Question

Proceed as in Example 4 in Section 11.3 to find a particular solution xo(t) of equation (11) in Section 11.3 d²x + kx = f(t) (11) dt2 m- when m = 1, k = 16, and the driving force f(t) is as given. Assume that when f(t) is extended to the negative t-axis in a periodic manner, the resulting function is odd. f(t) = 1 –t, 0 <t< 2; f(t + 2) = f(t) 0 xx(t) = 0 + n = 1

Answer 1

da diz + 16 x = f(t) flut) Let flut) = f(t)= |-t. oct 22 flt+2) = f(t) = 9ot Ean cas hit is fit). COS nat S 6-ot) dt = 6t-243) 2= n=1 2 nat at an 2 O 0o = 2 nat C1-t). Cos dt an 2 2 2 4 Gos 0 an ()-+). n32 o &in (1974) [1-6-1 )"}] an= 4 n372

4/5) = 0 + 2 Canar Coo (QN+l) At 4 nat f(t) = of [1- (-)n] cas 9 n=1 n² Now let ap 4 ne) Sep13) = $. Co + Ton cos 4+ do. Sin hoid + 16 = f(t) [Ch. cas had to do in mat] + 16. Got { 16 [on Cos ha t + dn sin ny ] = f(t) ] of I [)-(-)"). cas mat Comparing coefficient n= tt do sin 13² [ concos - 8 Cot n=1 4 n = n272 Coño, dn=o 4 Cn. [26 - ) nega [ 1 - (-in On 4 [1-(-gh nen [16 - 1 ng 4 8. Cant) * 120 +13 2 (801)[16- 4 9 n=o