337_pdfsam_math 54 differential equation solutions odd

337_pdfsam_math 54 differential equation solutions odd - D...

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Review Problems We look for a particular solution of the form x p ( t )= Ae t . Substituting this function into the equation, we obtain Ae t +9 Ae t = e t A = 1 10 x p ( t )= e t 10 , and so x ( t )= x h ( t )+ x p ( t )= c 1 cos 3 t + c 2 sin 3 t + e t 10 . To Fnd y , we multiply the Frst equation in (5.40) by 3 and add to the second equation. This yields 2( D +3)[ x ] 4 y =3 e t + e t . Thus y = 1 2 ( D +3)[ x ] 3 4 e t 1 4 e t = 3( c 1 + c 2 ) 2 cos 3 t 3( c 1 c 2 ) 2 sin 3 t 11 20 e t 1 4 e t . 5. Di±erentiating the second equation, we obtain y 0 = z 0 . We eliminate z from the Frst and the third equations by substituting y 0 for z and y 0 for z 0 into them: x 0 = y 0 y, y 0 = y 0 x x 0 y 0 + y =0 , y 0 y 0 + x =0 (5.41) or, in operator notation,
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Unformatted text preview: D [ x ] ( D 1)[ y ] = 0 , x + ( D 2 D )[ y ] = 0 . We eliminate y by applying D to the Frst equation and adding the result to the second equation: D 2 [ x ] D ( D 1)[ y ] + x + ( D 2 D )[ y ] = 0 ( D 2 + 1 ) [ x ] = 0 . This equation is the simple harmonic equation, and its general solution is given by x ( t ) = C 1 cos t + C 2 sin t. 333...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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