SampleTutorial4

SampleTutorial4 - MATH 3705AB Tutorial February 10 2006 1....

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MATH 3705AB Tutorial February 10 2006 1. Find the general solution of 2( x 3) 2 y II +( x 3) y I + y =0for x W =3 . Solution: x 0 = 3 is a singular point of this equation. This is not an Euler equation but it may be transformed into one by the change of variables t = x 3. Then y ( x )= z ( t ), y I = dy/dx = dz/dt · dt/dx = dz/dt = z I ,a n d y ”= d 2 z/dt 2 = z ”. Thus, the original equations becomes an Euler equation for z ( t ): 2 t 2 z ”+ tz I + z =0 . Weseektheso lut ionintheform z = t r ,t> 0, which leads to the indicial equation 2 r 2 r +1=0 . This quadratic equation has two complex conjugate roots r 1 , 2 = 1 4 ± 7 4 , and therefore the solution for t> 0i s z ( t )= C 1 t 1 / 4 cos( 7 4 ln t )+ C 2 t 1 / 4 sin( 7 4 ln t ) . Getting back to the original variable x yields y ( x )= C 1 | x 3 | 1 / 4 cos( 7 4 ln | x 3 | )+ C 2 | x 3 | 1 / 4 sin( 7 4 ln |
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SampleTutorial4 - MATH 3705AB Tutorial February 10 2006 1....

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