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# 285lect19 - 1 Lecture 19 For a number a consider the...

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1 Lecture 19 For a number a , consider the following operator: ( D & a ) [ f ] j x = f 0 ( x ) & af ( x ) : First, we prove the following property of some operators. We claim that, for every pair of real numbers, ( D & a ) ( D & b ) = ( D & b ) ( D & a ) : Indeed, ( D & a ) ( D & b ) [ f ] = ( D & a ) [( D & b ) [ f ]] = ( D & a ) [ f 0 & bf ] = ( f 0 & bf ) 0 & a ( f 0 & bf ) = f 00 & bf 0 & af 0 & abf: Likewise, ( D & b ) ( D & a ) [ f ] = ( D & b ) [ f 0 & af ] = ( f 0 & af ) 0 & b ( f 0 & af ) = f 00 & af 0 & bf 0 & abf: Hence, ( D & a ) ( D & b ) = ( D & b ) ( D & a ) : Now L [ y ] = 0 () ( D & r 0 ) m 0 ( D & r 1 ) m 1 ::: ( D & r l ) m l [ y ] = ( D & r 1 ) m 1 ::: ( D & r l ) m l ( D & r 0 ) m 0 [ y ] = 0 : Therefore, every solution of the equation ( D & r 0 ) m 0 [ y ] = 0 will be a solution of the equation L [ y ] = 0 . As a trial solution of the ODE ( D & r 0 ) m 0 [ y ] = 0 , take y ( x ) = u ( x ) e r 0 x ; where the function u ( x ) is to be found : We have ( D & r 0 ) [ u ( x ) e r 0 x ] = ( u ( x ) e r 0 x ) 0 & r 0 ( u ( x ) e r 0 x ) = u 0 e r 0 x + r 0 ue r 0 x & r 0 ( u ( x ) e r 0 x ) = u 0 e r 0 x : Likewise, ( D & r 0 ) 2 [ u ( x ) e r 0 x ] = ( D & r 0 ) [ u 0 ( x ) e r 0 x ] = u 00 ( x ) e r 0 x and ::: ( D & r 0 ) m 0 [ u ( x ) e r 0 x ] = ( D & r 0 ) & u ( m 0 & 1) e r 0 x ± = u m 0 e r 0 x ; 1

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whence u ( m 0 ) e r 0 x = 0 , whence u ( m 0 ) ( x ) & 0 ) u ( x ) = C 0 + C 1 x + ::: + C m 0 & 1 x m 0 & 1 : So, if r = r 0 is a root of multiplicity m 0 , then y ( x ) = & C 0 + C 1 x + ::: + C m 0 & 1 x m 0 & 1 ± e r 0 x is a solution of the ODE L [ y ] = 0 .
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285lect19 - 1 Lecture 19 For a number a consider the...

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