week07su - 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

week07su - Week 4 1 4.3 Undetermined Coef 2 4.5...

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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 0 so y p = A 0 e 2 x is a particular solution. y p = A 0 e 2 x , y p = A 0 (2 e 2 x ) , y p = 4 A 0 e 2 x , so 6 e 2 x = 4 A 0 e 2 x - 4(2 A 0 e 2 x ) + 3( A 0 e 2 x ) = - A 0 e 2 x , so A 0 = - 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 0 , a 1 and a 2 so y p = a 0 + 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 0 + a 1 x + a 2 x 2 ) all x . x 2 -coef: 3 = 3 a 2 , so a 2 = 1 . x -coef: 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 0 ) , so a 0 = 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 0 cos x + B 0 sin x. Before attempting to compute y p and y p to find A 0 and B 0 (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 0 cos x + B 0 sin x is the correct choice. —————- Now with y p = A 0 cos x + B 0 sin x, we have y p = - A 0 sin x + B 0 cos x, and y p = - A 0 cos x - B 0 sin x. Substitution in the DE gives 24 sin x = y p + 4 y p + 5 y p = ( - A 0 cos x - B 0 sin x ) + 4( - A 0 sin x + B 0 cos x )+ 5( A 0 cos x + B 0 sin x ) . Equating like terms coef of cos x : - A 0 + 4 B 0 + 5 A 0 = 0 , so B 0 = - A 0 . coef of sin x : - B 0 + 4( - A 0 ) + 5 B 0 = 24 , so - 4 A 0 + 4( B 0 ) = - 8 A 0 = 24 , gives A 0 = - 3 , B 0 = 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 transient solution and we get y = y H + y p y p as t → ∞ . We call y p the steady-state solution because of this property. Question: How does the behavior of y p differ in the usual trial solution, as compared with the modified solution? (resonance) —————- Problem 4b: Solve y + 16 y = 24 cos(4 x ) .

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