210_pdfsam_math 54 differential equation solutions odd

210_pdfsam_math 54 differential equation solutions odd - y...

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Chapter 4 We substitute y p and its derivatives into the original equation and equate the corresponding coefficients. This yields ±² A 1 t 2 +( A 0 +6 A 1 ) t +3 A 0 +6 A 1 ³ + ² A 1 t 2 +( A 0 +4 A 1 ) t +2 A 0 +2 A 1 ³ 2 ² A 1 t 2 + A 0 t ³´ e t 2 B = te t +1 [10 A 1 t +8 A 1 +5 A 0 ] e t 2 B = te t +1 10 A 1 =1 , 8 A 1 +5 A 0 =0 , 2 B =1 A 1 =1 / 10 , A 0 = 4 / 25 , B = 1 / 2 . Hence, a particular solution is y p ( t )= µ 1 10 t 4 25 te t 1 2 . 41. The characteristic equation in this problem is r 2 +2 r + 5 = 0, which has roots r = 1 ± 2 i . Therefore, a general solution to the corresponding homogeneous equation is given by y h ( t )=( c 1 cos 2 t + c 2 sin 2 t ) e t . (4.1) (a) For 0 t 3 π/ 2, g ( t ) 10, and so the equation becomes y 0 +2 y 0 +5 y =10 . Hence a particular solution has the form
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Unformatted text preview: y p ( t ) A . Substitution into the equation yields ( A ) + 2( A ) + 5( A ) = 10 5 A = 10 A = 2 , and so, on [0 , 3 / 2], a general solution to the original equation is y 1 ( t ) = ( c 1 cos 2 t + c 2 sin 2 t ) e t + 2 . We nd c 1 and c 2 by substituting this function into the initial conditions. 0 = y 1 (0) = c 1 + 2 , 0 = y 1 (0) = c 1 + 2 c 2 c 1 = 2 , c 2 = 1 y 1 ( t ) = (2 cos 2 t + sin 2 t ) e t + 2 . 206...
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