± cos x sin x sin x cos x ± ± ± ± ± = ( cos 2 x +sin 2 x ) =1 , W 2 ( x )=( 1) 3 2 W [1 , sin x ]( x )= ± ± ± ± ± 1s i n x 0c o s x ± ± ± ± ± = cos x, W 3
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Unformatted text preview: 1) 3 3 W [1 , cos x ]( x ) = 1 c o s x sin x = sin x. By using ormula (11) on page 340 o the text, we can now Fnd v 1 ( x ), v 2 ( x ), and v 3 ( x ). Since g ( x ) = tan x , we have (assuming that all constants o integration are zero) v 1 ( x ) = Z g ( x ) W 1 ( x ) W [1 , cos x, sin x ]( x ) dx = Z tan x dx = ln(sec x ) , v 2 ( x ) = Z g ( x ) W 2 ( x ) W [1 , cos x, sin x ]( x ) dx = Z tan x ( cos x ) dx = Z sin x dx = cos x, v 3 ( x ) = Z g ( x ) W 3 ( x ) W [1 , cos x, sin x ]( x ) dx = Z tan x ( sin x ) dx = Z sin 2 x cos x dx = Z 1 cos 2 x cos x dx = Z (cos x sec x ) dx = sin x ln(sec x + tan x ) . 378...
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