By Madisetti & Willians

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**Sample text**

Consequently, Eq. 5a has n possible solutions: c1 eλ1 t , c2 eλ2 t , . . , cn eλn t , with c1 , c2 , . . , cn as arbitrary constants. We can readily show that a general solution is given by the sum of these n solutions,1 so that yc (t) = c1 eλ1 t + c2 eλ2 t + · · · + cn eλn t 1 To prove this fact, assume that y (t), y (t), . , y (t) are all solutions of Eq. 5a. Then n 1 2 Q(D)y1 (t) = 0 Q(D)y2 (t) = 0 ······ ··· ······ Q(D)yn (t) = 0 Multiplying these equations by c1 , c2 , . . , cn , respectively, and adding them together yields Q(D) c1 y1 (t) + c2 y2 (t) + · · · + cn yn (t) = 0 This result shows that c1 y1 (t) + c2 y2 (t) + · · · + cn yn (t) is also a solution of the homogeneous Eq.

N) in the complementary solution are the characteristic modes (also known as modes or natural modes). There is a characteristic mode for each characteristic root, and the complementary solution is a linear combination of the characteristic modes. Repeated Roots The solution of Eq. 5a as given in Eq. 7 assumes that the n characteristic roots λ1 , λ2 , . . , λn are distinct. If there are repeated roots (same root occurring more than once), the form of the solution is modified slightly. By direct substitution we can show that the solution of the equation (D − λ)2 yc (t) = 0 is given by yc (t) = (c1 + c2 t)eλt In this case the root λ repeats twice.

Consider, for example, the input f (t) = at 2 + bt + c. The successive derivatives of this input are 2at + b and 2a. In this case, the input has only two independent derivatives. Therefore the particular solution can be assumed to be a linear combination of f (t) and its two derivatives. The suitable form for yp (t) in this case is therefore yp (t) = β2 t 2 + β1 t + β0 The undetermined coefficients β0 , β1 , and β2 are determined by substituting this expression for yp (t) in Eq. 11 and then equating coefficients of similar terms on both sides of the resulting expression.