Dynamical Systems: An Introduction with Applications in by Professor Dr. Pierre N. V. Tu (auth.)

By Professor Dr. Pierre N. V. Tu (auth.)

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1 = 2 = a. In this case using repeated integration gives xp = e2t [e(3-2)t(j e- 3t e2t dt) dt = e2t Jet( _e- t ) dt = _te 2t and the complete solution is x(t) = A1e2t + A 2e3t _ te 2t . It is easy to check that for xp = _te 2t , (LHS) = x- 5i + 6x = e2t ( -4 - 4t + 5 + lOt - 6t) = e2t = (RHS). The method of finding particular integrals by inverse operator is facilitated by use of Tables, available in most text books on differential equations. For example, for L(D)x = d(t), we have Xp = 1<'2) .

5. The second order linear· differential equation with constant coefficients of the form L(D)x where a equation f. 43) has the following solution (with arbitrary constants A, All A 2, Bll B 2. e. e. real and of multiplicity 1 (see Ch. eat(A cos (3t 4). e. Al = a+i(3, A2 == Xl = a-i(3 where of course, a == -b/2 = Re(A) and (3 == ~v'4c - b2 == Im(A) and i = V-T. IFrom BI = Al + A2 and B2 = i(AI - A 2) we have (~ _~) ( ~~ ) = ( ~~ ) , solving - iB2) d A2 are conjugate . b ( AI) A2 = 2I ( BI BI + iB2 .

This is case (ii), where Al A2 = -2 having each multiplicity 2. 23. x -3x + 7x C(A) = A3 - 5x = 0 3A 2 + 7A - 5 = 0 gives A = (1,1 ± 2i). This is the combination of case (i) and (iii). The solution is x(t) = clet + et (C2e 2it + C2e-2it) == clet + et(c2cos2t + C3 sin 2t). 24. e. one simple pair of complex roots -1 ± i and one pair of complex roots of multiplicity 2. This is case (iv). 25. L(D)x == (D5 + C2 sin t + C3t cos t + C4t sin t + 2D3 + D)x = 0 This is case (i) and (iv) combined. 4. e. limt-+oo x(t) = dlc: the system is by definition, stable.

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