285lect23 - 1 Lecture 24 c < ccr underdamped case 1.0.1 In...

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1 1.0.1 c < c cr General solution: x ( t ) = e Let ! 1 = p ! 2 0 & p 2 x ( t ) = e & pt ( C = q C 2 1 + = Ae & pt where Underdamped case. 1
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Example 1 m = 3 ; c = 30 ; k = 63 Solution. 3 x 00 3 r r 1 ; r 2 = & General solution: x 0 (0) = 2 ) C & 7 C Particular solution: 2
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1.1 Non-homogeneous equations with constant coe¢ - cients We consider the di/erential operator L = D n + p 1 D n & 1 + p 2 D n & 2 + ::: + p n & 1 D + p n D 0 ; where p j = const; j = 1 ; 2 ; 3 ; :::; n . Recall that a non-homogeneous ODE with constants coe¢ cients is de&ned by y ( n ) + p 1 y ( n & 1) + p 2 y ( n & 2) + ::: + p n y ( x ) = f ( x ) , or L [ y ] = f and that the complementary homogeneous equation is L [ y ] = 0 : If y c ( x ) is a complementary solution, L [ y c ] = 0 , and y p ( x ) is a particular solution of the non±homogeneous equation L [ y ] = f , L [ y p ] = f , then y ( x ) = y c ( x ) + y p ( x ) is a solution of the non-homogeneous equation; indeed, L [ y ] = L [ y c + y p ] = L [ y c ] + L [ y p ] = 0 + f = f: We already know how to &nd general solution of the complimentary equation. So, once we found a particular solution y p ( x ) , we know general solution of a non-homogeneous equation.
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285lect23 - 1 Lecture 24 c < ccr underdamped case 1.0.1 In...

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