12) Let y(x) = Enzoan be a solution of y" - 2xy' - 2y = 0. recurrence relation is given by (n+2)an+2 = 2an = 2an n ≥0, then A) ao (B) a1 (C) D (D) 200 (E) a1
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- Show that |exp(2³ + i) + exp(-iz²)| < e=*-3ry² + e²ry_dp„dp (27)? - exp (-it [b² – 2p*p¯] +[-ip &* (1 – e2nQL,") – ip*8¯ (1 – e-nQt.m) + 27Qt,ns]) . (1) I = Take I = e-itH(E€). Now ignore the terms e-itb Finally using the substitution and 27QL,ns in equation 1 . A = e2xt.Qn B = e-2#t,Qn show that | _dp„dp. exp(-2ip*p t – ip *A – iBp*§¯) (27)²- (,*,)* 4t -1 29. s 30. s = t + 1 15(15t – 1)³ |
- Find plz just B because i have already got A (b) recurrence relation and initial term.*n+1 + xn subject to initial conditions ro = 0, x1 = 1. Solve the recurrence relation In+2 = Once you found a closed-form solution, verify that it satisfies r2 = }. Then estimate the number of digits in r2300-Show that if (BC)_(AC) N N 8. and (AB)_( N (C) (A) N = X, = 2x, = 3x, N V = y, then the value of neither x nor y can exceed 4 12
- find/x(n) if XCZ) = Z² (5.0-2)(1-2) 2 Z² N 2Find L[e* tsin(2t) – e-2t+3 +e-2t cos 3t] 4s+4 s-2 A)- (s-1)2 +4)2 s-2 (s-2)2 +9 B)+ 4s+4 s-2 s-2 ((s-1)2 +4)2 (s-2)2 +9 4s+8 s+2 C) s+2 ((s-1)2 +4)2 (s+2)2 +4 4s-4 s+2 D) s+2 ((s-1)2+4)2 (s+2)2 +9Find the solution to the recurrence relation an=2a_(n−1)+a_(n−2)−2a_(n−3) with initial terms a0=1 ;a1=1 ;and a2=3.