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Unformatted text preview: Week 4: 1. 4.3 Undetermined Coef 2. 4.5 Springs (Homog case) 3. 5.4 Eigenvalues and Eigenvectors [5.5 Eigenspaces and Diagonalization ] start  Problem 3. Solve y 4 y + 3 y = 6 e 2 x . Solution: (a) Find the general solution y H of the homogeneous equation y 4 y + 3 y = 0 . (a) y H = c 1 e x + c 2 e 3 x . (b) Determine the value of the constant A so y p = A e 2 x is a particular solution. y p = A e 2 x , y p = A (2 e 2 x ) , y p = 4 A e 2 x , so 6 e 2 x = 4 A e 2 x 4(2 A e 2 x ) + 3( A e 2 x ) = A e 2 x , so A = 6 , y p = 6 e 2 x . 2 (c) Determine the general solution. y = y H + y p = c 1 e x + c 2 e 3 x 6 e 2 x .  Compare 3d: If F = 3 x 2 , find constants a , a 1 and a 2 so y p = a + a 1 x + a 2 x 2 . y p = a 1 + 2 xa 2 , y p = 2 a 2 gives polynomial equation 3 x 2 = 2 a 2 4( a 1 + 2 xa 2 ) + 3( a + a 1 x + a 2 x 2 ) all x . x 2coef: 3 = 3 a 2 , so a 2 = 1 . xcoef: 0 = 4(2 a 2 ) + 3 a 1 with a 2 = 1 , so a 1 = 8 3 . constant term: 0 = 2 a 2 4( a 1 ) + 3( a ) , so a = 26 9 . So y p = 26 9 + 8 3 x + x 2 .  We have one more linearity property: If y + a 1 y + a 2 y = F 1 and y + a 1 y + a 2 y = F 2 have particular solutions y p 1 and y p 2 , then y + a 1 y + a 2 y = F, for F = F 1 + F 2 , has particular solution y p = y p 1 + y p 2 . Example: y 4 y + 3 y = 6 e 2 x + 3 x 2 has solution y p = 6 e 2 x + 26 9 + 8 3 x + x 2 . Note: There is a problem on the HW that is easier when done using this method. (Which?)  Problem 4: Solve y + 4 y + 5 y = 24 sin x. 3 Solution: The usual trial function for F ( x ) = 24 sin x is y p = A cos x + B sin x. Before attempting to compute y p and y p to find A and B (the undetermined coefficients), we check y H for repetition. The homogeneous equation y + 4 y + 5 y = 0 has characteristic polynomial P ( r ) = r 2 + 4 r + 5 . The quadratic formula gives roots 2 i, say r 1 = 2 + i and r 2 = 2 i (why?). But the driving force F ( x ) = 24 sin x occurs as a solution of the DE with roots i, that is, a bi with a = 0 , b = 1 , which do not coincide with r 1 , r 2 : the usual y p = A cos x + B sin x is the correct choice.  Now with y p = A cos x + B sin x, we have y p = A sin x + B cos x, and y p = A cos x B sin x. Substitution in the DE gives 24 sin x = y p + 4 y p + 5 y p = ( A cos x B sin x ) + 4( A sin x + B cos x )+ 5( A cos x + B sin x ) . Equating like terms coef of cos x : A + 4 B + 5 A = 0 , so B = A . coef of sin x : B + 4( A ) + 5 B = 24 , so 4 A + 4( B ) = 8 A = 24 , gives A = 3 , B = 3 , then y p = 3 cos x + 3 sin x, and y = y H + y p = e 2 x ( c 1 cos x + c 2 sin x ) 3 cos x + 3 sin x.  4 Preliminaries on applications: The above solution y = y H + y p = e 2 t ( c 1 cos t + c 2 sin t ) 3 cos t + 3 sin t has a typical property found in our applications to springs and circuits. For physical reasons, y H usually has an exponential decay as t , so y H is referred to as the...
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 Spring '08
 zhang
 Linear Algebra, Algebra, Eigenvectors, Vectors

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